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victus00 [196]
3 years ago
14

A test tube containing CaCO₃ is heated until the entire compound decomposes. If the test tube plus calcium carbonate originally

weighed 30.08 grams and the loss of mass during the experiment was 4.40 grams, what was the approximate mass of the empty test tube?
A. 20.08 g
B. 21.00 g
C. 24.50 g
D. 25.08 g
E. 25.68 g
Can anyone explain this one for me?
Chemistry
1 answer:
Marysya12 [62]3 years ago
8 0
Find the molar mass of CaCO3 then subtract the molar mass what it originally weighed and the loss of mass. Hopefully this works!
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2 years ago
What is the total number of oxygen atoms represented in the formula na2c03.10h2o?
Wewaii [24]

As per the given chemical formula- Na2CO3.10H2O, one mole of the chemical compound contains 13 moles of oxygen atoms. Hence

Number of moles of oxygen atoms in one mole of Na2CO3.10H2O = 13

number of moles of oxygen atoms in 0.2 moles of Na2CO3.10H2O = 13 X 0.2 = 2.6

Now, one mole of a substance contains 6.022 X 10^23 particles of the substance. Thus

number of atoms of oxygen in one mole of oxygen atom = 6.022 X 10^23

number of moles of oxygen atoms in 2.6 moles of oxygen atoms = 2.6 X 6.022 X 10^23 = 15.657 X 10^23

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3 years ago
How many grams of hydrogen chloride can be produced from 1.00g of hydrogen and 55.0g of chlorine? what is the limiting reactant?
zloy xaker [14]

Answer:

m_{HCl}=36.1gHCl

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required grams of HCl by firstly identifying the limiting reactant via the moles of each reactant as they are in a 1:1 mole ratio:

n_{H_2}=1.00gH_2*\frac{1molH_2}{2.02gH_2}=0.500molH_2\\\\ n_{Cl_2}=55.0gCl_2*\frac{1molCl_2}{70.9gCl_2}=0.776molCl_2

Thus, we infer the hydrogen is the limiting reactant and therefore we use its 1:2 mole ratio with HCl whose molar mass is 36.46 g/mol:

m_{HCl}=0.500molH_2*\frac{2molHCl}{1molH_2}*\frac{36.46gHCl}{1molHCl}\\\\m_{HCl}=36.1gHCl

Regards!

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3 years ago
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Answer:

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