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2 years ago

Will give brainliest! if 32.0 g of hcl is to be diluted to make a 4.80 m solution, how much water should be added?

Chemistry

1 answer:

nadezda [96]2 years ago

6 0

The required volume of water is 0.18 liters.

<h3>What is molarity?</h3>

Molarity of any solution is define as the number of moles of solute present in per liter of solution as;

M = n/V

Moles of solute will be calculated as:

n = W/M, where

W = given mass of HCl = 32g

M = molar mass of HCl = 36.4g/mol

n = 32 / 36.4 = 0.88 mole

Given molarity of solution = 4.80M

On putting all values in the above equation, we get

V = (0.88) / (36.4) = 0.18 L

Hence required volume of water is 0.18L.

To know more about volume & concentration, visit the below link:

brainly.com/question/26762947

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Answer:

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11. How many elements do all the "P" orbital span (go across) in each period? (circle your answer)

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2 years ago

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) What is the max

jolli1 [7]

<u>Answer:</u> The mass of aluminium chloride that can be formed are 46.3 g

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

<u>For Aluminium:</u>

Given mass of aluminium = 32 g

Molar mass of aluminium = 26.98 g/mol

Putting values in above equation, we get:

$\text{Moles of aluminium}=\frac{32g}{26.98g/mol}=1.186mol$

<u>For Chlorine:</u>

Given mass of chlorine = 37 g

Molar mass of chlorine = 71 g/mol

Putting values in above equation, we get:

$\text{Moles of chlorine gas}=\frac{37g}{71g/mol}=0.521mol$

For the given chemical equation:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)$

By Stoichiometry of the reaction:

3 moles of chlorine gas is reacting with 2 moles of aluminium.

So, 0.521 moles of chlorine gas will react with = $\frac{2}{3}\times 0.521=0.347moles$ of aluminium.

As, given amount of aluminium is more than the required amount. Thus, it is considered as an excess reagent.

So, chlorine gas is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

3 moles of chlorine gas is producing 2 moles of aluminium chloride

So, 0.521 moles of chlorine gas will react with = $\frac{2}{3}\times 0.521=0.347moles$ of aluminium chloride.

Now, calculating the mass of aluminium chloride by using equation 1, we get:

Moles of aluminium chloride = 0.347 moles

Molar mass of aluminium chloride = 133.34 g/mol

Putting all the values in equation 1, we get:

$0.347mol=\frac{\text{Mass of aluminium chloride}}{133.34g/mol}\\\\\text{Mass of aluminium chloride}=46.3g$

Hence, the mass of aluminium chloride that can be formed are 46.3 g

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3 years ago

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3 years ago

Hydrobromic acid solution of unknown concentration is titrated with a 0.500M LiOH solution.

hram777 [196]

Answer:

1.00 M

Explanation:

Step 1: Write the balanced equation

HBr + LiOH ⇒ LiBr + H₂O

Step 2: Calculate the reacting moles of lithium hydroxide

40.00 mL of 0.500 M solution react. The reacting moles of LiOH are:

$40.00 \times 10^{-3} L \times \frac{0.500mol}{L} = 0.0200 mol$

Step 3: Calculate the reacting moles of hydrobromic acid

The molar ratio of HBr to LiOH is 1:1. The reacting moles of hydrobromic acid are 1/1 × 0.0200 mol = 0.0200 mol.

Step 4: Calculate the molarity of hydrobromic acid

0.0200 moles of HBr are in 20.00 mL of the solution. The molarity of the hydrobromic acid solution is:

$M= \frac{0.0200 mol}{20.00 \times 10^{-3} L } =1.00 M$

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3 years ago