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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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2 years ago

2. What are the different areas of an electron cloud called?

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5 0

3 years ago

When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin

loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.2^0C$ = Depression in freezing point

$K_f$ = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality = $\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9$

$8.2^0C=1\times K_f\times 1.9$

$K_f=4.32^0C/m$

Now Depression in freezing point for sodium chloride is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=20.0^0C$ = Depression in freezing point

$K_f$ = freezing point constant

m= molality = $\frac{136g\times 1000}{950g\times 58.5g/mol}=2.45$

$20.0^0C=i\times 4.32^0C\times 2.45$

$i=1.9$

Thus vant hoff factor for sodium chloride in X is 1.9

3 0

2 years ago

If a sample containing 18.1 g of NH3 is reacted with 90.4 g of

USPshnik [31]

Answer:

3.64g

Explanation:

Given parameters:

Mass of NH₃ = 18.1g

Mass of Cu₂O = 90.4g

Unknown:

Limiting reactant = ?

Mass of N₂ formed = ?

Solution:

The reaction equation is given as:

Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol

Molar mass of NH₃ = 14 + 3(1) = 17g/mol

Number of moles of Cu₂O = $\frac{18.1}{143.2}$ = 0.13moles

Number of moles of NH₃ = $\frac{90.4}{17}$ = 5.32moles

From this reaction;

1 mole of Cu₂O combines with 2 mole of NH₃

So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃

= 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

Mass = number of moles x molar mass

1 mole of Cu₂O will produce 1 mole of N₂

0.13 mole of Cu₂O will produce 0.13 mole of N₂

Mass = 0.13 x (2 x 14) = 3.64g

5 0

3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago