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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

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 (NH4)2CO3 ( 1 mol (NH4)2CO3 / 96.09 g (NH4)2CO3) ( 2 mol NH4 / 1 mol (NH4)2CO3 ) = 0.65 mol ammonium ions

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1.00 M CaCl2 Density = 1.07 g/mL

Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

1) The value of %(m/M):

$\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%$

2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

$Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}$

$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

$n=\frac{40 g/mol}{2}=20$

$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

$n=\frac{35.5 g/mol}{1}=35.5$

$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0

3 years ago

Using the fact that at 0.10 M, the absorbance was 0.357; and at 0.20 M, the absorbance was 0.714, the absorptivity (slope) is

olga_2 [115]

Answer:

The answer is "3.57 and 0.07".

Explanation:

Using the slop formula:

$=\frac{n_2-n_1}{n_2-n_1}\\\\ =\frac{0.714-0.357}{0.20-0.10} \\\\ = ABSORPTIVITY, \ \ a=3.57 \\\\A= a\times b \times c$

Given:

$A=0.250$

length path $b=1$

from calibration it is found that

$a=3.57\\\\consentation \ c=\frac{A}{a\times b} \\\\C=\frac{0.250}{3.57\times 1}$

$=0.07 \ M$

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3 years ago

Identify the reaction type of the chemical reaction below. Check all that apply.

Ilya [14]

Answer:

The type of reaction is a single-replacement reaction.

Explanation:

Mg switches places with H, leaving H by itself.

8 0

3 years ago

What is one result of a chemical change?

Phoenix [80]

A new material is formed in result of a chemical change. Typically, the chemical changes always make the new material.

3 0

3 years ago

A chemical reaction produced 10.1 cm3 of nitrogen gas at 23 °C and 746 mmHg. What is the volume of this gas if the temperature a

adell [148]

Answer:

volume of gas = 9.1436cm³

Explanation:

We will only temperature from °C to K since the conversion is done by the addition of 273 to the Celsius value.

Its not necessary to convert pressure and volume as their conversions are done by multiplication and upon division using the combined gas equation, the factors used in their conversions will cancel out.

V1 =10.1cm³ , P1 =746mmHg, T1=23°C =23+273=296k

V2 =? , P2 =760mmmHg , T2=0°C = 0+273 =273K

Using the combined gas equation to calculate for V2;

$\frac{V1P1}{T1}=\frac{V2P2}{T2} \\ re-arranging, \\V2 =\frac{V1P1T2}{P2T1}$

$V2 =\frac{10.1*746*273}{760*296}$

V2=9.1436cm³

6 0

4 years ago