What is the name of this molecule? (will give BRAINLIEST)


A straight chain of four carbons. There is a triple bond between the second and third carbons when counting from left to right or right to left.

Answer:

That its small pointed. Pink(Himalayan salt)or white(normal salt)

Explanation:

Summa dees questions are so stupid, deys makin me salty.

Answer:

I am not sure, but I think this is the answer 72.987 g/mol

According to the ideal gas law, when the temperature of a gas doubles, its volume doubles as well (Option B).

The ideal gas law relates the pressure, volume, number of moles and temperature of an ideal gas.

Let's consider the equation of the ideal gas law.

P . V = n . R .T

V = n . R . T / P

As we can see, there is a direct relationship between the volume and the temperature. Thus, if the temperature doubles, the volume will double as well.

According to the ideal gas law, when the temperature of a gas doubles, its volume doubles as well (Option B).

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Answer:

1. density = 0.89 g/cm3

2. Yes is possible to identify the liquid

3. ethanolamine

Explanation:

Data:

mass = 682 g

volume = 0.767 L = 767 mL or cm3

1.

To calculate the density of the liquid it is necessary to know that the density formula is:

[tex]density=\frac{mass(g)}{volume(cm^{3}) }[/tex]

The data obtained is replaced in the formula:

[tex]density=\frac{682g)}{767(cm^{3}) }=0.89\frac{g}{cm^{3} }[/tex]

2.

With the given data it is possible to identify the liquid, this because the density value is a basic property of each liquid.

3.

It is possible to determine what liquid it is, since when comparing the value obtained with those reported in the collection of Material Safety Data Sheets (MSOS), the value that agrees is that of ethanolamine.

Answer : The net ionic equation will be,

[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

[tex]Ba(OH)_2(aq)+H_2SO_4(aq)\rightarrow 2H_2O(aq)+BaSO_4(s)[/tex]

The ionic equation in separated aqueous solution will be,

[tex]Ba^{2+}(aq)+2OH^-(aq)+2H^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2H^+(aq)+2OH^{-}(aq)[/tex]

In this equation, [tex]H^+\text{ and }OH^-[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]

Answer:

The correct answer is 2.016 x 10⁻¹⁷

Explanation:

We have the following chemical reactions and their equilibrium constants (K):

(1) H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq)  K₁= 4.20×10⁻⁷

(2) HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq)    K₂= 4.80×10⁻¹¹

And we have to obtain K for the following reaction:

H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq)

If we add equations (1) and (2) we obtain the the desired equation. Remember that when we add chemical equations, the global equilibrium constant is the product of the constants.

H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq)     K₁= 4.20×10⁻⁷

+

HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq)      K₂= 4.80×10⁻¹¹

-------------------------------------------------------------

H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq)    K= K₁ x K₂

K = K₁ x K₂ = (4.20×10⁻⁷) x (4.80×10⁻¹¹) = 2.016 x 10⁻¹⁷

Answer:

Explanation:

the terms strong and weak as applied to acids. As a part of this it defines and explains what is meant by pH, Ka and pKa.

It is important that you don't confuse the words strong and weak with the terms concentrated and dilute.

As you will see below, the strength of an acid is related to the proportion of it which has reacted with water to produce ions. The concentration tells you about how much of the original acid is dissolved in the solution.

It is perfectly possible to have a concentrated solution of a weak acid, or a dilute solution of a strong acid.

Answer: The molecular formula for the given organic compound is [tex]C_{18}H_{20}O_2[/tex]

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2=39.61g[/tex]

Mass of [tex]H_2O=9.01g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, [tex]\frac{12}{44}\times 39.61=10.80g[/tex] of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, [tex]\frac{2}{18}\times 9.01=1.00g[/tex] of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = [tex]\frac{0.9}{0.10}=9[/tex]

For Hydrogen = [tex]\frac{1}{0.10}=10[/tex]

For Oxygen = [tex]\frac{0.10}{0.10}=1[/tex]

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is [tex]C_9H_{10}O[/tex]

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

[tex]n=\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

[tex]n=\frac{268.34g/mol}{134g/mol}=2[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2[/tex]

Thus, the molecular formula for the given organic compound is [tex]C_{18}H_{20}O_2[/tex].

Answer:

6.65dm³

Explanation:

Equation of reaction,

4C3H5(NO3)3(s) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g)

From the equation of reaction, 4 moles of Nitroglycerin gave 29 moles of various gases.

Molar mass of nitroglycerin C₃H₅(NO₃)₃ = 908g

Since all the product of the reaction are in gaseous phase, let's assume that law of conservation of matter is held hence there's no loss in mass.

908g of C₃H₅(NO₃)₃ = 908g of products

2.70×10²g of C₃H₅(NO₃)₃ = 2.70×10²g of products

Number of moles = mass / molar mass

Molar mass of C₃H₅(NO₃)₃ = 908g/mol

Number of moles = 2.70×10² / 908

Number of moles = 0.297 moles

But 1 mole = 22.4dm³

0.297mole = x dm³

x = (0.297 × 22.4) / 1

x = 6.65dm³

The volume of gas that'll be produced when 2.70×10²g of C₃H₅(NO₃)₃ would be 6.65dm³

Answer:

93.15 %

Explanation:

We have to start with the chemical reaction:

[tex]CaCl_2~+~Na_2CO_3~->~CaCO_3~+~NaCl[/tex]

Now, we can balance the reaction:

[tex]CaCl_2~+~Na_2CO_3~->~CaCO_3~+~2NaCl[/tex]

Our initial data are the 15.71 g of [tex]CaCl_2[/tex], so we have to do the following steps:

1) Convert from grams to moles of [tex]CaCl_2[/tex] using the molar mass (110.98 g/mol).

2) Convert from moles of [tex]CaCl_2[/tex] to moles of [tex]CaCO_3[/tex] using the molar ratio. ( 1 mol [tex]CaCl_2[/tex]= 1 mol of [tex]CaCO_3[/tex]).

3) Convert from moles of [tex]CaCO_3[/tex] to grams of [tex]CaCO_3[/tex] using the molar mass. (100 g/mol).

[tex]15.71~g~CaCl_2\frac{1~mol~CaCl_2}{110.98~g~CaCl_2}\frac{1~mol~CaCO_3}{1~mol~CaCl_2}\frac{100~g~CaCO_3}{1~mol~CaCO_3}=14.16~g~CaCO_3[/tex]

Finally, we can calculate the yield percent:

[tex]%~=~\frac{13.19~g~CaCO_3}{14.16~g~CaCO_3}*100=93.15~%[/tex]

I hope it helps!

The percentage yield obtained when excess sodium carbonate, Na₂CO₃, is added to a solution containing 15.71 g CaCl₂ is 93.2%

Molar mass of CaCl₂ = 40 + (35.5×2) = 111 g/mol

Mass of CaCl₂ from the balanced equation = 1 × 111 = 111 g

Molar mass of CaCO₃ = 40 + 12 + (16×3) = 100 g/mol

Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g

SUMMARY

From the balanced equation above,

111 g of CaCl₂ reacted to produce 100 g of CaCO₃

From the balanced equation above,

111 g of CaCl₂ reacted to produce 100 g of CaCO₃.

Therefore,

15.71 g of CaCl₂ will react to produce = [tex]\frac{15.71 * 100}{111} \\\\[/tex] = 14.15 g of CaCO₃.

Thus, the theoretical yield of of CaCO₃ is 14.15 g

Actual yield of CaCO₃ = 13.19 g

Theoretical yield of CaCO₃ = 14.15 g

[tex]Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{13.19}{14.15} * 100\\\\[/tex]

Therefore, the percentage yield of the reaction is 93.2%

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Answer:

ammonia

Explanation:

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  ammonia

Explanation:

The mixture contains two base compound which are

           ammonia,

and     diethylamine

Now the addition of HCl which is  a strong acid in step 1  will cause the protonation of  the  two base compound , which makes the soluble hence resulting in them being extracted to the aqueous layer as represented in below

       [tex]NH_3 + HCl\to NH_4 ^{+} + Cl^-[/tex]

and

     [tex](CH 3CH 2) 2NH + HCl \to (CH 3CH 2) 2NH_2^{+} + Cl[/tex]

Answer:

Explanation:

The given chemical reaction is:

[tex]2H_2O_{(l)} \to^{I^-}} 2H_2O_{(l)}+O_2_{(g)}[/tex]

From above equation  [tex]I^-[/tex] serves as catalyst which is not consumed by the reaction and also it is completely recovered; as a result to that , the full volume of KI will definitely react with AgNO₃.

Given that :

the volume of potassium iodide [tex]V_{KI} = 15 \ ml[/tex]

the molarity of potassium [tex]M_{KI} = 0.1 \ M[/tex]

the volume of distilled water [tex]V_W = 15 \ mL[/tex]

The volume of 3% [tex]H_2O_2 \ \ V_{H_2O_2} = 5 \ mL[/tex]

Molarity of AgNO₃ [tex]M_{AgNO_3} = 0.1 \ M[/tex]

Let take an integral look with the reaction between KI and AgNO₃; we have

[tex]KI + AgNO_3 \to KNO_3 + AgI[/tex]

At the end point; the moles of KI will definitely be equal to the moles of AgNO₃

So;

[tex]M_{KI}V_{KI}= M_{AgNO_3}V_{AgNO_3} \\ \\ V_{AgNO_3} = \dfrac{M_{KI}V_{KI}}{M_{AgNO_3}} \\ \\ \\ V_{AgNO_3} = \dfrac{ 0.1*15}{0.1}[/tex]

[tex]V_{AgNO_3} = 15 \ ml[/tex]

Thus; the volume of 0.1 M AgNO₃  needed to reach the end point is 15 mL

Answer:

Canonical structures of a chemical specie explain its observed properties from a valence bond theory perspective.

Explanation:

Resonance is a valence bond concept introduced by Linus Pauling to explain the observed properties of certain chemical species such as bond lengths, bond angles, bond order , etc.

There are certain chemical species for which a single chemical structure does not suffice in explaining its observed properties. For instance, the bond order in CO3^2- is about 1.33. Its bond length, shows that the C-O bond present in CO3^2- is neither a pure C-O single bond nor a pure C-O double bond. Hence the structure of CO3^2- is 'somewhere in between' three contributing canonical structures as shown in the image attached to this answer. The resonance structures of NO3^- are also shown.

Answer:

Mass of Al2S3 remaining is 17.212 g

Explanation:

Equation of the reaction is given below:

Al2S3 + 6H2O -----> 2Al(OH)3 + 3H2S

From the balanced equation above

6 mole of H20 reacts with 1 mole of Al2S3

i.e. 6 * 18.02 g of H2O reacts with 1 * 150.71 g of Al2S3

= 108.12 g of H2O reacts with 150.71 g of Al2S3

Therefore 2.0 g of water will react with 2.0 * (150.71/108.12) g of Al2S3

= 2.788 g of Al2S3

Mass of Al2S3 remaining = 20.0 g - 2.788 g = 17.212 g

According to the properly balanced chemical equation, the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction is 17.22 grams.

Given the following data:

To calculate the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction:

First of all, we would write a properly balanced chemical equation for this chemical reaction.

                        [tex]Al_2S_3 + 6H_2O ---> 2Al(OH)_3 + 3H_2S[/tex]

By stoichiometry:

1 mole of [tex]AL_2S_3[/tex] reacts with 6 moles of [tex]H_2O[/tex]

Next, we would calculate the mass of each compound.

For [tex]AL_2S_3[/tex]:

[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 1 \times 150.17[/tex]

Mass = 150.17 grams

For [tex]H_2O[/tex]:

[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 6 \times 18.02[/tex]

Mass = 108.12 grams

108.12 grams of [tex]H_2O[/tex] = 150.17 grams of [tex]AL_2S_3[/tex]

2.00 grams of [tex]H_2O[/tex] = X grams of [tex]AL_2S_3[/tex]

Cross-multiplying, we have:

[tex]108.12 \times X = 150.17 \times 2\\\\108.12X = 300.34\\\\X = \frac{300.34}{108.12}[/tex]

X = 2.78 grams of [tex]AL_2S_3[/tex]

Remaining mass = [tex]20.00 - 2.78[/tex]

Remaining mass = 17.22 grams of [tex]AL_2S_3[/tex]

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Answer:

[Cd²⁺] = 2.459x10⁻⁶M

Kf = 9.96x10⁶

Explanation:

Solubility of CdC₂O₄ is 6.1x10⁻³M and ksp is 1.5x10⁻⁸

The ksp of CdC₂O₄ is:

CdC₂O₄(s) ⇄ Cd²⁺(aq) + C₂O₄²⁻(aq)

ksp = [Cd²⁺] [C₂O₄²⁻] = 1.5x10⁻⁸

As solubility is 6.1x10⁻³M, concentration of C₂O₄²⁻ ions is 6.1x10⁻³M. Replacing:

[Cd²⁺] = 1.5x10⁻⁸ / [6.1x10⁻³M]

All Cd²⁺ in solution is 6.1x10⁻³M and exist as Cd²⁺ and as Cd(NH₃)₄²⁺. That means concentration of Cd(NH₃)₄²⁺ is:

[Cd(NH₃)₄²⁺] + [Cd²⁺] = 6.1x10⁻³M

[Cd(NH₃)₄²⁺] = 6.1x10⁻³M - 2.459x10⁻⁶M = 6.098x10⁻³M

In the same way, the whole concentration of NH₃ in solution is 0.150M, as you have 4ₓ6.098x10⁻³M = 0.024M of NH₃ producing the complex, the concentration of the free NH₃ is:

[0.150M] = [NH₃] + 0.024M

The equilibrium of the complex formation is:

Cd²⁺ + 4 NH₃ → Cd(NH₃)₄²⁺

The kf, formation constant, is defined as:

Kf = [Cd(NH₃)₄²⁺] / [Cd²⁺] [NH₃]⁴

Replacing:

Kf = [6.098x10⁻³M] / [2.459x10⁻⁶M] [0.1256M]⁴

Answer:

Manganese(II) phosphate | Mn3(PO4)2 - PubChem

Answer:

Magnese(ll) posphate M23 (p042) Molecular weight.

That is what the leters stand for!

Answer:

E) 1.85 M

Explanation:

M(C12H22O11) = 342.3 g/mol

22.5 g * 1mol/342.3 g = 0.0657 mol

35.5 mL = 0.0355 L

Molarity = mol solute/L solution = 0.0657 mol/0.0355L =1.85 mol/L = 1.85 M

The molarity of the aqueous solution is 1.85 M. The correct option is E) 1.85 M

From the question,

We are to determine the molarity (that is, concentration) of the given sucrose solution

First, we will determine the number of moles present in the given mass of sucrose

Mass of sucrose = 22.5 g

Using the formula

[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]

Molar mass of sucrose = 342.2965 g/mol

∴ Number of moles of sucrose present = [tex]\frac{22.5}{342.2965}[/tex]

Number of moles of sucrose present = 0.0657325 moles

Now, for the molarity (concentration) of the sucrose solution

From the formula

Number of moles = Concentration × Volume

Then,

[tex]Concentration = \frac{Number\ of\ moles}{Volume}[/tex]

From the question,

Volume = 35.5 mL = 0.0355 L

∴ [tex]Concentration = \frac{0.0657325}{0.0355}[/tex]

Concentration = 1.85 M

Hence, the molarity of the aqueous solution is 1.85 M. The correct option is E) 1.85 M

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Answer:

ΔH = - 44.0kJ

Explanation:

H2O(l)→H2O(g), ΔH =44.0kJ

In the reaction above, liquid water changes to gaseous water. This occurs through a process known as boiling. This process requires heat, hence the ΔH  is positive.

If he reaction is reversed, we have;

H2O(g)→H2O(l)

In this reaction, gaseous water changes to liquid water. This process is known as condensation. The water vapor loses heat in this reaction. Hence ΔH would be negative but still  have the same value.

Answer:

0.077M is the concentration of the hydroxyl ion

Explanation:

If 7.35 cm3 of this base is take and diluted to 147 cm3, then what is the concentration of the hydroxyl ion?

Use the dilution equation:

M1V1 = M2V2

M1 * 147cm³ = 1.5 M * 7.35 cm³

M1 = 1.5 M * 7.35 cm³ / 147 cm³

M1 = 0.077 M

How many moles of hydroxyl ion are there in the 7.35 cm3?

1000 cm³ contains 1.5 mol OH- ions

7.35 cm³ contains : 7.35 cm³ / 1000 cm³ *1.5 mol

= 0.011025 mol

2 i hope this helps

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Answer:

Part A

The natural abundance of Si-30 = 3.1

Part B

The mass of Si-30 = 29.9551 amu

Explanation:

Part A

The sum of the natural abundances = 100

Si-28 has a mass of 27.9769 amu and a natural abundance of 92.2%.

Si-29 has a mass of 28.9765 amu and a natural abundance of 4.67%

Si-30's mass and natural abundance are unknown.

Natural abundance for Si-30 = 100% - 92.2% - 4.67% = 3.13% = 3.1% to 2 s.f.

Part B

The total atomic mass of an element is an addition combination of the mass and natural abundances of all the isotopes of that element.

Molar mass of Silicon normally = 28.0855 amu

Let the mass of Si-30 be m

28.0855 = (27.9769×0.922) + (28.9765×0.0467) + (m×0.0313)

28.0855 = 27.14790435 + 0.0313m

0.0313m = 28.0855 - 27.14790435 = 0.93759565

m = (0.93759565/0.0313) = 29.9551325879 amu = 29.9551 amu

Hope this Helps!!

Part A:  The natural abundance of Si-30 = 3.1

  Lets solve the question:

Natural abundance (NA) refers to the abundance of isotopes of a chemical element as naturally found on a planet.

Natural abundance for Si-30 = [tex]100\% - 92.2\% - 4.67\% = 3.13\% = 3.1\%[/tex] in significant figures.

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Answer:

D.  0.4 (mol/L)/S

Explanation:

You simply have to plug in the given values into the rate law.

Rate = k[A][B]

Rate = (0.1)(1)²(2)²

Rate = (0.1)(1)²(4)²

Rate = 0.4

Answer:

ΔH°f C₂H₅O₂N(s)  = -537.2kJ

Explanation:

Based on the reaction:

4 C₂H₅O₂N(s) + 9O₂(g) → 8CO₂(g) + 10H₂O(l) + 2N₂(g)

ΔHrxn = ΔH°f products - ΔH°f reactants.

As:

ΔH°fO₂(g) = 0

ΔH°fCO₂(g) = -393.5kJ/mol

ΔH°fH₂O(l) = -285.8kJ/mol

ΔH°fN₂(g) = 0

The ΔHrxn is:

ΔHrxn = (8×-393.5kJ/mol + 10×-285.8kJ/mol) - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol

-6006kJ/mol - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol

-4×ΔH°fC₂H₅O₂N(s) = 2149kJ/mol

ΔH°fC₂H₅O₂N(s) = 2149kJ/mol / -4

Answer:

1). strong nuclear force 2). nuclear binding energy 3), mass defect

Explanation:

Right on Edge

1. Strong nuclear force the force that keeps the nucleons bound inside the nucleus of an atom.

2. Nuclear binding energy the amount of energy needed to split the nucleus into individual protons and neutrons.

3. Mass defect the difference between the mass of the nucleons and the mass of an Atom.

The term strong nuclear force is defined as the force that binds protons and neutrons together. It also binds them all together in a nucleus and is responsible for the energy released in nuclear reactions.

The examples of strong nuclear force are the force that hold protons and neutrons in nuclei of atoms. The elements' greater than the hydrogen atom. The fusion of hydrogen into helium in the sun's core.

Thus, 1. option B, 2. option B and 3. option B is correct.

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Answer:

Enzyme is carbonic anhydrase

Substrate is [tex]CO_2[/tex]

Turnover number is [tex]10^{7}[/tex]

Explanation:

An enzyme is used by a living organism as a catalyst to perform a specific biochemical reaction.

A substrate is a molecule upon which an enzyme acts.

Turnover number refers to the number of substrate molecules transformed by a single enzyme molecule per minute. Here, the enzyme is the rate-limiting factor.

Here,

Enzyme is carbonic anhydrase

Substrate is [tex]CO_2[/tex]

Turnover number is [tex]10^{7}[/tex]

Answer:

CH4 (g) + O2 (g) → CO2 (g) + H2O (g)

This is a combustion reaction.

Explanation:

The combination of methane (CH4) and oxygen (O2) yields carbon dioxide (CO2) and water (H2O).

CO2 is typically a gas, and water, in this case, is in a gas form because it evaporated.

The reaction is combustion because the methane reacts with the oxygen to produce carbon dioxide and water. Combustion reactions must involve O2 as one reactant.