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3 years ago

A 31.3-g sample of ammonium carbonate contains ________ mol of ammonium ions.

Chemistry

2 answers:

Elan Coil [88]3 years ago

8 0

Answer:

$n_{NH_4}=0.652molNH_4$

Explanation:

Hello,

Ammonium carbonate has the following formula:

$(NH_4)_2CO_3$

In this manner, the moles are computed by knowing its molecular mass:

$M_{(NH_4)_2CO_3}=14*2+1*8+12*1+16*3=96g/mol$

Thus, by applying a mass-mole-ions relationship, one obtains:

$ions_{(NH_4)_2CO_3}=31.3g(NH_4)_2CO_3*\frac{1mol(NH_4)_2CO_3}{96g(NH_4)_2CO_3}*\frac{2molNH_4}{1mol(NH_4)_2CO_3}\\n_{NH_4}=0.652molNH_4$

Best regards.

Leona [35]3 years ago

7 0

<span>We need to calculate the equivalent amount in units of moles of ammonium ions from the mass units. For this we need the molar mass of the substances involved. We calculate as follows:

31.3 g </span>(NH4)2CO3 ( 1 mol (NH4)2CO3 / 96.09 g (NH4)2CO3) ( 2 mol NH4 / 1 mol (NH4)2CO3 ) = 0.65 mol <span>ammonium ions</span>

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2 years ago

Read 2 more answers

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

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