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Bezzdna [24]
3 years ago
13

Please help last assignment of the day

Chemistry
1 answer:
Andrej [43]3 years ago
3 0

In this qn, you use the mole formula,

amount (mol) = mass (g) ÷ molar mass

(molar mass is the mass number of an element, which u can find by referring to the periodic table)

so by subbing in the values given to you, you can find the answer

5. amount of Cu = 55.9 ÷ 63.5 = 0.880mol (3sf)

6. amount of N2 = 340 ÷ 2(14.0) = 12.1mol (3sf)

(note: you multiply the molar mass of N by 2 as there are 2 N atoms)

for qn 7 and 8, youre finding the mass and not the amount. By bringing the molar mass over to the other side of the formula, you get

mass = amount (mol) × molar mass

thus,

7. mass of Fe = 0.063 × 55.8 = 3.52g (3sf)

8. mass of H2 = 94.7 × 2(1.0) = 47.4g (3sf)

extra things to take note of:

- likewise, you can also find the molar mass of an element by using

molar mass = mass(g) ÷ amount(mol)

^^ this is commonly only needed when you need to find the molar mass of an isotope. the question will give you the amount and mass of the element, and you just need to factor them in to find your answer.

- typically, you use 3 sf in your final answer and 5sf in your working. remember to write (3sf) at the end of your number statement when necessary.

- when you get an answer of (eg.) 0.6mol, you still need to write it in 3sf, so your answer would be 0.600mol

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Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the fi
Savatey [412]

The question is incomplete, the complete question is;

Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the first heating, your crucible and contents weighs 17.51 g. After the second heating, your crucible and contents weighs 17.50 g.

What is the theoretical yield of sodium carbonate?

What is the experimental yield of sodium carbonate?

What is the percent yield for sodium carbonate?

Which errors could cause your percent yield to be falsely high, or even over 100%?

Answer:

See Explanation

Explanation:

We have to note that water is driven away after the second heating hence we are concerned with the weight of the pure dry product.

Hence;

From the reaction;

2 NaHCO3 → Na2CO3(s) + H2O(l) + CO2(g)

Number of moles of  sodium bicarbonate = 18.56 - 15.98 = 2.58 g/87 g/mol

= 0.0297 moles

2 moles of sodium bicarbonate yields 1 mole of sodium carbonate

0.0297 moles of 0.015 moles  sodium bicarbonate yields 0.0297 * 1/2 = 0.015 moles

Theoretical yield of sodium carbonate = 0.015 moles * 106 g/mol = 1.59 g

Experimental yield of sodium bicarbonate = 17.50 g - 15.98 g = 1.52 g

% yield = experimental yield/Theoretical yield * 100

% yield = 1.52/1.59 * 100

% yield = 96%

The percent yield may exceed 100% if the water and CO2 are not removed from the system by heating the solid product to a constant mass.

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