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3 years ago

Convert 8.876 × 10^12 m^2 to units of km^2.

Chemistry

1 answer:

Darya [45]3 years ago

7 0

Answer:

$8.876\times 10^{18}\ km^2$

Explanation:

In this problem, we need to convert $8.876 \times 10^{12}\ m^2$ to km².

We know that,

1 km = 1000 m

⇒ 1 km² = 10⁶ m²

So,

$8.876 \times 10^{12}\ m^2=8.876 \times 10^{12}\times 10^6\ km^2\\\\=8.876\times 10^{18}\ km^2$

So, $8.876 \times 10^{12}\ m^2$ is equal to $8.876\times 10^{18}\ km^2$ .

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dimaraw [331]

Answer:

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2 years ago

Calculate the molarity of 29.9 g of MgS in 911 mL of solution??

LUCKY_DIMON [66]

0.58 is the molarity of 29.9 g of MgS in 911 mL of solution.

Define the molarity of a solution.

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

Given data:

Volume of solution = 911 mL=0.911L

Given the mass of MgS = 29.9 g

Molar mass of MgS = 56.38 g/mol

Moles of MgS = 0.5303299042

$Molality = \frac{Moles \;solute}{Volume \;of \;solution \;in \;litre}$

$Molality = \frac{0.5303299042}{0.911L}$

Molality = 0.5821403998=0.58

Hence, 0.58 is the molarity of 29.9 g of MgS in 911 mL of solution.

Learn more about Molarity here:

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3 0

2 years ago

A chemist adds 26.5g of ammonium chloride to 10g of sodium hydroxide, which follows the reaction below. Assuming the reaction go

4vir4ik [10]

Answer : The amount of reactant left in excess is 13.1075 grams.

Explanation : Given,

Mass of $NH_4Cl$ = 26.5 g

Mass of $NaOH$ = 10 g

Molar mass of $NH_4Cl$ = 53.5 g/mole

Molar mass of $NaOH$ = 40 g/mole

First we have to calculate the moles of $NH_4Cl$ and $NaOH$ .

$\text{Moles of }NH_4Cl=\frac{\text{Mass of }NH_4Cl}{\text{Molar mass of }NH_4Cl}=\frac{26.5g}{53.5g/mole}=0.495moles$

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{10g}{40g/mole}=0.25moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$NH_4Cl+NaOH\rightarrow NH_4OH+NaCl$

From the balanced reaction we conclude that

As, 1 mole of $NaOH$ react with 1 mole of $NH_4Cl$

So, 0.25 moles of $NaOH$ react with 0.25 moles of $NH_4Cl$

From this we conclude that, $NH_4Cl$ is an excess reagent because the given moles are greater than the required moles and $NaOH$ is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 0.495 - 0.25 = 0.245 moles

Now we have to calculate the mass of $NH_4Cl$ .

$\text{Mass of }NH_4Cl=\text{Moles of }NH_4Cl\times \text{Molar mass of }NH_4Cl$