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2 years ago

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When hydrogen peroxide (H2O2) is added to potassium iodide (KI) solution, the hydrogen peroxide decomposes into water (H2O) and

oxygen (O2). The chemical equation for this decomposition reaction is:
2H2O2 → 2H2O + O2

In this reaction, the potassium iodide is needed for the reaction to happen, but it isn’t changed in the reaction and can be used again. What is the role of potassium iodide in this reaction?

A. Reactant

B. Product

C. Precipitate

D. Catalyst

1 answer:

riadik2000 [5.3K]2 years ago

5 0

Answer:

D. Catalyst

Explanation:

Because a Catalyst is a substance that causes or accelerates a chemical reaction without itself being affected.

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What is the pressure of 1.27 L of a gas at 288°C, if the gas had a volume of 875 ml at

lyudmila [28]

The pressure of 1.27 L of a gas at 288°C, if the gas had a volume of 875 ml at 145 kPa and 176°C is 1.195 atm.

<h3>What is ideal gas equation?</h3>

Ideal gas equation of any gas will be represented as:
PV = nRT, where

P = pressure

V = volume

n = moles

R = universal gas constant

T = temperature

First we calculate the moles of gas, when the volume of gas 875 ml at

145 kPa and 176°C as:

n = (1.431atm)(0.875L) / (0.082L.atm/K.mol)(449.15K)

n = 1.252 / 36.83 = 0.033 moles

Now we measure the pressure of 0.033 moles of gas of 1.27 L of a gas at 288°C as:

P = (0.033mol)(0.082L.atm/K.mol)(561K) / (1.27L) = 1.195 atm

Hence required pressure of gas is 1.195 atm.

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2 years ago

Discuss/ Define the law of multiple proportions and provide an example.

scoundrel [369]

Answer:

Explanation:

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

For example:

In given photosynthesis reaction:

6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂

there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.

Law of multiple proportion:

When two elements combine to form two or more compounds with different proportions, the weight of on element that combine with other elements in fixed proportion is in the ratio of small whole number.

For example:

Consider the example of carbon dioxide and carbon monoxide.

CO and CO₂

we are given with 1 g carbon on both case while 1.3 g oxygen for carbon monoxide and 2.6 for carbon dioxide. It means the ratio of oxygen is 1:2.

There is 1.3 g of oxygen in carbon monoxide for one g of carbon while in case of carbon dioxide there is 2.6 g of oxygen for one gram of carbon.

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3 years ago

Which event would be impossible to explain by using John Dalton's model of the atom<br>

lisov135 [29]

Answer:

we need the model to answer

Explanation:

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Mass of the element Cr

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Answer:

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3 years ago

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

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

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

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

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

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

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

3 years ago