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Free_Kalibri [48]
4 years ago
9

Can someone explain to me what I'm doing wrong? 68mL was wrong as well. Selecting "tenths place" for 3a and "hundredths place" f

or 3b and then putting a decimal was wrong as well. I'm lost as to what I should do.

Chemistry
1 answer:
ella [17]4 years ago
4 0

Answer:

Tens place

Ones place

69.5 ml

Explanation:

Hi! Sorry about your frustration.

Looking at the attachment again, I can then say that

3a) the answer would be only tens place. You may think ones place is correct, but then again look at the beaker. Are there any measurement in ones graduated on it? The question says "certain", while it's possible to calculate in ones, you most likely would be guessing and not certain about it.

3b) from the explanation I did in 3a above, it is thus clear ones place is the answer here. Calculating in tens is very certain, calculating in tenth and hundredth place is quite hard, if not impossible. Therefore, ones place is possible, but uncertain.

3c) If both 68 & 69 are incorrect, I'd advise you try something even much closer to 70, in 69.5

Kindly vote brainliest, thanks

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A compound that contains only carbon, hydrogen, and oxygen is 58.8% C and 9.87% H by mass. What is the empirical formula of this
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<u>Answer:</u> The empirical formula of the compound becomes C_5H_{10}O_2

<u>Explanation:</u>

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

Let the mass of the compound be 100 g

Given values:

% of C = 58.8%

% of H = 9.87%

% of O = [100 - 58.8 - 9.87] = 31.33%

Mass of C = 58.8 g

Mass of H = 9.87 g

Mass of O = 31.33 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of C}=\frac{58.8g}{12g/mol}=4.9 mol

\text{Moles of H}=\frac{9.87g}{1g/mol}=9.87 mol

\text{Moles of O}=\frac{31.33g}{16g/mol}=1.96mol

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 1.96 moles

\text{Mole fraction of C}=\frac{4.9}{1.96}=2.5

\text{Mole fraction of H}=\frac{9.87}{1.96}=5.03\approx 5

\text{Mole fraction of O}=\frac{1.96}{1.96}=1

Converting the mole fraction into whole numbers by multiplying them with 2.

\text{Mole fraction of C}=2.5\times 2=5

\text{Mole fraction of H}=5\times 2=10

\text{Mole fraction of O}=1\times 2=2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 5 : 10 : 2

Hence, the empirical formula of the compound becomes C_5H_{10}O_2

8 0
3 years ago
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