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3 years ago

How many grams of K2CO3 are needed to prepare a 400.0 mL of a 4.25M solution?

Chemistry

1 answer:

Basile [38]3 years ago

7 0

Answer:

235 g

Explanation:

From the question;

Volume is 400.0 mL
Molarity of a solution is 4.25 M

We need to determine the mass of the solute K₂CO₃,

we know that;

Molarity = Number of moles ÷ Volume

Therefore;

First we determine the number of moles of the solute;

Moles = Molarity × volume

Moles of K₂CO₃ = 4.25 M × 0.4 L

= 1.7 moles

Secondly, we determine the mass of K₂CO₃,

We know that;

Mass = Moles × Molar mass

Molar mass of K₂CO₃, is 138.205 g/mol

Therefore;

Mass = 1.7 moles × 138.205 g/mol

= 234.9485 g

= 235 g

Thus, the mass of K₂CO₃ needed is 235 g

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castortr0y [4]

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3 years ago

PLEASE HELP!!!!!!!!!

KonstantinChe [14]

Answer:

Option "C" is the correct answer to the following question.

Explanation:

Given:

Pressure in an automobile tire (P) = 1.88 atm

Temperature (K) = 25°C = 273 + 25 = 298 Kelvin

New temperature (K1) = 37°C = 273 + 37 = 310 Kelvin

Find:

New pressure in an automobile tire (P1) = ?

Computation:

$\frac{P}{K}= \frac{P1}{K1} \\\\\frac{1.88}{298}= \frac{P1}{310}\\\\1.9557$

New pressure in an automobile tire (P1) = 1.9557

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5 0

3 years ago

Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016Na+ ions arrive at the negative electrode and 3.9

EleoNora [17]

Answer : The current passing between the electrodes is, $1.056\times 10^{-2}A$

Explanation :

First we have to calculate the charge of sodium ion.

$q=ne$

where,

q = charge of sodium ion

n = number of sodium ion = $2.68\times 10^{16}$

e = charge on electron = $1.6\times 10^{-19}C$

Now put all the given values in the above formula, we get:

$q=(2.68\times 10^{16})\times (1.6\times 10^{-19}C)=4.288\times 10^{-3}C$

Now we have to calculate the charge of chlorine ion.

$q'=ne$

where,

q' = charge of chlorine ion

n = number of chlorine ion = $3.92\times 10^{16}$

e = charge on electron = $1.6\times 10^{-19}C$

Now put all the given values in the above formula, we get:

$q'=(3.92\times 10^{16})\times (1.6\times 10^{-19}C)=6.272\times 10^{-3}C$

Now we have to calculate the current passing between the electrodes.

$I=\frac{q}{t}+\frac{q'}{t}$

$I=\frac{4.288\times 10^{-3}}{1.00}+\frac{6.272\times 10^{-3}}{1.00}$

$I=1.056\times 10^{-2}A$

Thus, the current passing between the electrodes is, $1.056\times 10^{-2}A$

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3 years ago

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Anettt [7]

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