A student has an unknown sample of solution X. This solution is placed in a 1.00 cm wide cuvet and inserted into the spectrometer, producing an absorbance reading of 0.275 at a wavelength of 525.0 nm. What is the concentration of solution X in the unknown sample

Answers

Answer 1

Answer:

The concentration of the sample is 3.564x10⁻³M.

Explanation:

Using Lambert-Beer law, absorbance of a sample is directely proportional to its concentration.

The general graph of the absorbance of the standards with different concentrations is:

Y = 75.9X + 0.0045

R² = 0.9946

Where Y is the absorbance of the sample and X its concentration in mole/L.

If a solution has an absorbance of 0.275:

0.275 = 75.9X + 0.0045

0.2705 = 75.9X

3.564x10⁻³M = X → The concentration of the sample.


Related Questions

A new non-electrolyte molecule is discovered. When 241 mg of the molecule is dissolved in 250.0 mL of water, it has an osmotic pressure of 0.072 atm at 25 oC.What is the molar mass of the molecule

Answers

Answer:

327.89g/mol

Explanation:

Step 1:

The following data were obtained from the question:

Van 't Hoff factor (i) = 1 (since the molecule is non-electrolyte)

Temperature (T) = 25°C = 25°C + 273 = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Mass of molecule = 241mg

Volume of water = 250mL

Molarity (M) =?

Osmotic pressure (Π) = 0.072 atm

Step 2:

Determination of the molarity of the molecule.

This can be obtained as follow:

Π = iMRT

0.072 = 1 x M x 0.0821 x 298

Divide both side by 0.0821 x 298

M = 0.072 / (0.0821 x 298)

M = 2.94×10¯³ mol/L

Step 3:

Determination of the number of mole of the molecule. This can be obtained as follow:

Molarity = 2.94×10¯³ mol/L

Volume of water = 250mL = 250/1000 = 0.25L

Mole of molecule =..?

Molarity = mole /Volume

2.94×10¯³ = mole / 0.25

Cross multiply

Mole of molecule = 2.94×10¯³ x 0.25

Mole of molecule = 7.35×10¯⁴ mole.

Step 4:

Determination of the molar mass of the molecule.

Mole of molecule = 7.35×10¯⁴ mole.

Mass of molecule = 241mg = 241×10¯³g

Molar mass of molecule =..?

Mole = Mass /Molar Mass

7.35×10¯⁴ = 241×10¯³/ Molar Mass

Cross multiply

7.35×10¯⁴ x molar mass = 241×10¯³

Divide both side by 7.35×10¯⁴

Molar Mass = 241×10¯³/7.35×10¯⁴

Molar Mass = 327.89g/mol

Therefore, the molar mass of the molecule is 327.89g/mol

Identify the Lewis acids and Lewis bases in the following reactions:
1. H+ + OH- <-> H2O Lewis acid: Lewis base:
2. Cl- + BCl3 <-> BCl4- Lewis acid: Lewis base:
3. K+ + 6H2O <-> K(H2O)6+ Lewis acid: Lewis base:

Answers

Answer: 1. [tex]H^++OH^-\rightarrow H_2O[/tex]  Lewis acid : [tex]H^+[/tex], Lewis base : [tex]OH^-[/tex]

2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex] Lewis acid : [tex]BCl_3[/tex], Lewis base : [tex]Cl^-[/tex]

3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex] Lewis acid : [tex]K^+[/tex], Lewis base : [tex]H_2O[/tex]

Explanation:

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

1. [tex]H^++OH^-\rightarrow H_2O[/tex]

As [tex]H^+[/tex] gained electrons to complete its octet. Thus it acts as lewis acid.[tex]OH^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]H^+[/tex].

2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex]

As [tex]BCl_3[/tex] is short of two electrons to complete its octet. Thus it acts as lewis acid. [tex]Cl^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]BCl_3[/tex].

3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex]

As [tex]K^+[/tex] is short of electrons to complete its octet. Thus it acts as lewis acid. [tex]H_2O[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]K^+[/tex].

Which process is a physical change

Answers

Answer:

a physical change is something that has not been modified chemically and can possibly be changed back to the state it was once before. A physical change keeps all the same atoms and none of them is modified.

Example:

When a block of clay is morphed into a giraffe statue, it can be morphed back to its original state. If someone burnt the block of clay, the atoms would be modified and it would be unable to go back to its previous state.

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(If you're referring to a question with these answers)

A. iron rusting

B. milk turning to curd

C. water boiling

D. paper burning

E. hard water staining pipes

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Answer:

C. Water Boiling

(If you are referring to a question with these answers I think this is the correct answer if not I do apologize)

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When 106 g of water at a temperature of 21.4 °C is mixed with 64.3 g of water at an unknown temperature, the final temperature of the resulting mixture is 46.8 °C. What was the initial temperature of the second sample of water? (The specific heat capacity of liquid water is 4.184 J/g ⋅ K.)

Answers

Answer:

THE INITIAL TEMPERATURE OF THE SECOND SAMPLE IS 4.93 C OR 277.93 K

Explanation:

Mass of first sample of water = 106 g

Initial temp of first sample = 21.4  °C = 21.4 + 273 K = 294.4 K

Mass of second sample = 64.3 g

Final temp of theresulting mixture = 46.8  °C = 46.8 + 273 K = 319.8 K

Specific heat capacity of water = 4.184 J/g K

It is worthy to note that;

Heat gained by the first sample = Heat lost by the second sample

Since heat = mass * specific heat capacity * change in temperature, we have

Mass * specific heat * change in temp of the first sample  = Mass * specific heat * change in temp. of the second sample

MC (T2 - T1) = MC (T2-T1)

106 * 4.184 * ( 319.8 - 294.4) = 64.3 * 4.184 * ( 319.8 - T1)

106 * 4.184 * 25.4 = 269.0312 ( 319.8 - T1)

11 265.0016 = 269.0312 (319.8 - T1)

Since the change in temperature = 319.8 -T1

Change in temperature =11265.0016 / 269.0312

Change in temperature =  41.87

Change in temperature = 319.8 -T1

41.87 = 319.8 - T1

T1 = 319.8 - 41.87

T1 = 277.93 K

T1 = 4.93  °C

So therefore, the initial temperature of the sacond sample is 4.73  °C or 277.93 K

what is the equation for "acid dissociation constant" of "carbonic acid"

Answers

Answer:

H2CO3 = 2H+ + CO3-

Explanation:

It is simply what carbonic acid breaks down into when placed in water. Since carbonic acid is made up of H and CO3, these are the products.

A researcher placed 25.0 g of silver chloride, AgCl, in sunlight and allowed the substance to decompose completely to form silver, Ag, with the release of chlorine gas, Cl2. The gas was collected in a container during the decomposition. The researcher determined that the mass of the silver formed was 18.8 g, and the mass of the chlorine gas formed was 6.2 g. The equation for the reaction is:

Answers

Answer:

A. The law of definite proportions states that all pure samples of a particular chemical compound contain the same elements combines in the same proportion by mass.

B. The law of conservation of mass states that during ordinary chemical reactions, matter can neither be created or destroyed.

Note: The full question is as follows;

A researcher placed 25.0 g of silver chloride, AgCl, in sunlight and allowed the substance to decompose completely to form silver, Ag, with the release of chlorine gas, Cl2. The gas was collected in a container during the decomposition. The researcher determined that the mass of the silver formed was 18.8 g, and the mass of the chlorine gas formed was 6.2 g. The equation for the reaction is:

2AgCl ----> 2Ag + Cl2

a. State the law of definite proportions. Then use the researcher's data to confirm the law of definite proportions. Show your calculations.

b. State the law of conservation of matter. Then use the researcher's data to confirm the conservation of matter. Show your calculations.

Explanation:

A. Mass of silver obtained from AgCl = 18.8g.

Percentage mass of silver in the chloride = (18.8/25.0) * 100 = 75.2 %

Mass of chlorine obtained from AgCl = 6.2

Percentage mass of chlorine = (6.2/25) * 100 = 24.8 %

In one mole of AgCl with a molar mass of 143.3 g/mol; mass of silver = 107.8, mass of Cl = 35.5

Percentage mass of Ag = (107.8/143.3) * 100 = 75.2%

Percentage mass of Cl = (35.5/143.3) * 100 = 24.8%

Since the percentages by mass of Ag and AgCl obtained from the sample is the same to that obtained from a mole of AgCl, the law of definite proportions which states that all pure samples of a particular chemical compound contain the same elements combined in the same proportion by mass is verified.

B. Mass of reactant; AgCl sample = 25.0

Mass of products; At = 18.8 g; Cl = 6.2 g

Sum of products masses = 18.8 + 6.2 = 25.0 g

Therefore mass of reactant = mass of products.

This is in accordance with the law of conservation of mass which states that during ordinary chemical reactions, matter is neither created nor destroyed.

g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Answers

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899        0.01714        0           0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714    0.01714 (end)

0.00185  | 0 → 0.01714    0.01714 (end)  

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

Answer:

[tex]\rm [Na^{+}]= \text{0.055 12 mol/L}[/tex]

[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \text{0.018 99 mol/L}[/tex]

Explanation:

The overall equation for the reaction is  

Na₂HAsO₄ + NaOH ⟶ Na₃AsO₄ + H₂O

1. Mass balance for Na

All the Na⁺ comes from the Na₂HAsO₄ and the NaOH.

The mass balance equation for Na is  

[tex]\rm c_{Na^{+}} = 2[Na^{+}]_{Na_{2}HAsO_{4}} + [Na^{+}]_{NaOH}[/tex]

At the moment of mixing and before the reaction started, the total volume had doubled, so the concentrations of each component were halved.

[Na₂HAsO₄] = ½ × 0.037 98 =0.018 99 mol·L⁻¹

[NaOH]         = ½ × 0.034 28 = 0.017 14 mol·L⁻¹

[tex]\rm c_{Na^{+}} = 2\times 0.01899 + 0.01714 = 0.03798 + 0.01714\\c_{Na^{+}}= \textbf{0.055 12 mol/L}[/tex]

2. Mass balance for arsenate species

All the arsenate species come from the Na₂HAsO₄.

The reactions involved are

HAsO₄²⁻+ OH⁻ ⇌ AsO₄³⁻ + H₂O

HAsO₄²⁻ + H₂O ⇌ H₂AsO₄⁻ + OH⁻

H₂AsO₄⁻ + H₂O ⇌ H₃AsO₄ + OH⁻

The mass balance equation for arsenate species is

[tex]\rm c_{\text{arsenate}} = [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}][/tex]

At the moment of mixing, the concentration of Na₂HAsO₄ had halved.

[Na₂HAsO₄] = ½ × 0.039 78 = 0.018 99 mol·L⁻¹

[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \textbf{0.018 99 mol/L}[/tex]

Consider the reaction in a commercial heat pack: 4 Fe (s) + 3 O2(g) ® 2 Fe2O3 (s) DH = -1652 kJ a) How much heat is released when 1.00 g iron is reacted with excess O2? b) What mass of O2 must react with iron in order to generate 2150 kJ of heat?

Answers

Answer:

a) -7.395kJ of energy are released.

b) 125g of O₂ must react.

Explanation:

Based on the reaction:

4 Fe (s) + 3 O₂(g) → 2 Fe₂O₃ (s) ΔH = -1652 kJ

4 moles of iron with an excess of oxygen release -1652kJ of energy

a) The heat released is:

1.00g Fe (molar mass: 55.845g/mol)

1.00g × (1mol / 55.845g) = 0.0179 moles de Fe.

As 4 moles release -1652kJ, 0.0179 moles release:

0.0179 mol Fe × (-1652kJ / 4mol Fe) = -7.395kJ of energy are released.

b) As 3 moles of oxygen produce -1652kJ, 2150kJ are released when react:

2150kJ × (3 mol O₂ / 1652kJ) = 3.9 moles of O₂

As molar mass of O₂ is 32g/mol, mass of 3.9 moles of O₂ is:

3.9 mol O₂ × (32g / mol) = 125g of O₂ must react.

Predict the order of elution if you were to attempt to separate using column chromatography on silica gel and with a gradient eluent system starting with petroleum ether and slowly increasing MTBE % over time.

Answers

Answer:

Follows this order: B=> A => C.

Explanation:

NB: kindly check the attachment for the diagram of compounds A, B and C.

Elution is a very important concept in chromatography separation techniques. It deals with the use of eluent in the removal of an adsobate from an adsorbent. The principle behind Elution is just about how polar the solvent is.

So, in this question Compound B will go with the Elution first because of its polarity. Compound B has lesser polarity as compared to Compounds A and B.

Compound A will then elutes second because of its polarity too as resonance increases its polarity.

Last, compound C elutes because it has the highest polarity which is caused by electronegative atoms.

Consider this reaction:

2Cl2O5 —> 2Cl2 + 5O2

At a certain temperature it obeys this rate law.
rate = (2.7.M^-1•s^-1) [Cl2O5]^2

Suppose a vessel contains Cl2O5 at a concentration of 0.600M. calculate how long it takes for the concentration of Cl2O5 to decrease by 94%. you may assume no other reaction is important. round your answer to two digits

Answers

Answer:

[tex]t=9.7s[/tex]

Explanation:

Hello,

In this case, we have a second order kinetics given the second power of the concentration of chlorine (V) oxide in the rate expression, thus, the integrated equation for the concentration decay is:

[tex]\frac{1}{[Cl_2O_5]}=kt+\frac{1}{[Cl_2O_5]_0}[/tex]

Thus, the final concentration for a 94% decrease is:

[tex][Cl_2O_5]=0.600M-0.600M*0.94=0.036M[/tex]

Therefore, we compute the time for such decrease:

[tex]kt=\frac{1}{[Cl_2O_5]}-\frac{1}{[Cl_2O_5]_0}=\frac{1}{0.036M}-\frac{1}{0.60M} =26.1M^{-1}[/tex]

[tex]t=\frac{26.1M^{-1}}{k}= \frac{26.1M^{-1}}{2.7M^{-1}*s^{-1}}\\\\t=9.7s[/tex]

Regards.

A semipermeable sac containing 4% NaCl, 9% glucose, and 10% albumin is suspended in a solution with the following composition: 10% NaCl, 10% glucose, and 40% albumin. Assume that the sac is permeable to all substances except albumin. State whether each of the following will (a) move into the sac, (b) move out of the sac, or (c) not move.

glucose: a. moves into sac
water: b. moves out of sac
albumin: c; does not move
NaCl: a; moves into sac

Answers

Answer:

Glucose: (a) moves into sac

Water: (b) moves out of sac

Albumin: (c) does not move

NaCl: (a) moves into sac

Explanation:

A semi-permeable membrane is a membrane that allow certain molecules or ions to pass through by diffusion.

Diffusion is the movement of molecules from a region of higher concentration to a region of lower concentration until equilibrium is attained. A special form of diffusion known as osmosis transports water molecules across a semi-permeable membrane.

Osmosis is the movement of water molecules from a region lower solute concentration (high water molecules concentration) to a region of higher solute concentration (low water molecules concentration) until equilibrium is attained.

From the above definitions and the given data;

Glucose concentration is higher in solution outside the sac, thus, glucose molecules will move into the sac.

Water molecules are higher in the sac as a result of the lower concentration of solutes, therefore, water molecules will move out of the sac into the solution outside.

Since the sac is impermeable to albumin, it does not move.

NaCl concentration is lower in the sac, therefore, it will move ro  the solution outside into the sac.

Glucose will move into the sac, water will move out of the sac, albumin will neither move in nor out, and NaCl will move into the sac.

The molecules of each of the substances will move by diffusion from the region of higher concentration to the region of lower concentration of each substance as long as there is a permeable membrane.

Water, on the other hand, will move by osmosis from the region of high to low water potential through a permeable membrane. Regions with higher concentrations of substances usually have low water potentials and vice versa.

Thus, both glucose and NaCl molecules will diffuse from the solution into the sac, and water molecules will move from the sac into the surrounding solution. Since the sac is not permeable to albumin, then the movement in or out is inhibited.  

More on osmosis and diffusion can be found here: https://brainly.com/question/19867503?referrer=searchResults

Choose the slope and Y intercept that correspond with the graph

Answers

y=3x + 4

Explanation:

Where 3 is the slope and 4 is the y-intercept

1- A volumen constante un gas ejerce una presión de 880 mmHg a 20º Celsius dentro de una olla a presión ¿Qué temperatura habrá si el marcador de presión muestra un valor de 1050 mmHg?

Answers

Answer:

In this problem the correct thing would be to use the ideal gas equation.

Explanation:

Well in this exercise we will use the following equation:

(P x V) / T = (p x v) / t

On the right side of the equation we will find the initial values, that is, the values ​​with which the reaction begins and in general they are always the first to write in the problems.

Instead on the left side of the equation, the letters that are in lowercase are the final values, that is to say at the end of the reaction that the values ​​of pressure, temperature and volume are reached.

P is pressing, just like p, T and t are temperature, and V and v are volume.

We use this equation so we consider the behavior of said gas to be an IDEAL gas, a constant volume.

That is why the given pressures require an atmosphere to pass, which is another unit used to press the pressure ... Much needed in this equation! An atmosphere is equivalent to 760 millimeters of mercury ...

Then the final and initial pressures would be:

initial pressure: 1.15 atm

final pressure: 1.38 atm

In this way you already have the values ​​to be able to solve in the equation your unknown that would be the final temperature:

Considering that the volume is constant, we cancel it from the equation, 1.15 atm would be in the value of P and 1.38 in the value of p ... In this way it considers that 20 degrees Celsius is the initial temperature or ses T, we would only have to clear the t.

A gas has a volume of 6.6 L at a temperature of 40 C. What is the volume of

the gas if the temperature changes to 15 C?

Answers

Answer:

6.07 L

Explanation:

It appears that the reading has been made at constant pressure .

At constant pressure , the gas law formula is

V/T = constant  V is volume and T is temperature of the gas.

V₁ / T₁ = V₂ / T₂

V₁ = 6.6 L ,

T₁ = 40°C

= 273 + 40

= 313 K

T₂ = 15+ 273

= 288K

V₂ = ?

Putting the values in the formula above

6.6 / 313  = V₂ / 288

V₂ = 6.07 L.

4-Methylphenol, CH3C6H4OH (pKa 10.26), is only slightly soluble in water, but its sodium salt, CH3C6H4O-Na , is quite soluble in water. Describe the solubility of 4-methylphenol in solutions of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na2CO3). The pKa values for the conjugate acids of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na2CO3) are 15.7, 6.36, and 10.33, respectively.

Answers

Explanation:

We know that more is the [tex]pK_{a}[/tex] value, weaker will be the acid. Also, an acid completely dissociates into ions in an aqueous base solution when [tex]pK_{a}[/tex] of conjugate acid of base is greater than acid.

4-methylphenol [tex](CH_{3}C_{6}H_{4}OH)[/tex] ([tex]pK_{a} = 10.26[/tex]) is quite soluble in its sodium salt. In NaOH, the dissociation will be as [tex]Na^{+}[/tex] and [tex]OH^{-}[/tex] ions as NaOH is a strong base.

Therefore, 4-methylphenol will readily dissolve in NaOH solution.

As, [tex]NaHCO_{3}[/tex] is not a strong base but as 4-Methylphenol forms a sodium salt hence, it will have a low solubility as compared to NaOH.

Whereas [tex]Na_{2}CO_{3}[/tex] is not a base but when dissolved in water it shows basic character as it produces NaOH (strong base) and [tex]H_{2}CO_{3}[/tex] (weak acid). As a result, the solution gets basic. Hence, 4-methylphenol will readily dissolve in [tex]Na_{2}CO_{3}[/tex].

1. Reaccionan 9.7 Kg de un mineral de níquel al 70% con 8L de una solución de ácido fosfórico al 60% y con una densidad de 1.36g/ml.

Answers

Answer:

The reaction produces 201.4 g of hydrogen gas and 12.2 kg of Nickel Phosphate.

Explanation:

English Translation

9.7 Kg of a 70% nickel mineral react with 8L of a 60% phosphoric acid solution and with a density of 1.36g / ml.

Solution

The problem doesn't seen to be complete as it doesn't ask a question in the end. But, we will just calculate the amount of each product expected to cover the grounds.

The balanced chemical reaction between Nickel and Phosphoric acid is given as

3Ni + 2H₃PO₄ → 3H₂ + Ni₃(PO₄)₂

We need to first obtain the limiting reagent, that is, the reagent that is used up during the reaction and is in short supply. This reagent determines the amount of products that will be formed.

Mass of nickel that is present at the start = 70% of 9.7 kg = 6.79 kg

Mass of Phosphoric acid present at the start of the reaction = 60% of (8000 mL × 1.36 g/mL) = 6528 g = 6.528 kg

Converting both of these to number of moles

Number of moles = (mass)/(Molar mass)

For nickel,

Mass = 6.79 kg = 6790 g

Molar mass = 58.6934 g/mol

Number of moles at the start = (6790/58.6934) = 115.7 moles

For Phosphoric acid

Mass = 6528 g

Molar mass = 97.994 g/mol

Number of moles = (6528/97.994) = 66.6 moles

3 moles of Ni reacts with 2 moles of H₃PO₄

From the number of moles present initially, shows that Phosphoric acid is in limited supply and is the limiting reagent.

From the stoichiometric balance of the reaction

2 moles of H₃PO₄ gives 3 moles of H₂

66.6 moles of H₃PO₄ will give (66.6×3/2) of H₂, that is, 99.9 moles of H₂.

Mass of H₂ liberated from the reaction = (Number of moles) × (molar mass) = 99.9 × 2.016 = 201.3984 g = 201.4 g

2 moles of H₃PO₄ gives 1 mole of Ni₃(PO₄)₂

66.6 moles of H₃PO₄ will give (66.6×1/2) of Ni₃(PO₄)₂, that is, 33.3 moles of Ni₃(PO₄)₂.

Mass of Ni₃(PO₄)₂ produced from the reaction = (Number of moles) × (molar mass) = 33.3 × 366.02 = 12,188.466 g = 12.2 kg

Hope this Helps!!!

Which of the following errors could cause your percent yield to be falsely high, or even over 100%?

Select ALL that apply.

A.) Heating the sample too vigorously.
B.) Handling the crucible directly with your hands.
C.) Failing to completely decompose the sodium bicarbonate sample.
D.) Taking the mass of the empty crucible without the lid, but including the lid in all other mass measurements.
E.) Taking the mass of all samples with the lid included.

Answers

Answer:

B.Handling the crucible directly with your hands.

D.Taking the mass of the empty crucible without the lid, but including the lid in all other mass measurements.

E.Taking the mass of all samples with the lid included.

Explanation:

When observed critically , the measures associated with the errors which could cause your percent yield to be falsely high, or even over 100% are those which increase the weight of the substance with the individual neglecting.

Handling the crucible directly with your hands,Taking the mass of the empty crucible without the lid, but including the lid in all other mass measurements and taking the mass of all samples with the lid included will all increase the weight of the substance. Instead the substance should be placed alone without any form of support or contamination.

which element causes burning when me mix it with oxygen

Answers

Answer:

Hydrogen peroxide is ans

Alkyl derivatives of mercury are highly toxic and can cause mercury poisoning in humans. Dimethylmercury is one of the strongest known neurotoxins. Although it is said to have a slightly sweet smell, inhaling enough to discern this would be hazardous.
Give the empirical formula of dimethylmercury.

Answers

Answer:

The empirical formula of dimethylmercury is C2H6Hg

Explanation:

Dimethylmercury, as it says in the name, presents not only the mercury metal in its structure (Hg) but also two radical groups called methyl, which is why its name begins with the prefix DI, referring to the fact that there are two methyl.

Why is tape attracted to my skin? Give explanation

Answers

Answer:

Since the tape has extra electrons, it has a negative charge. When you move your finger close to the tape, electrons in your skin are repelled and move away. This makes the skin on your finger tip have a slight positive charge. Since positive and negative attract, the tape moves toward your finger.

What will be the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat? The specific heat of Al is 0.900 J/g·C.

Answers

LET'S PUT IN WHAT WE KNOW!!!
Q
=
725 J
m
=
55.0 g
c
=
0.900 J/(°C⋅g)
Δ
T
=
final temperature - initial temperature
Δ
T
=
(
x

27.5
)
°C
We solve for
Δ
T
.
725 J
=
55.0 g

0.900 J/(°C⋅g)
(
x

27.5
)
°C
NOW IT'S JUST BASIC ALGREBRA
725
=
49.5
x

1361
2086
=
49.5
x
42.1
=
x
The final temperature is 42.1 °C.

42.1°C is the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat.

What is temperature?

The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points or thermometric substances.

The most popular scales include the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the scale of degrees Fahrenheit (°F), or the Kelvin scale (K), with the latter being mostly used for scientific purposes. One of the United Nations System of Units' (SI) seven base units is the kelvin.

Q=725 J

m=55.0 g

c=0.900 J/(°C⋅g)

ΔT=final temperature - initial temperature

ΔT=(x−27.5)°C

725 J=55.0 g⋅0.900 J/(°C⋅g)(x−27.5)

X=42.1°C

Therefore, 42.1°C is the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat.

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WILL GIVE BRAINLIEST!!!!! will give brainliest!!!! ******According to the reaction in question I above, how many grams of solid copper will theoretically be

produced when 14.4 g of aluminum are reacted with 14.4 g of copper (II) sulfate? Which reactant is the

limiting reactant? Show your work. Be sure to include units!

Answers

Answer:

5.73 g Cu

Explanation:

M(CuSO4) = 159.6 g/mol

M(Al) = 27.0 g/mol

M(Cu) = 63.5 g/mol

14.4 g Al * 1 mol/27.0 g = 0.5333 mol Al

14.4 g CuSO4 * 1 mol/159.6 g = 0.0902 mol CuSO4

                                2 Al         +     3CuSO4 ------->  3Cu + Al2(SO4)3

from reaction           2 mol               3 mol

given                     (0.5333 mol )x       0.0902 mol

needed                   0.0601 mol

x= 2*0.0902/3 = 0.0601 mol   Al

Al is excess, CuSO4 is a limiting reactant.

                               2 Al         +     3CuSO4 ------->  3Cu + Al2(SO4)3

from reaction                               3 mol                  3 mol

given                                            0.0902 mol         x mol

x = 0.0902 mol Cu

0.0902 mol Cu * 1 mol/63.5 g Cu = 5.73 g Cu

how many grams of H2 are needed to produce 14.34 g of NH3?

please help, this hw is due in a few hours

Answers

Answer:

im pretty sure its 2.54g H2

Explanation:

14.34gNH3 / 17.03gNH3 <-- molar mass

.842g x 3 mol <-- mols of H2

2.52 / 2 mol <-- mols of NH3

1.26 x 2.016gH2= 2.54gH2

Precision can be defined as the?

Answers

Answer:Precision can be defined as the. reproducibility of a measured value. Precision is how close the measured values are to each others. In contrast with accuracy, accuracy is the agreement between a measured value and an accepted value.

Explanation:

Consider a cobalt-silver voltaic cell that is constructed such that one half-cell consists of the cobalt, Co, electrode immersed in a Co(NO3)3 solution, and the other half-cell consists of the silver, Ag, electrode immersed in a AgNO3 solution. The two electrodes are connected by a copper wire. The Co electrode acts as the anode, and the Ag electrode acts as the cathode. To maintain electric neutrality, you add a KNO3 salt bridge separating the two half-cells. Use this information to solve Parts B, C, and D.
A. The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell is the site of the reduction reaction.
Type the half-cell reaction that takes place at the anode for the cobalt-silver voltaic cell. Indicate the physical states using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. Use (aq) for an aqueous solution. Do not forget to add electrons in your reaction.
B. The half-cell is a chamber in the voltaic cell where one half-cell is the site of an oxidation reaction and the other half-cell is the site of a reduction reaction.
Type the half-cell reaction that takes place at the cathode for the cobalt-silver voltaic cell. Indicate physical states using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. Use (aq) for an aqueous solution. Do not forget to add electrons in your reaction.

Answers

Answer:

Anode half reaction;

Co(s) ----> Co^2+(aq) + 2e

Cathode half reaction;

2Ag^+(aq) + 2e-------> 2Ag(s)

Explanation:

A voltaic cell is an electrochemical cell that spontaneously produces electrical energy from chemical reactions. A voltaic cell comprises of an anode (where oxidation occurs) and a cathode (where reduction occurs). The both electrodes are connected with a wire . A salt bridge ensures charge neutrality in the anode and cathode compartments. Electrons flow from anode to cathode.

For the cell referred to in the question;

Anode half reaction;

Co(s) ----> Co^2+(aq) + 2e

Cathode half reaction;

2Ag^+(aq) + 2e-------> 2Ag(s)

Hydrogen bonds can be found between molecules of which substance? NH3 H2 HI CH4

Answers

Answer:

All except ch4

Explanation:

NH3 N H 3 and HF can form hydrogen bonds as they have a hydrogen atom bonded to fluorine and nitrogen atoms.

Hydrogen bonds are formed between hydrogen and a highly electronegative atom such as oxygen, halogens etc.  Among the given compounds HI form hydrogen bond.

What is hydrogen bond?

Hydrogen bond a strong bond type formed between hydrogen and an electronegative atom. Water, hydrogen halides, hydrogen sulphide etc are formed by hydrogen bonds.

Hydrogen is an electropositive atom and will easily lose an electron to a electronegative atom. Thus hydrogen bonds with atoms by sharing electrons each other where, the shared pair of electrons are attractively pulled to the electronegative atom.

Therefore, all the hydrogen bonded compounds are polar in nature.  Hydrogen bonds are strong bonds and it can be seen in proteins, DNA, and in other biomolecules.

HI or hydrogen iodide forms hydrogen bond because iodine is comparatively electronegative.

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If a jet’s cruising altitude is 32,200ft(to three significant figures),the distance in km is :(1 mile=1.61km;1 mile=5280 ft)

Answers

Answer:

9.82 km.

Explanation:

Hello,

In this case, given the conversion factors from miles to kilometres and from miles to feet, we can directly compute the jet’s cruising altitude in kilometres as shown below:

[tex]32,200ft\times \frac{1mile}{5280ft}\times \frac{1.61km}{1mile} \\\\=9.82km[/tex]

Best regards.

Why don't we discuss the results during the results section of the project?

Answers

Answer: The result section of the project contains your findings while carrying out your research or study.

Explanation:

The Results section of a research or study usually contains only the findings of your study or research.The findings which usually include

1. Data presented in tables, charts, graphs, and other figures.

2. A contextual analysis of this data explaining its meanings. Usually in sentences.

Our result gotten is not discussed in result section because every project or research work has a discussion page where every results or findings are discussed. The result section is expected to carry what you found.

A student sets up the following equation to convert a measurement. (The ? stands for a number the student is going to calculate.) Fill in the missing part of this equation. Note: your answer should be in the form of one or more fractions multiplied together. (23. Pa cm^3)____?kPa . m^3

Answers

Answer:

The correct answer will be "-6.7 × 10¹⁰ kg.m/s".

Explanation:

The required conversions are:

⇒  [tex]1 \ kg=1000 \ g[/tex]

⇒  [tex]1 \ m=100 \ cm[/tex]

Now,

The complete conversion will be:

=  [tex][-6.7\times 10^5 \ \frac{kg \ m}{s} ]\times [\frac{10^3 \ g}{kg}\times \frac{10^2 \ cm}{1 \ m}][/tex]

On cancelling the terms, we get

=  [tex]-6.7\times 10^{10} \frac{kg \ m}{s}[/tex]

So that the missing terms will be [tex][\frac{10^3 \ g}{kg}\times \frac{10^2 \ cm}{1 \ m}][/tex] and [tex][-6.7\times 10^{10}\frac{kg \ m}{s}][/tex]

Which compound has the lowest melting point? KCl CaCl2 Na2O C6H12O6

Answers

It is called ethane.
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