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babymother [125]

3 years ago

9

7.20g of potassium sulfate reacts with the excess oxygen gas in a single displacement reaction, how many grams of oxygen are con

sumed in this reaction

1 answer:

skad [1K]3 years ago

5 0

The balanced chemical reaction:

K2SO4 + O2 = 2KO2 + SO2

Assuming that the reaction is complete, all of the potassium sulfate is consumed. We relate the substances using the chemical reaction. We calculate as follows:

7.20 g K2SO4 ( 1 mol / 174.26 g) ( 1 mol O2 / 1 mol K2SO4 ) ( 32 g / 1 mol ) = 1.32 g O2 consumed in the reaction.

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Answer:

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3 years ago

If you have 250 mL of 0.75 M LINO2.<br> How many grams of this substance<br> would you have?

Vesna [10]

Mass of this substance = 9.928 g

<h3>Further explanation</h3>

Molarity is a way to express the concentration of the solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

$\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}$

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V = Volume of solution

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4 0

3 years ago

If the value for ΔS is postive, and the value for ΔH is negative, thr reaction will be

dmitriy555 [2]

It follows that the reaction is spontaneous at high temperatures Option A.

<h3>What is ΔS ?</h3>

The term ΔS is referred to as the change in the entropy of the system. Now recall that entropy is defined as the degree of disorderliness in a system. If a system is highly disorderly then it means that it has a high entropy. Also, ΔH has to do with the heat change that accompanies a reaction.

We know that both the entropy and the heat change can both either be positive or negative. Now we know that the equation ΔG = ΔH - TΔS can be used to ascertain whether or not a reaction will be spontaneous. If the result is negative, then the reaction will be spontaneous.

As such, when then it follows that the reaction is spontaneous at high temperatures Option A.

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1 year ago

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3 years ago

One mole of oxygen gas is at a pressure of 6.00 atm and a temperature of 27.0°C. (a) If the gas is heated at constant volume unt

Umnica [9.8K]

Answer: a) 900 K

b) 1200 K

Explanation:

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 6.00 atm

V= Volume of the gas = ?

T= Temperature of the gas = 27°C = 300 K $0^00C=273K$

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas = 1

$V=\frac{nRT}{P}=\frac{1\times 0.0821\times 300}{6.00}=4.10L$

a) To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=6.00atm\\T_1=300K\\P_2=3\times 6.00=18.0atm\\T_2=?$

Putting values in above equation, we get:

$\frac{6.00}{300K}=\frac{18.0}{T_2}\\\\T_2=900K$

The final temperature is 900 K

b) The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 6.00 atm

$P_2$ = final pressure of gas = $2\times 6.00atm=12.0atm$

$V_1$ = initial volume of gas = 4.10 L

$V_2$ = final volume of gas = $2\times 4.10 L=8.20L$

$T_1$ = initial temperature of gas = $300K$

$T_2$ = final temperature of gas =?

Now put all the given values in the above equation, we get:

$\frac{6.00\times 4.10}{300}=\frac{12.0\times 8.20}{T_2}$

$T_2=1200K$

The final temperature is 1200 K

5 0

3 years ago