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3 years ago

How many grams are in0.465 moles of HCI

Chemistry

1 answer:

MArishka [77]3 years ago

3 0

Answer:

Explanation:

Amount HCl = 0.455mole

Molar mass = 1 + 35.5 = 36.5g/mol

Amount = mass/molar mass

:- mass = Amount × Molar mass

Mass = 16.97g

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What factors does the climate include?

rosijanka [135]

Answer:

A. amount of precipitation, average temperature

Explanation:

Precipitation and average temperature are factors that climate includes. These factors are determined by other factors such as location of an area(like how far a place is from large bodies of water like the sea), ocean currents, lattitude (distance from the equator), winds (prevailing winds) , topography (such as mountains) and the like.

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4 years ago

The molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution is ________.

Nonamiya [84]

Answer:

1.86 M

Explanation:

From the question given above, the following data were obtained:

Mass of sucrose (C12H22O11) = 22.5 g

Volume of solution = 35.5 mL

Molarity of solution =?

Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

Mass of sucrose (C12H22O11) = 22.5 g

Molar mass of C12H22O11 = (12×12) + (22×1) + (16×11)

= 144 + 22 + 176

= 342 g/mol

Mole of C12H22O11 =?

Mole = mass /Molar mass

Mole of C12H22O11 = 22.5 /342

Mole of sucrose (C12H22O11) = 0.066 mole

Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

35.5 mL = 0.0355 L

Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

Mole of sucrose (C12H22O11) = 0.066 mole

Volume of solution = 0.0355 L.

Molarity of solution =?

Molarity = mole /Volume

Molarity of solution = 0.066/0.0355

Molarity of solution = 1.86 M

Therefore, the molarity of the solution is 1.86 M.

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3 years ago

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3 years ago

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide Al2O3 dissolved in molten c

gtnhenbr [62]

Answer:

1.13 × 10⁶ g

Explanation:

Let's consider the reduction of aluminum (III) from Al₂O₃ to pure aluminum.

Al³⁺ + 3 e⁻ → Al

We can establish the following relations:

1 Ampere = 1 Coulomb / second
The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
1 mole of Al is produced when 3 moles of electrons circulate
The molar mass of Al is 26.98 g/mol.

The mass of aluminum produced under these conditions is:

$90.0 s \times \frac{1s}{620c} \times \frac{96,468c}{1mole^{-} } \times \frac{3mole^{-}}{1molAl} \times \frac{26.98gAl}{1molAl} =1.13 \times 10^{6} g Al$

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3 years ago

How many grams are in0.465 moles of HCI​

How many grams are in0.465 moles of HCI