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4 years ago

9

Hydrogen peroxide, which decomposes in a first order reaction, has a half-life of 10 hours in air. how long will it take for hyd

rogen peroxide to decompose to 10% of its original concentration?

1 answer:

VLD [36.1K]4 years ago

4 0

Hello! Let me try to answer this :)

Thanks and please correct if there are any mistakes ^ ^

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How large would the state of New York (235 miles across) be?

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235 miles equals 5,280 feet

Explanation:

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3 years ago

The temperature of evaporation is much higher for water than for alcohol. Without knowing more about the chemistry of alcohol, w

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Answer:

Fewer hydrogen bonds form between alcohol molecules. As a result, less heat is needed for alcohol molecules to break away from solution and enter the air.

Explanation:

Hydrogen bonding is a kind of intermolecular interaction that occurs when hydrogen is bonded to a highly electronegative atom.

Both water and alcohols exhibit hydrogen bonding. However, alcohols exhibit fewer hydrogen bonds than water.

As a result of this, the temperature of evaporation is much higher for water than for alcohol because hydrogen bonds hold water molecules more closely than alcohol molecules are held.

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3 years ago

You are an intermediate product of an industrial process which intends to separate iron from its ore. A well known iron ore is h

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Answer:

Malachite

Explanation:

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3 years ago

The stopcock connecting a 3.06 L bulb containing methane gas at a pressure of 9.61 atm, and a 6.65 L bulb containing oxygen gas

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Answer : The final pressure in the system is 4.22 atm.

Explanation :

First we have to calculate the moles of methane.

$PV=n_1RT$

where,

P = pressure of gas = 9.61 atm

V = volume of gas = 3.06 L

T = temperature of gas = T

$n_1$ = number of moles of methane gas = ?

R = gas constant

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

$(9.61atm)\times (3.06L)=n_1\times RT$

$n_1=\frac{29.4}{RT}$

Now we have to calculate the moles of oxygen gas.

$PV=n_2RT$

where,

P = pressure of gas = 1.75 atm

V = volume of gas = 6.65 L

T = temperature of gas = T

$n_2$ = number of moles of oxygen gas = ?

R = gas constant

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

$(1.75atm)\times (6.65L)=n_2\times RT$

$n_2=\frac{11.6}{RT}$

Now we have to determine the final pressure in the system after mixing the gases.

$P_{total}=(n_1+n_2)\times \frac{RT}{V_{total}}$

where,

$P_{total}$ = final pressure of gas = ?

$V_{total}$ = final volume of gas = (3.06 + 6.65)L = 9.71 L

T = temperature of gas = T

R = gas constant

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

$P_{total}=(\frac{29.4}{RT}+\frac{11.6}{RT})\times \frac{RT}{9.71L}$

$P_{total}=4.22atm$

Therefore, the final pressure in the system is 4.22 atm.

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3 years ago

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Think D is the most efficient out of the bunch

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4 years ago

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