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3 years ago

If 110g of copper sulphate is present in 550g of solution calculate the concentration of solution

Chemistry

1 answer:

Bingel [31]3 years ago

7 0

Answer:

20%

Explanation:

mass by mass percentage of a solution =(mass of solute)/(mass of solution)

mass of solute=550g

therefore 110×100/550=20%

hope u will understand .:") credit to the owner

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Which of the following was originally a tenet of Dalton's atomic theory, but had to be revised about a century ago

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4 years ago

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2567 milliliters to liters

ollegr [7]

Answer:

2.567

Explanation:

To change milliliters into liters you divide by 1000

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When 100.g Mg3N2 reacts with 75.0 g H2O, what is the maximum theoretical yield of NH3?

Temka [501]

Answer : The correct option is, 23.6 g

Explanation : Given,

Mass of $Mg_3N_2$ = 100.0 g

Mass of $H_2O$ = 75.0 g

Molar mass of $Mg_3N_2$ = 101 g/mol

Molar mass of $H_2O$ = 18 g/mol

First we have to calculate the moles of $Mg_3N_2$ and $H_2O$ .

$\text{Moles of }Mg_3N_2=\frac{\text{Given mass }Mg_3N_2}{\text{Molar mass }Mg_3N_2}$

$\text{Moles of }Mg_3N_2=\frac{100.0g}{101g/mol}=0.990mol$

and,

$\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}$

$\text{Moles of }H_2O=\frac{75.0g}{18g/mol}=4.17mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$Mg_3N_2(s)+6H_2O(l)\rightarrow 3Mg(OH)_2(aq)+2NH_3(g)$

From the balanced reaction we conclude that

As, 6 moles of $H_2O$ react with 1 mole of $Mg_3N_2$

So, 4.17 moles of $H_2O$ react with $\frac{4.17}{6}=0.695$ moles of $Mg_3N_2$

From this we conclude that, $Mg_3N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2O$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 6 moles of $H_2O$ react to give 2 moles of $NH_3$

So, 4.17 moles of $H_2O$ react to give $\frac{2}{6}\times 4.17=1.39$ mole of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

Molar mass of $NH_3$ = 17 g/mole

$\text{ Mass of }NH_3=(1.39moles)\times (17g/mole)=23.6g$

Therefore, the maximum theoretical yield of $NH_3$ is, 23.6 grams.

4 0

3 years ago

Ethylene glycol is used as an antifreeze in car. if 400 g of ethylene glycol is added to 4.00 kg of water, what is the molaity?

Illusion [34]

Data:

solute: ethylene glicol => not ionization

molar mass of ethylene glicol (from internet) = 62.07 g/mol

solute = 400 g

solvent = water = 4.00 kg

m =?

ΔTf = ?

Kf = 1.86 °C/mol

Formulas:

m: number of moles of solute / kg of solvent

ΔTf = Kf*m

number of moles of solute = mass in grams / molar mass

Solution

number of moles of solute = 400 g / 62.07 g/mol = 6.44 moles

m = 6.44 mol / 4 kg = 1.61 m <-------- molality (answer)

ΔTf = 1.86 °C / m * 1.61 m = 2.99 °C <---- lowering if freezing point (answer)

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3 years ago

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sattari [20]

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