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3 years ago

N2 + 3H2 + 2NH3 .

Chemistry

1 answer:

HACTEHA [7]3 years ago

7 0

Answer:

Explanation:

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What mass of H2 is needed to react with 8.75 g of O2 according to the following equation: O2(g) + H2(g) → H2O(g)?

Alika [10]

Answer:- B: $1.10g H_2$ is the right answer.

Solution:- The balanced equation is:

$O_2(g)+2H_2(g)\rightarrow 2H_2O(g)$

We have been given with 8.75 grams of oxygen and asked to calculate the grams of hydrogen needed to react with given grams of oxygen according to the balanced equation.

From balanced equation, 1 mole of oxygen reacts with 2 moles of hydrogen.

So, let's convert grams of oxygen to moles and multiply it by the mole ratio to calculate the moles of hydrogen that are easily converted to grams on multiplying by it's molar mass.

The complete set up looks as:

$8.75g O_2(\frac{1mole}{32g})(\frac{2mole H_2}{1mole O_2})(\frac{2.02g}{1mole})$

= $1.10g H_2$

Hence, the right option is B: $1.10g H_2$ .

7 0

3 years ago

Read 2 more answers

which has a higher entropy for the reaction: 2nh3(g)→n2(g) 3h2(g) which has a higher entropy for the reaction: reactants product

krok68 [10]

The one that has a higher entropy for the reaction is products.

<h3>What is entropy?</h3>

Entropy is a measureable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty. It is the measurement of the amount of thermal energy per unit of temperature in a system that cannot be used for productive work. It is a measure of a system's molecular disorder or unpredictability since work is produced by organized molecular motion.

It should bm be noted that the entropy of gas is more than entropy of aqueous which is more than the entropy of liquid and the entropy of solid.

On the product side there are more gas than the reactant side. Therefore, product has more entropy.

Learn more about entropy on:

brainly.com/question/6364271

#SPJ1

3 0

1 year ago

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

5 0

3 years ago

PLEASE HELP ME

ki77a [65]

Answer:

1.15 atm

Explanation:

According to Dalton's law of partial pressures, the total pressure is the sum of all the partial pressures of the gases present in the mixture.

Therefore we have:

Total pressure = partial pressure of carbon monoxide + partial pressure of oxygen + partial pressure of carbon dioxide

We were given the following:

Total pressure = 2.45 atm

Pressure of oxygen = 0.65 atm

Pressure of carbon monoxide = x

Pressure of carbon dioxide = 0.65 atm

Therefore:

2.45 = x + 0.65 + 0.65

2.45 = x + 1.3

x = 2.45 - 1.3

x = 1.15 atm

6 0

3 years ago

Explain why is it better to use images on hazard labels, rather than words

enot [183]

In the case of an emergency where you might not have enough time to read several lines of writing, not to mention trying to find the hazard warnings when the whole bottle is probably covered in writing, it is much easier to locate and read universal hazard symbols.

5 0

3 years ago

Read 2 more answers