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3 years ago

Given the atomic weights of carbon, 12.01; hydrogen, 1.01; and oxygen, 16.0, what is the molar mass of glucose (C6H12O6)?

2 answers:

wel3 years ago

5 0

Multiply them, then add total.

C 6 * 12.01 =

H 12 * 1.01 =

O 6 * 16.0 =
+ ------------
molar mass =

Add them together to get molar mass

nirvana33 [79]3 years ago

5 0

its D 180.18 .............................

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Perform the calculation and record the answer with the correct number of significant figures.

kotykmax [81]

The question is incomplete, here is the complete question:

A. (6.5-6.10)/3.19

B. (34.123 + 9.60) / (98.7654 - 9.249)

<u>Answer:</u>

<u>For A:</u> The answer becomes 0.1

<u>For B:</u> The answer becomes 0.4884

<u>Explanation:</u>

Significant figures are defined as the figures present in a number that expresses the magnitude of a quantity to a specific degree of accuracy.

Rules for the identification of significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant. For example: 664, 6.64 and 66.4 all have three significant figures.
All zeros between the integers are always significant. For example: 5018, 5.018 and 50.18 all have four significant figures.
All zeros preceding the first integers are never significant. For example: 0.00058 has two significant figures.
All zeros after the decimal point are always significant. For example: 2.500, 25.00 and 250.0 all have four significant figures.
All zeroes used solely for spacing the decimal point are not significant. For example: 10000 has one significant figure.

<u>Rule applied for addition and subtraction:</u>

The least precise number present after the decimal point determines the number of significant figures in the answer.

<u>Rule applied for multiplication and division:</u>

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits

<u>For A:</u> (6.5-6.10)/3.19

This a a problem of subtraction and division.

First, the subtraction is carried out.

$\Rightarow \frac{6.5-6.10}{3.19}=\frac{0.4}{3.19}$

Here, the least precise number after decimal was 1.

$\Rightarrow \frac{0.4}{3.19}=0.125$

Here, the least precise number of significant digit is 1. So, the answer becomes 0.1

<u>For B:</u> (34.123 + 9.60) / (98.7654 - 9.249)

This a a problem of subtraction, addition and division.

First, the subtraction and addition is carried out.

$\Rightarow \frac{34.123+9.60}{98.7654-9.249}=\frac{43.723}{89.5164}=\frac{43.72}{89.516}$

Here, the least precise number after decimal in addition are 2 and in subtraction are 3

$\Rightarrow \frac{43.72}{89.516}=0.48840$

Here, the least precise number of significant digit are 4. So, the answer becomes 0.4884

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3 years ago

How many grams of nicl2∙6h2o will be used to prepare a 0. 0350 m, 500. 0 ml of nicl2 solution?

marin [14]

The mass of NiCl₂•6HO₂ needed to prepare a 0.035 M 500 mL solution of NiCl₂•6HO₂ is 4.165 g

<h3>What is molarity? </h3>

This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

<h3>How to determine the mole of NiCl₂•6HO₂</h3>

Molarity = 0.035 M
Volume = 500 mL = 500 / 1000 = 0.5 L

Mole of NiCl₂•6HO₂ =?

Mole = Molarity × Volume

Mole of NiCl₂•6HO₂ = 0.035 × 0.5

Mole of NiCl₂•6HO₂ = 0.0175 mole

<h3>How to determine the mass of NiCl₂•6HO₂</h3>

Mole of NiCl₂•6HO₂ = 0.0175 mole
Molar mass of NiCl₂•6HO₂ = 238 g/mol

Mass of NiCl₂•6HO₂ =?

Mass = mole × molar mass

Mass of NiCl₂•6HO₂ = 0.0175 × 238

Mass of NiCl₂•6HO₂ = 4.165 g

Thus, 4.165 g of NiCl₂•6HO₂ is needed to prepare the solution

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