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3 years ago

An isotope of what element is used as the standard for the

Chemistry

1 answer:

balandron [24]3 years ago

7 0

Answer:

Since 1961 the standard unit of atomic mass has been one-twelfth the mass of an atom of the isotope carbon-12. An isotope is one of two or more species of atoms of the same chemical element that have different atomic mass numbers (protons + neutrons).

isotope carbon-12

Explanation:

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Why is water called a universal solvent?

kondor19780726 [428]

Answer: B

Explanation:

Water is called the "universal solvent" because it dissolves more substances than any other liquid.

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3 years ago

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Identify the relationship between <2 and <4*

dexar [7]

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denis23 [38]

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3 years ago

1. In order to make 1 salad it requires 1 head of lettuce, two tomatoes and three carrots. What would be the coefficients for th

pogonyaev

Answer:

3 salad = 3 lettuce + 6 tomatoes + 9 three carrots

Coefficients: 3, 6, 9

Explanation:

1 salad = 1 lettuce + 2 tomatoes + 3 three carrots

<em>Multiply all the coefficients of 1 salad by 3:</em>

3(1 salad) = 3(1 lettuce + 2 tomatoes + 3 three carrots)

<em>Expand the equation:</em>

3 salad = 3 lettuce + 6 tomatoes + 9 three carrots

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3 years ago

The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of

timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = $1.83g/cm^3=1.83g/ml$

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml$

Now we have to calculate the molarity of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L$

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg$

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

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3 years ago