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3 years ago

Write a balanced equation for decomposition of H2CO3

Chemistry

1 answer:

nikdorinn [45]3 years ago

6 0

Use the rule AB --> A + B for decomposition

H2CO3 --> H2 + CO3

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If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams

Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of $KCl_3$ = 2.0 moles

Molar mass of $O_2$ = 32 g/mole

Now we have to calculate the moles of $MgO$

The balanced chemical reaction is,

$2KClO_3\rightarrow 2KCl+3O_2$

From the balanced reaction we conclude that

As, 2 mole of $KClO_3$ react to give 3 mole of $O_2$

So, 2.0 moles of $KClO_3$ react to give $\frac{2.0}{2}\times 3=3.0$ moles of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g$

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}$

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = $214.0^oC=273+214.0=487K$

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

w = mass of oxygen gas

$\rho$ = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

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

$P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}$

$P=1.78atm$

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

7 0

3 years ago

An atom of chromium has 24 protons, 28 neutrons, and 24 electrons. A chromium atom has_____Subatomic particles in the necleus

navik [9.2K]

3 subatomic particles in the nucleus
which is proton(24),neutron (28) and electrons(24)

5 0

3 years ago

If 14.5 g of MnO4- (permanganate) react with manganese (II) hydroxide how many grams of manganese (IV) oxide will be produced? T

Veronika [31]

Answer:

$m_{MnO_2}=21.2gMnO_2$

Explanation:

Hello,

In this case, given the balanced reaction:

$2MnO_4^-+2Mn(OH)_2\rightarrow 4MnO_2+2OH^-+H_2O$

We can see a 2:4 mole ration between permanganate ion (118.9 g/mol) and manganese (IV) oxide (86.9 g/mol), that is why the resulting mas of this last one turns out:

$m_{MnO_2}=14.5gMnO_4^-*\frac{1molMnO_4^-}{118.9gMnO_4^-}*\frac{4MnO_2}{2molMnO_4^-} *\frac{86.9gMnO_2}{1molMnO_2} \\\\m_{MnO_2}=21.2gMnO_2$

Best regards.

5 0

3 years ago

I need help please with this chemistry work

Marysya12 [62]

Answer:

question 1: 3

question 2: the number of Valence electrons in the atom

hope it helps

8 0

2 years ago

You have two sealed jars of water at the same temperature. In the first jar there is a large amount of water. In the second jar

melisa1 [442]

since both the jars are kept at the same temperature the vapor pressure will be same in both the cases.

3 0

3 years ago