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3 years ago

Consider the balanced equation:

Chemistry

2 answers:

Mashutka [201]3 years ago

6 0

Answer:

A). 3 Mg, 2 P, 14 O, 12 H

Explanation:

Mamont248 [21]3 years ago

3 0

The answer is 6 moles of water will be produced.

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Calculate the molarity when 135 g of KCl is dissolved to make a 2.50 L solution. Round to two significant digits.

inysia [295]

Explanation:

$\textrm{Molarity of a solution}=\frac{\textrm{Number of moles of solvent}}{\textrm{Volume of solution in Liters}}$

We are given a $2.50L$ solution which contains $135g$ of $KCl$ .

To calculate number of moles of $KCl$ present, we need to find it's molar mass from charts. Molar Mass of $KCl$ is known to be $74.55\frac{g}{Mol}$ .

Number of moles of $KCl$ present = $\frac{\textrm{Given weight}}{\textrm{Molar Mass}}\textrm{ = }\frac{135g}{74.55\frac{g}{Mol}}\textrm{ = }1.81087mol$

Molarity = $\frac{1.81087mol}{2.50L}=0.7243\frac{mol}{L}$

∴ Molarity of given $KCl$ solution = $0.72\frac{mol}{L}$

7 0

3 years ago

Calcutale Grxn for the following equation at 25°C: <br><br> 4KClO3(s) → 3KClO4(s) KCl(s)

Step2247 [10]

Answer:

-133.2 kJ

Explanation:

Let's consider the following balanced equation.

4 KClO₃(s) → 3 KClO₄(s) + KCl(s)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression.

ΔG°rxn = 3 mol × ΔG°f(KClO₄(s)) + 1 mol × ΔG°f(KCl(s)) - 4 mol × ΔG°f(KClO₃(s))

ΔG°rxn = 3 mol × (-303.1 kJ/mol) + 1 mol × (-409.1 kJ/mol) - 4 mol × (-296.3 kJ/mol)

ΔG°rxn = -133.2 kJ

5 0

3 years ago

A diver has 3,400 J of gravitational potential energy after climbing

Zanzabum

Answer:

57.8kg

Explanation:

Potential energy is given by:

$U =mgh$

Where U is potential energy, m is mass, g is acceleration due to gravity, and h is height. Using this equation:

$3400=m(9.8)(6)\\3400=58.8m\\m=57.8kg$

3 0

3 years ago

How good are you with chemistry

Ugo [173]

Answer:

<h2>━☆ﾟ.*･｡ﾟFine, but not as good</h2>

7 0

3 years ago

Read 2 more answers

A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k

Sergeeva-Olga [200]

<u>Answer:</u> The partial pressure of oxygen gas is 2.76 bar

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = $0.0831\text{ L bar }mol^{-1}K^{-1}$

n = Total number of moles = ?

Putting values in above equation, we get:

$5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol$

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ........(1)

where,

$p_A$ = partial pressure of carbon dioxide = 0.318 bar

$p_T$ = total pressure = 5.57 bar

$\chi_A$ = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

$0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, $\chi_{N_2}=\frac{0.221}{0.493}=0.448$

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

$p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar$

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0

3 years ago