HURRY!!

You have a 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?

lions [1.4K]

Answer:

$\boxed {\boxed {\sf 0.4 \ mol \ HCl}}$

Explanation:

Molarity is concentration measured in moles per liters. It is the number of moles of solute per liters of solution. The formula is:

$molarity= \frac{moles \ of \ solute}{liters \ of \ solution}}$

We know the solution of HCl has a molarity of 0.5 molar and there are 0.8 liters of solution.

1 molar (M) is equal to 1 mole per liter.
Let's convert the molarity of 0.5 M HCl to 0.5 mol HCl per liter. This will make unit cancellation easier.

The moles of solute or HCl are unknown, so we can use x. Now, we can substitute all known values into the formula.

$0.5 \ mol \ HCl /L = \frac {x}{0.8 \ L}$

Since we are solving for the moles of solute (x), we must isolate the variable. It is being divided by 0.8 liters. The inverse of division is multiplication, so we multiply both sides by 0.8 L.

$0.8 \ L *0.5 \ mol \ HCl /L = \frac {x}{0.8 \ L} *0.8 \ L$

$0.8 \ L *0.5 \ mol \ HCl /L=x$

The units of liters (L) cancel.

$0.8 * 0.5 \ mol \ HCl= x$

$0.4 \ mol \ HCl=x$

This solution contains <u>0.4 moles of HCl.</u>

6 0

3 years ago

2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electrons

Julli [10]

Answer :

The oxidation state of tin changes from (+2) to (+4).

The number electrons transferred in the reaction are, 10 electron.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in basic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$2BrO_3^-+5SnO_2^{2-}+H_2O\rightarrow 5SnO_3^{2-}+Br_2+2OH^-$

The oxidation-reduction half reaction will be :

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}$

Reduction : $BrO_3^-\rightarrow Br_2$

First balance the main element in the reaction.

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}$

Reduction : $2BrO_3^-\rightarrow Br_2$

Now balance oxygen atom on both side.

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}+H_2O$

Reduction : $2BrO_3^-+H_2O\rightarrow Br_2$

Now balance hydrogen atom on both side.

Oxidation : $SnO_2^{2-}+OH^-\rightarrow SnO_3^{2-}+H_2O$

Reduction : $2BrO_3^-+3H_2O\rightarrow Br_2+12OH^-$

Now balance the charge.

Oxidation : $SnO_2^{2-}+OH^-\rightarrow SnO_3^{2-}+H_2O+2e^-$

Reduction : $2BrO_3^-+3H_2O+10e^-\rightarrow Br_2+12OH^-$

The charges are not balance on both side of the reaction. We are multiplying oxidation reaction by 5 and then added both equation, we get the balanced redox reaction.

Oxidation : $5SnO_2^{2-}+5OH^-\rightarrow 5SnO_3^{2-}+5H_2O+10e^-$

Reduction : $2BrO_3^-+3H_2O+10e^-\rightarrow Br_2+12OH^-$

The balanced chemical equation in basic medium will be,

$2BrO_3^-+5SnO_2^{2-}+H_2O\rightarrow 5SnO_3^{2-}+Br_2+2OH^-$

The number electrons transferred in the reaction are, 10 electron.

7 0

3 years ago

A solution is made by adding 20 g table salt to 100 mL water. The solubility of salt is 36 g/100 mL water. Which term best descr

marysya [2.9K]

Answer:

We say that the solution is unsaturated.

Explanation:

If the salt solubility is 36 g in 0.1 L of water then we can dissolve 360 g of salt in 1 L of water.

Because the solution contains 200 g of salt in 1 L of water, the solution is unsaturated because more salt can be added until we reach the saturation point.

We call the solution dilute when we compare the concentration of a solution with the concentration of another solution, but here we do not compare different solutions.

3 0

2 years ago

Read 2 more answers

How do you make 20 mL of 25 M HCl given 1.00 M HCI?<br><br> Add_mL of acid

Norma-Jean [14]

Answer:

To prepare 20 mL of 0.25 M HCl from 1.00 M HCl, 5 mL of the 1.00 M HCl is measured and transferred in a volumetric flask or beaker containing about 10 mL of water. Then, water is added to make it up to the 20mL mark.

Explanation:

The question is not correct, because the molarity of HCl cannot be greater than 12.2 M.

Also, a higher concentration of an acid can not be prepared by any dilution method, but can only be prepared by distillation procedures requiring great care and expertise.

Therefore, the following assumptions are made about the question :

The initial acid concentration, M1 = 1.00 M

Required acid concentration, M2 = 0.25 M

Volume of required acid solution, V2 = 20 mL

Using the dilution formula: M1V1 = M2V2

Volume of initial acid solution, V1 must be found then.

Making V1 subject of the formula; V1 = M2V2/M1

V1 = 0.25 × 20 / 1.00

V1 = 5 mL

To prepare 20 mL of 0.25 M HCl from 1.00 M HCl, 5 mL of the 1.00 M HCl is measured and transferred in a volumetric flask or beaker containing about 10 mL of water. Then, water is added to make it up to the 20mL mark.

5 0

3 years ago

Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid wat

Arada [10]

here's the answer to your question

8 0

3 years ago