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dangina [55]
3 years ago
10

How would you prepare 100.0 mL of 0.400 M CaCl2 from a stock solution of 2.00 M CaCl2?

Chemistry
2 answers:
aleksley [76]3 years ago
7 0

Answer:

Which of these did you include in your answer?

Use the dilution equation to calculate the volume of the stock solution required.

Stock solution volume required is 20.0 mL.

Add water to concentrated solution to reach the desired volume.

The volume of water to add is 80.0 mL.

Explanation:

the answers for edge 2020

Sever21 [200]3 years ago
4 0

Answer:

Measure 20mL of the stock

Explanation:

All we need to do is to find the volume of the stock solution needed. This is illustrated below:

C1 = 2M

V1 =?

C2 = 0.400 M

V2 = 100mL

C1V1 = C2V2

2 x V1 = 0.4 x 100

Divide both side by 2

V1 = ( 0.4 x 100) /2

V1 = 20mL

Therefore, to prepare the solution, we will measure 20mL of the stock solution and make it up to the 100mL mark.

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What are three other elements that have 8 valance electrons.
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8 valence electrons means that the element is stable. When we hear stable, we know the group to look at! Noble gases all have 8 valence electrons making them stable and have a full octet. 

Here are some:-

Krypton
Argon
Neon
5 0
3 years ago
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What is the word equation for atoms of magnesium reacting with molecules of oxygen to produce magnesium oxide?
Kazeer [188]

Answer:

The "word equation" is Magnesium + Dioxygen → Magnesium Oxide

Explanation: This is also the same as Mg + O₂ → MgO.

5 0
2 years ago
A 13.5 g sample of an unknown gas occupies 5.10 L at 149.83 kPa and 301 K. What is the molar mass of the gas ?
alisha [4.7K]

Answer:

The molar mass of the gas is 44.19 g/mol

Explanation:

Amount of sample of gas = m = 13.5 g

Volume occupied by the gas = V = 5.10 L

Pressure of the gas = P = 149.83 KPa

1 KPa = 0.00986 atm

P = 149.83 \textrm{ KPa} \times 0.00986 \textrm{ atm/KPa} = 1.48 \textrm{ atm}

Assuming M g/mol to be the molar mass of the gas

Assuming the gas is behaving as an ideal gas

\textrm{PV} =\textrm{nRT} \\\textrm{PV} = \displaystyle \frac{m}{M}\textrm{ RT } \\1.48 \textrm{ atm}\times 5.10 \textrm{ L} = \displaystyle \frac{13.5 \textrm{ g}}{M}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 301\textrm{K} \\M = 44.19 \textrm{ g/mol}

The molar mass of gas is 44.19 g/mol

7 0
4 years ago
Empirical Formula of P3O4H2?
puteri [66]

Answer:

H2 P4 O1. Explanation: In order to calculate the Empirical formula , we will assume that we have started with 10 g of the compound.

Explanation:

3 0
3 years ago
Use the periodic table to find the molar mass of
Naddika [18.5K]

I can give you a clue of getting mass. if it is the atomic number is even multiply by 2 but if the atomic number is odd multiply by 2 and add 1

sodium atomic number = 11 so mass = 11*2 +1 = 23

oxygen. atomic number = 8 so mass = 8*2 = 16

carbon atomic number = 6 so mass = 6*2 = 12

3 0
3 years ago
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