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Tanya [424]
3 years ago
5

What is the molarity of 0.50 liter of an aqueous solution that contains 0.20 mole of NaOH (gram-formula mass = 40. g/mol)?

Chemistry
1 answer:
monitta3 years ago
7 0

Answer:

0.4mol/L

Explanation:

Data obtained from the question include:

Number of mole of NaOH(solute) = 0.20 mole

Volume of solution = 0.50 L

Molarity =.?

Molarity is simply mole of solute per unit litre of a solution. This can represented mathematically as:

Molarity = mole of solute /Volume of solution

Molarity of NaOH = 0.2mol/0.5L

Molarity of NaOH = 0.4mol/L

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The fuel used to power the booster rockets on
Effectus [21]

Answer:

746 moles of H2O are been produced from 373 moles of Al.

Explanation:

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3 years ago
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

8 0
3 years ago
(20 Points!!!!!)
trasher [3.6K]

i am pretty sure it would be a chemical change so A

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How is the oxidation state of a transition metal determined from the chemical formula ?
Vera_Pavlovna [14]

Answer:

Explanation:

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We know that for compounds to be formed, atoms would either lose, gain or share electrons between one another.

The oxidation state is usually expressed using the oxidation number and it is a formal charge assigned to an atom which is present in a molecule or ion.

To ascertain the oxidation state, we have to comply with some rules:

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For example, let us find the oxidation state of Cr in Cr₂O₇²⁻

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We see that the oxidation number of Cr, a transition metal in the given ion is +6.

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