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Gekata [30.6K]
3 years ago
12

Round 0.0011675cg to four significant figures

Chemistry
1 answer:
atroni [7]3 years ago
8 0

Answer:

answer is 0.001168 is correct

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Read 2 more answers
A compound is found to contain 42.88 % carbon and 57.12 % oxygen by weight. To answer the questions, enter the elements in the o
Ipatiy [6.2K]

Answer:

The empirical formule is CO

Explanation:

Step 1: Data given

Suppose the mass of a compound is 100 grams

Suppose the compound contains:

42.88 % C = 42.88 grams C

57.12 % O = 57.12 grams O

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles C = mass C / molar mass C

Moles C = 42.88 grams / 12.01 g/mol

Moles C = 3.57 moles

Moles O = 57.12 grams / 16.0 g/mol

Moles O = 3.57 moles

Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

C: 3.57 moles / 3.57 moles = 1

O: 3.57 moles / 3.57 moles = 1

The empirical formule is CO

3 0
3 years ago
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