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ivolga24 [154]
3 years ago
6

A student attempted to identify an unknown compound by the method used in the experiment. She found that when she heated the sam

ple weighing 0.4862g, the mass barely changed, dropping to 0.4855g. When the product was converted to a chloride, the mass went up to 0.5247g. Is the sample a carbonate? What are the two compounds that might be the unknown?
Chemistry
2 answers:
Brrunno [24]3 years ago
6 0

Answer:

The sample is a carbonate.

The unknown samples are Na2CO3 and K2CO3.

Explanation:

Let's assume ,the sample is Na2CO3.

Na2CO3 produces 2 moles of NaCl via formation of Na2O

Number of moles of NaCO3= 0.48262g/106gmol= 4.586×10^-3mol

Mass of NaCl= 2(4.586×10^-3)×58.5=0.536g

Let's assume the sample is K2CO3

K2CO3 produces2 moles of KCl

Number of moles of K2CO3= 0.4862g/138.21gmol

Number of moles= 3.517×10^-3mol

Mass of KCl formed= 2(3.517×10^-3)×74.55=0.5246g

yanalaym [24]3 years ago
3 0
Na2CO3 + 2Cl- ⇒ 2NaCl + CO3^-2 
<span>
1 mole of Na2CO3 = 106 g </span>
<span>2 moles of NaCl     = 2 x 58.4
                               = 116.8 g 
</span>Na2CO3 would increase by 116.8 / 106 = 1.10 to form 2NaCl.
<span>0.4862 g x 1.10 = 0.515 grams of NaCl.
</span>
K2CO3 + 2Cl- ⇒ 2KCl + CO3^-2 
<span>1 mole of K2CO3 = 138.2 g </span>
<span>2 moles of KCl = 149.1 </span>
<span>
K2CO3 would increase by </span>149.1 /138.2 = 1.079 <span>to form 2KCl
</span>
<span> 0.4862 x 1.079 = 0.5246 g</span>


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Answer:

0.143 g of KCl.

Explanation:

Equation of the reaction:

AgNO3(aq) + KCl(aq) --> AgCl(s) + KNO3(aq)

Molar concentration = mass/volume

= 0.16 * 0.012

= 0.00192 mol AgNO3.

By stoichiometry, 1 mole of AgNO3 reacts with 1 mole of KCl to form a precipitate.

Number of moles of KCl = 0.00192 mol.

Molar mass of KCl = 39 + 35.5

= 74.5 g/mol

Mass = molar mass * number of moles

= 74.5 * 0.00192

= 0.143 g of KCl.

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3 years ago
What is the name of the compound N2(g)
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N2 stands for nitrogen.
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PLEASE HELP 20 points
777dan777 [17]

  n = 1.5atm (15L) / .0821 (280k) = .98 mol NaCl

  NaCl = 22.99g Na + 35.45g Cl = 58.44g NaCl

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How many molecules of CO2 are present
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When butane burns completely, only water and carbon dioxide gas are produced. If 11.6 g of butane and 40.0 L of oxygen at 22.0o
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19.7 litre volume of carbon dioxide gas at 22.0o C and 102 kPa can be collected over water.

<h3>What is vapour pressure?</h3>

Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature.

Moles of Butane = mass in grams / molar mass = 11.6 / 58.12 = 0.2

Volume of O_2 (V) = 40 liter

Temperature (T) = 22°C = 22 + 273 = 295 K

Pressure (P) = 102 kPa = 102 / 101.325 = 1.007 atm

Moles of O_2 (n) can be calculated by ideal gas equation.

PV = nRT

n = 1.007 40 ÷ 0.0821 295 = 1.663

Balanced chemical reaction;

2C_4H_10 + 13O_2 ---> 8CO_2 + 10H_2O

From reaction;

13 moles O_2 require 2 moles C_4H_10

So, 1.663 moles O_2 will require = 2 x 1.663 ÷13 = 0.256 moles of C_4H_10

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Moles of CO_2 produced = (8/2) 0.2 = 0.8 moles

Pressure of CO_2 (P) = 102 - 2.24 = 99.76 kPa = 99.76  ÷ 101.325 = 0.985 atm

Applying the ideal gas equation for CO_2,

PV = nRT

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V = 19.7 liter

The volume of CO_2 produced = 19.7 liter.

Learn more about the vapour pressure here:

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