Answer:
26.982
Explanation:
Use the periodic table
Please help I will give brainiest
Answer:
Answer
Answer is option B coz da negative charge r more thn positive
Answer:
B.
Explanation:
On the hand in the option B there are 3 positive and 5 negative signs so the answer is B.
(a) Given that the path length of the cuvette is 1 cm, what is the extinction coefficient of the 0.020 mM Yellow 5 dye at lambda max?(b) Given the data in the table below, what is the concentration of the diluted solution?Stock solution concentration 0.075 MStock solution absorbance 1.84Diluted solution absorbance 0.78
Answer:
(a). Kindly check the explanation section.
(b). 24.5 M^-1 cm^-
(c). 0.0318 M
Explanation:
So, we are given the following data parameters which is going to aid us in solving the above Question;
The path length = 1cm, extinction coefficient = unknown, absorbance is unknown and concentration= 0.02mM.
Using the formula below; we can determine the extinction coefficient as;
Absorbance = extinction coefficient × concentration × path length. -----------(1).
(a). Since, the absorbance is not given in the Question it won't be possible to determine the value of the extinction coefficient. Thus, say the value of absorbance = A.
Then, extinction coefficient = A/ 0.02 × 1 cm.
(b). Making use of the formula above, the extinction coefficient can be Determine as follows;
extinction coefficient = 1.84/0.075 × 1 = 24.5 M^-1 cm^-1.
(c). The concentration can be Determine by also making use of the formula above and making the concentration the subject of the formula;
Concentration = absorbance/ extinction coefficient × path length.
Concentration = 0.78/24.5 M^-1 cm^-1. × 1cm = 0.0318 M.
Which unit of measurement would best be used to measure the volume of a points a DROP of water? .
Answer:
milliliter
Explanation:
smallest form of measuring volume from all the available choices
An atom of an element has two electrons in the first energy level and five electrons in the second energy level. Write the electron configuration and name the element. How many unpaired electrons does an atom of this element have?
Answer:
electronic configuration: 1s^2,2s^2,2p^3
name: nitrogen
unpaired electron: 3
Explanation:
The atom with 2 electrons and 5 electrons in the first and second energy level is NITROGEN. The element has 3 unpaired electron.
The electronic configuration of the atom is 1s² 2s² 2p³Since the atom has 7 Electrons, the element which has 7 electrons is NITROGEN. The 2p orbital has 3 unpaired electrons in it's second energy level.Therefore, the 2p orbital has 3 unpaired electrons in it's outermost shell.
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What elements make up water?
Answer:
1 oxygen and 2 hydrogen
Why can’t you ice skate on a lake when it is not frozen?
What is the mole fraction of each component if 3.9 g of benzene (C6H6) is dissolved in 4.6 g of toluene (C7H8)
Answer:
Step 1 of 6
(a)
The mass of benzene is  , so calculate the moles of benzene as follows:

The mass of toluene is, so calculate the moles of toluene as follows:

Now, calculate the mole fraction as follows:


Therefore, the mole fraction of benzene and toluene is  and  respectively.
Step 2 of 6
(b)
The formula to calculate the partial pressure is as follows:

Here,  is the partial pressure of benzene,  is the vapour pressure of pure benzene and  is the mole fraction of benzene.
Vapour pressure of pure benzene at  is.
Substitute the values in the equation as follows:

Therefore, the partial pressure is  .
Step 3 of 6
(c)
Vapor pressure of the solution at 1 atm is  .
When the total pressure of the vapour pressure of the mixture is  at a temperature, then, the solution boils. It corresponds to the boiling point of the solution.
Calculate the total pressure of the solution at  as follows:

Since, the total pressure is less than the atmospheric pressure, the solution will not boil at  .
Calculate the total pressure of the solution at  as follows:

Since, the total pressure is greater than the atmospheric pressure, the solution will boil at  .
Therefore, the boiling point of the solution is  .
Step 4 of 6
(d)
Mole fraction of benzene at  is calculated as follows:

Mole fraction of toluene at  is calculated as follows:

Therefore, the mole fractions of benzene and toluene are  and  respectively.
Step 5 of 6
(e)
Vapor pressure of benzene at  is  .
Partial pressure of benzene is calculated as follows:

Vapor pressure of toluene at  is  .
Partial pressure of toluene is calculated as follows:

Step 6 of 6
Weight composition of the vapour that is in equilibrium with the solution is calculated as follows:

Weight composition of the vapour that is in equilibrium with the solution is calculated as follows:

Explanation:
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a solution is dilute when?
A. it has a lot of solute
B. has little solvent
C. it has a lot of solvent
D. has maximum solute
Washing machines use a large amount of water. A student suggested that old pairs of stained jeans which have to be washed more frequently should be replaced by new pairs of jeans to conserve water. Which of these statements best describes the suggestion made by the student? Question 9 options: It is not practical because an old pair of jeans needs less water to be washed than a new pair of jeans. It is practical because a huge amount of water can be conserved by this method. It is practical because it is easy to implement. It is not practical because it takes a huge amount of water to produce a new pair of jeans.
Answer:
C |||| It is not practical because it takes a huge amount of water to produce a new pair of jeans
Explanation:
If you're doing flvs then it's C.
Answer:
C!
Explanation:
i got it right on the test UwU
Using the equations
2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)∆H° = -6271 kJ/mol
2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.6 kJ/mol
C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/mol
Determine the enthalpy (in kJ/mol) for the reaction
6 C (s) + 3 H₂ (g) → C₆H₆ (l).
The enthalpy : 49.1 kJ/mol
Further explanationThe change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
Based on the principle of Hess's Law, the change in enthalpy of a reaction will be the same even though it is through several stages or ways
Reaction
1. 2C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)∆H° = -6271 kJ/mol
Reverse
12 CO₂ (g) + 6 H₂O (g) ⇒ 2C₆H₆ (l) + 15 O₂ (g) ∆H° = 6271 kJ/mol : 2
6CO₂ (g) + 3H₂O (g) ⇒ C₆H₆ (l) + 15/2 O₂ (g) ∆H° = 3135.5 kJ/mol
2. 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.6 kJ/mol x 3/2
3H₂ (g) + 3/2O₂ (g) → 3H₂O (g) ∆H° = -725.4 kJ/mol
3. C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/mol x 6
6C (s) + 6O₂ (g) → 6CO₂ (g) ∆H° = -2361 k/j/mol
-------------------------------------------------------------------------------------
6 C (s) + 3 H₂ (g) → C₆H₆ (l) ∆H° = 49.1 kJ/mol
We add up and the same compound that is on different sides we eliminate
Are light and other forms of energy are made of atoms
Answer:
yes they are
Explanation:
QUESTION 24
Calculate [H 30+] of a solution with a pH of 2.14.
Answer:
The hydronium ion concentration can be found from the pH by the reverse of the mathematical operation employed to find the pH. [H3O+] = 10-pH or [H3O+] = antilog (- pH).
I hope what I just explained is enough .
Which of these compounds do not have a pH dependent solubility at less than pH 7.00?
Answer:
tbh idekkk
Explanation:
Which element is most likely suspected?
iron
Sulfur
Chromium
Nickel
Answer: sulfur
Explanation:
Answer:
its sulfur babes
Explanation:
What is the mass, in grams, of 2.3 moles of gold?
Answer:
453.031g
Explanation:
atomic mass (molar mass)- 196.97g/m
2.3(196.97)=453.031
These questions are from an experiment where we had a mixture of Ferrocene, acetylferrocene and diacetyl ferrocene and we separated each by using column chromatography. Before doing the column chromatography we did TLC analysis to ensure that we did, in fact, have a mixture of these 3 compounds.
During the column chromatography we first eluted the ferrocene using pure hexane. Then we eluted the acetylferrocene using a 1:1 mixture of hexane and methylene chloride. Then we eluted the diacetylferrocene using a 9:1 mixture of hexane and methylene chloride.
1. Rank ferrocene, acetylferrocene, and diacetylferrocene in order of increasing polarity. Do the TLC results from your fractions support this ranking? Explain.
2. Rank the solvents used in the experiment in order of increasing polarity.
3. Why do you start with the least polar solvent/solvent mixture and progress to increasing polar solvent/solvent mixtures when eluting the ferrocene compounds from the column rather than starting with more polar solvent system and progressing to less polar solvents?
4. How do the thin layer and column chromatography for this experiment compare in regard to stationary and mobile phases?
5. What key feature in the IR spectrum of acetylferrocene distinguishes it from the spectrum of ferrocene?
Answer:
Explanation:
The polarity of the 3 compounds would be in the order of
Ferrocene < Acetylferrocene < Diacetylferrocene
Your TLC data has to also support this observation . This can be checked by measuring the values of Rf ( Retention factor = distance travelled by solute/solvent ) .The Rf values also has to follow this particular order: -
Ferrocene > acetylferrocene > diacetylferrocene
2) Hexane happens to be a non-polar solvent. The polarity of hexane can be increased if some polar solvents for example, ethyl and methylene chloride etc are added
Therefore, in the increasing order of solvents polarity, we have
Hexane < 1:1 mixture of hexane: methylene chloride < 9:1 mixture of methylene chloride:
3) Chromatographic techniques all have a stationary phase in addition to a mobile phase. In the case of column chromatography, the silica gel will be the stationary phase and the solvent that will be poured will be the mobile phase.
4) The TLC and column chromatography both happen to have the same stationary phase which is the silica gel. Also, the same solvent mixture is used in both the techniques. This makes the result of the 2 to be almost the same. The difference seen between them is that, TLC works against the gravity while on the other hand column chromatography works in the direction of the gravity.
5) The key feature in the IR spectra of the acetylferrocene that will be absent in the spectra of ferrocene is the presence of carbonyl stretching frequency at close to 1700 per cm(cm-1). This peak is easily differentiated between both acetyl ferrocene and ferrocene.
What mass of a solution labeled 6.3% sucrose (C12H22O11, 342 g/mol) by mass contains 11.0 g of sucrose?
Answer:
174.6g of solution contains the 11.0g of sucrose
Explanation:
A solution of 6.3% by mass of sucrose contains 6.3g of sucrose in 100g of solution, that is, 6.3g of sucrose / 100g of solution.
As we want to find the mass contains 11.0g of sucrose, we can write:
11.0g of sucrose * (100g of solution / 6.3g of solution) =
174.6g of solution contains the 11.0g of sucroseDetermine the number of Au atoms found in 5.18 grams of gold.
A.1.02x1022 atoms
B. 3.84 x 1022 atoms
C.2.13 x 1022 atoms
D.4.58x1022 atoms
1.58 x 1022. atoms
Please let me know
Answer:
1.58× 10²² atoms of Au
Explanation:
Given data:
Mass of gold = 5.18 g
Number of atoms of gold = ?
Solution:
First of all we will calculate the number of moles of gold.
Number of moles = mass/molar mass
Number of moles = 5.18 g/196.96 g/mol
Number of moles = 0.0263 mol
Number of atoms:
Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms of Au
0.0263 mol × 6.022 × 10²³ atoms of Au / 1mol
0.158× 10²³ atoms of Au
1.58× 10²² atoms of Au
Calculate Kb for a 0.1 M solution of cyanide ion, given that Ka for hydrocyanic acid is 6.2 x 10â10.a) 8.4 x 10â9b) 7.87 x 10â3c) 1.34 x 10â3d) 1.8 x 105e) 7.87 x 10â6f) 1.61 x 10â5g) 7.4 x 10â8
Answer:
Kb = 1.6 × 10⁻⁵
Explanation:
Step 1: Given data
Acid dissociation constant of hydrocyanic acid (Ka): 6.2 × 10⁻¹⁰
Concentration of cyanide ion (Cb): 0.1 M
Step 2: Calculate the basic dissociation constant (Kb) of cyanide ion
We have the Ka of HCN. We can calculate the Kb of its conjugate base using the following expression.
Ka × Kb = Kw = 1.0 × 10⁻¹⁴
Kb = 1.0 × 10⁻¹⁴/Ka
Kb = 1.0 × 10⁻¹⁴/6.2 × 10⁻¹⁰
Kb = 1.6 × 10⁻⁵
The Kb for a 0.1 M solution of cyanide ion is :
- 1.6 × 10⁻⁵
Base dissociation constantGiven:
Acid dissociation constant of hydrocyanic acid (Ka)= 6.2 × 10⁻¹⁰
Concentration of cyanide ion (Cb)= 0.1 M
Base dissociation constant (kb)=?
Ka × Kb = Kw = 1.0 × 10⁻¹⁴
Kb = 1.0 × 10⁻¹⁴/Ka
Kb = 1.0 × 10⁻¹⁴/6.2 × 10⁻¹⁰
Kb = 1.6 × 10⁻⁵
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Are transition metals reactive?
Answer: Yes
Explanation:
How something works is related to its structure
Answer:
Forms and Functions
Explanation:
It refers to the direct relationship between the structure of a thing and the way it functions.
What is the temperature 0 Kelvin called?
What happens at that temperature?
Answer:
0 Kelvin is called Absolute Zero. Particles will stop moving, and lose all energy.
Part A Calculate the [H3O+] of the following polyprotic acid solution: 0.300M H3PO4. Express your answer using two significant figures.
Part B Calculate the pH of this solution.
Express your answer using one decimal place.
Part C Calculate the [H3O+] and pH of the following polyprotic acid solution: 0.310M H2C2O4.
Express your answer using two significant figures.
Part D Calculate the pH of this solution.
Express your answer using two decimal places.
Answer:
A. [H₃O⁺] = 0.900 M
B. pH = 0.0 to one decimal place
C. [H₃O⁺] = 0.620 M
D. pH = 0.21 to two decimal places
Explanation:
A. Equation for the complete dissociation of H₃PO₄ is given below:
H₃PO₄(s) + 3H₂O(l) -----> PO₄³⁻(aq) + 3H₃O⁺(aq)
From the equation of reaction, one mole of H₃PO₄ produces 3 moles of H₃O⁺
Therefore, [H₃O⁺] = 3 * 0.300M = 0.900 M
B. pH = -log[H₃O⁺]
pH = - log(0.900)
pH = -log (9 * 10⁻¹)
pH = 0.0 to one decimal place
C. Equation for the complete dissociation of H₂C₂O₄ is given below:
H₂C₂O₄(s) + 2H₂O(l) ----> C₂O₄⁻(aq) + 2H₃O⁺
From the equation of reaction, one mole of H₂C₂O₄ produces 3 moles of H₃O⁺
Therefore, [H₃O⁺] = 2 * 0.310 M = 0.620 M
D. pH = -log[H₃O⁺]
pH = -log(0.620)
pH = -log(0.62 * 10⁻¹)
pH = 0.21 to two decimal places
The concentration of [[tex]\rm H_3O^+[/tex]] has been 0.9 M, and the pH has been 0.045.
The polyprotic acid has been able to donate more than one proton in an acid-base reaction.
The balanced chemical reaction can be:
[tex]\rm H_3PO_4\;+\;3\;H_2O\;\rightarrow\;PO_4^3^-\;+\;3\;H_3O^+[/tex]
(a) According to the equation, 1 mole of [tex]\rm H_3PO_4[/tex] gives 3 moles of hydronium ions. The molarity has been defined as moles per liter. Assuming the volume of reaction to be 1 liter.
1 mole [tex]\rm H_3PO_4[/tex] = 1 M [tex]\rm H_3PO_4[/tex]
1 mole [tex]\rm H_3PO_4[/tex] = 3 moles [[tex]\rm H_3O^+[/tex]]
0.3 mole [tex]\rm H_3PO_4[/tex] = 3 [tex]\times[/tex] 0.3 moles [[tex]\rm H_3O^+[/tex]].
0.3 mole [tex]\rm H_3PO_4[/tex] = 0.9 moles [[tex]\rm H_3O^+[/tex]]
The concentration of [[tex]\rm H_3O^+[/tex]] has been 0.9 mole/L, or 0.9 M.
(b) pH of the solution can be defined as:
pH = -log [[tex]\rm H_3O^+[/tex]]
pH = -log (0.9)
pH = 0.045
The pH of the solution has been 0.045.
(c) The polyprotic acid have the balanced equation:
[tex]\rm H_2C_2O_4\;+\;2\;H_2O\;\rightarrow\;C_2O_4^-\;+\;2\;H_3O^+[/tex]
According to the equation, 1 M [tex]\rm H_2C_2O_4[/tex] gives 2 M [tex]\rm H_3O^+[/tex].
The concentration of hydronium ions from 0.310 M [tex]\rm H_2C_2O_4[/tex] has been:
1 M [tex]\rm H_2C_2O_4[/tex] = 3 M [tex]\rm H_3O^+[/tex]
0.310 M [tex]\rm H_2C_2O_4[/tex] = 2 [tex]\times[/tex] 0.310 moles [[tex]\rm H_3O^+[/tex]]
0.310 M [tex]\rm H_2C_2O_4[/tex] = 0.62 M [[tex]\rm H_3O^+[/tex]].
The [[tex]\rm H_3O^+[/tex]] concentration of the solution has been 0.62 M.
(d) pH of the solution can be defined as:
pH = -log [[tex]\rm H_3O^+[/tex]]
pH = -log (0.62)
pH = 0.20
The pH of the solution has been 0.20.
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Look at the picture below. Name the isotope
Calcium-19
Calcium-39
Potassium-20
Potassium-39
Blood takes about 1.55 s to pass through a 2.00 mm long capillary. If the diameter of the capillary is 5.00 μm and the pressure drop is 2.45 kPa, calculate the viscosity ???? of blood. Assume laminar flow.
Answer:
The viscosity [tex]\mathbf{\eta = 7.416 \times 10^{-4} \ N.s/m^2}[/tex]
Explanation:
From the given information:
Time t = 1.55 s
The radius of capillary = 5.00 μm /2
The pressure drop ΔP = 2.45 kPa
The length of the capillary = 2.00 mm
∴
The viscosity of the blood flow can be calculated by using the formula:
[tex]\eta = \dfrac{r^2 \Delta P }{8Lv}[/tex]
where;
v = L/t
Then;
[tex]\eta = \dfrac{r^2 \Delta P }{8L(\dfrac{L}{t})}[/tex]
[tex]\eta = \dfrac{(\dfrac{5 \times 10^{-6} \ m}{2})^2(2.45 \times 10^3 \ Pa) }{8(2.0 \times 10^{-3} \ m ) (\dfrac{2.0 \times 10^{-3} \ m }{1.55 \ s })}[/tex]
[tex]\eta = 7.416 \times 10^{-4} \ Pa.s[/tex]
To (N.s/m²)
[tex]\mathbf{\eta = 7.416 \times 10^{-4} \ N.s/m^2}[/tex]
Roughly how many water molecules are ionized at any given time in a sample of room-temperature water? a. one in a thousand b. one in a million c. one in a billion d. one in a trillion
Answer:
one in a billion
Explanation:
We have it known that Pure water is entirely made up of H2O molecules. Just 1 in 10⁹, that is a billion molecules are ionized when it is at room temperature.
The Ionization of water has this chemical equation:
H2O <--> H+ + OH-
When At equilibrium and at room temperature, we have the rate of forward reaction and the backward reaction to be very fast such that no water molecule would eventually remain ionized.
Therefore our answer is roughly one in a billion water.
How many valence electrons are found in P3- ?
Answer:
5 valence elctrons
Explanation:
the outer most orbitals, 3s2 and 3p3 contains 5 electrons, thus valences electrons for P is 5
How would you synthesize 1-hexanol starting from 1-propanol, incorporating a Grignard reagent in the synthesis?
Answer:
hjhjhjhhhhh
Explanation:
A(n)___ is the smallest neutral unit of two or more atoms held together by a covalent bond?
Answer:
A(n)___ is the smallest neutral unit of two or more atoms held together by a covalent bond?
=>MOLECULE
The indicator phenolphthalein is added to the _____________.
a. Analyte
b. Water
c. Solute
d. Titrant
Answer:
A
Explanation:
I took the test