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kaheart [24]
4 years ago
8

Calculate the percentage composition of (NH4)CO3

Chemistry
2 answers:
ad-work [718]4 years ago
6 0

Answer:

%N = 29.16 %

%H = 8.39 %

%C = 12.50 %

%O = 49.95%

Explanation:

Step 1: Data given

Molar mass of (NH4)2CO3 = 96.086 g/mol

Molar mass of N = 14.007 g/mol

Molar mass of H = 1.008 g/mol

Molar mass of C = 12.011 g/mol

Molar mass of O = 15.999 g/mol

Step 2: Calculate the percentage of the elements

% N = (2*14.007 / 96.086) *100 %

%N = 29.16 %

%H = (8*1.008 / 96.086) *100%

%H = 8.39 %

%C = (12.011 / 96.086) *100%

%C = 12.50 %

%O = (3*15.999/96.086) *100%

%O = 49.95%

faltersainse [42]4 years ago
4 0

Explanation: In order to determine percent composition of the elements in a compound, the molar mass (MM) of each element is divided by the MM of the compound and multiplied by 100. The molar mass of ammonium carbonate ((NH4)2CO3) is 96.086 g/mol .

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Which elements above will form cations? List them below.
cricket20 [7]

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Halogens always form anions, alkali metals and alkaline earth metals always form cations. Most other metals form cations (e.g. iron, silver, nickel), whilst most other nonmetals typically form anions (e.g. oxygen, carbon, sulfur).

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A helium-filled balloon at 310.0 K and 1 atm, contains 0.05 g He, and has a volume of 1.21 L. It is placed in a freezer (T = 235
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Answer : The value of \Delta E of the gas is 2.79 Joules.

Explanation :

First we have to calculate the moles of helium.

\text{Moles of helium}=\frac{\text{Mass of helium}}{\text{Molar mass of helium}}

Molar mass of helium = 4 g/mole

\text{Moles of helium}=\frac{0.05g}{4g/mole}=0.0125mole

Now we have to calculate the heat.

Formula used :

q=nc_p\Delta T\\\\q=nc_p(T_2-T_1)

where,

q = heat

n = number of moles of helium gas = 0.0125 mole

c_p = specific heat of helium = 20.8 J/mol.K

T_1 = initial temperature = 310.0 K

T_2 = final temperature = 235.0 K

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

q=nc_p(T_2-T_1)

q=(0.0125mole)\times (20.8J/mol.K)\times (235.0-310.0)K

q=-19.5J

Now we have top calculate the work done.

Formula used :

w=-p\Delta V\\\\w=-p(V_2-V_1)

where,

w = work done

p = pressure of the gas = 1 atm

V_1 = initial volume = 1.21 L

V_2 = final volume = 0.99 L

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

w=-p(V_2-V_1)

w=-(1atm)\times (0.99-1.21)L

w=0.22L.artm=0.22\times 101.3J=22.29J

conversion used : (1 L.atm = 101.3 J)

Now we  have to calculate the value of \Delta E of the gas.

\Delta E=q+w

\Delta E=(-19.5J)+22.29J

\Delta E=2.79J

Therefore, the value of \Delta E of the gas is 2.79 Joules.

3 0
3 years ago
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