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3 years ago

The conversion factor 100mL/1 L would be used to change what

Chemistry

1 answer:

ValentinkaMS [17]3 years ago

8 0

Answer:

Liters to milliliters.

Explanation:

Conversion factors are used to convert from one unit to another. The unit you are converting from goes in the denominator, and the unit you are converting to goes in the numerator. Since liters is in the denominator, you are converting from liters to another unit. Since milliliters is in the numerator, you are converting to that unit. Therefore, you would be converting from liters to milliliters.

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What quantity of energy would be necessary to boil water with a mass of

anygoal [31]

Answer:

i think 554 I think

Explanation:

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3 years ago

A chemist prepares a solution of copper(II) fluoride by measuring out of copper(II) fluoride into a volumetric flask and filling

Simora [160]

The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

Answer: The concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

Explanation:

To calculate the molarity of solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of copper (II) fluoride = 0.0498 g

Molar mass of copper (II) fluoride = 101.54 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L$

Hence, the concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

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3 years ago

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krok68 [10]

A and d is physical, b and c is chemical

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3 years ago

What is the correct mole to mole ratio for Aluminum to Aluminum chloride in the following reaction: 2Al + 3Cl2 --> 2AlCl3

RoseWind [281]

Answer:

3;2

Explanation:

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4 years ago

Write electron configurations for each of the following. the cations: Mg2+,Sn2+,K+,Al3+,Tl+,As3+

Eddi Din [679]

Answer:

Mg⁺² ⇒ 1s² 2s² 2p⁶

Sn²⁺ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

K⁺ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶

Al³⁺ ⇒ 1s² 2s² 2p⁶

Ti⁺ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 4f¹⁴ 6s² 5d¹⁰

As⁺³ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²

Explanation:

The electron configuration indicates the way the electrons of an atom or ion are structured. In the case of cations, by knowing the electronic configuration of the atom (which is neutral), we can find out the cations' configuration by substracting n outermost electrons, where n is the charge of the cation.

Mg⁰ ⇒ [Ne] 3s² = 1s² 2s² 2p⁶ 3s². Thus

Mg⁺² ⇒ [Ne] = 1s² 2s² 2p⁶.

In a similar fashion, the answers are:

Sn²⁺ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

K⁺ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶

Al³⁺ ⇒ 1s² 2s² 2p⁶

Ti⁺ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 4f¹⁴ 6s² 5d¹⁰

As⁺³ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²

3 0

4 years ago