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Aliun [14]
3 years ago
9

How many liters of a 1.5 M solution can you make if you have .50 mol of KCl?

Chemistry
1 answer:
vaieri [72.5K]3 years ago
7 0

Answer:

The answer to your question is 0.33 liters

Explanation:

Data

Volume = ?

Molarity = 1.5 M

number of moles = 0.5

Formula

Molarity = \frac{number of moles }{volume}

Solve for V

Volume = \frac{number of moles}{molarity}

Substitution

Volume = \frac{0.5}{1.5}

Simplification and result

       Volume = 0.33 l

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Por qué razón se fomenta la inmigración Europea?​
Morgarella [4.7K]

Hola aquí va la respuesta!

Se fomentó la inmigración europea porque la economía no era buena y vinieron a Paraguay para mejorar su situación.

8 0
3 years ago
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Which is TRUE of the products in this combustion reaction? 4Fe + 3O2 2Fe2O3
ycow [4]
The only true answer is A. 

products are on the right side of the reaction

products and reactants don't necessarily have same physical or chemical properties

It is the reactants that <span>are the atoms, molecules, or compounds that participate in the reaction.</span>
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3 years ago
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Which graph shows the pressure-temperature relationship for a gas at a fixed volume?
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C. Graph A

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3 years ago
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How many atoms of carbon are in a container that has 0.450 mol of CO2 and 2.55 mol of CaC2?
Fantom [35]

Answer:

The number of carbon atoms in the container is 1.806 × 10²⁴ or the container contains 1.806 × 10²⁴ atoms of carbon

Explanation:

By Avogadro's number, 1 mole of a substance contains 6.02 × 10²³ particles of the substance

Here we have 0.45 mole of CO₂ contains

0.45 × 6.02 × 10²³ particles of CO₂ that is 2.709 × 10²³ particles of CO₂ or equivalent to 2.709 × 10²³ atoms of Carbon

Similarly, 2.55 moles of CaC₂ contains 2.55 × 6.02 × 10²³ particles of CaC₂ or 1.5351 × 10²⁴ atoms of Carbon

The total number of carbon atoms is therefore;

2.709 × 10²³ + 1.5351 × 10²⁴ = 1.806 × 10²⁴ atoms of carbon.

5 0
4 years ago
The recommended daily allowance (rda of calcium is 1.2 g. calcium carbonate contains 12.0% calcium by mass. how many grams of ca
defon
1) Calcium carbonate contains 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
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2) Mass fraction of this is excessive data.

3) The solution is:

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m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)

m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g
4 0
3 years ago
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