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3 years ago

What is the concentration of an alcl3 solution if 150. ml of the solution contains 550. mg of cl- ion?

Chemistry

1 answer:

valina [46]3 years ago

6 0

The concentration of AlCl3 solution if 150 ml of the solution contains 550 mg of cl- ion is 0.0344 M

calculation

concentration = moles /volume in liters

volume in liters = 150 /1000= 0.15 L

number of moles calculation

write the equation for dissociation of Al2Cl3

that is AlCl3 ⇔ Al^3+ + 3 Cl ^-

find the moles of Cl^- formed

moles =mass/molar mass

mass in grams= 550/ 1000 =0.55 grams

molar mass of Cl^- =35.5 g/mol

moles is therefore= 0.55/35.5 =0.0155 moles

by use of mole ration betweem AlCl3 to Cl^- which is 1:3 the moles of AlCl3 is =0.0155 x 1/3= 5.167 x10^-3 moles

concentration of AlCl3 is therefore= 5.167 x10^-3/ 0.15 =0.0344 M

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How many grams of nitric acid HNO₃, are required to neutralize (completely react with) 4.30 grams of Ca(OH)2 according to the ac

Brrunno [24]

Answer:

7.32g of HNO3 are required.

Explanation:

1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.

From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.

2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:

• starting with the 4.30 grams of Ca(OH)2.

• using the molar mass of Ca(OH)2 (74g/mol).

• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .

• using the molar mass of HNO3 (63.02g/mol).

$4.30g\text{ Ca\lparen OH\rparen}_2*\frac{1\text{ mol Ca\lparen OH\rparen}_2}{74g\text{ Ca\lparen OH\rparen}_2}*\frac{2\text{ moles HNO}_3}{1\text{ mole Ca\lparen OH\rparen}_2}*\frac{63.02g\text{ HNO}_3}{1\text{ mole HNO}_3}=7.32g\text{ HNO}_3$

So, 7.32g of HNO3 are required.

4 0

1 year ago

Ultra Clorox bleach is 6.68% sodium hypochlorite in water. How much of each is needed to make an 800.mL bottle?

sweet-ann [11.9K]

If we have 6.68% NaClO, it is the same as saying--> 6.68 grams NaClO= 100 mL of solution. we can use this as a conversion.

800. mL (6.68 mL/ 100 mL)= 53.4 mL

solution = solute + solvent

solute= NaClO
solvent= H2O

solvent= 800-53.4= 747 mL of H2O

so, we you need 53.4 mL of NaClO and 747 mL of water or 53.4 grams of NaClO and 747 mL of water

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3 years ago

How many moles are there in 3.612 x 1024 formula units of Na2SO4?

vitfil [10]

Answer:

<h3>Theanswer is 6 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.612 \times {10}^{24} }{6.02 \times {10}^{23} } \\$

We have the final answer as

<h3>6 moles</h3>

Hope this helps you

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3 years ago

Consider two different ions. The anion has a valence of -2. The cation has a valence of +2. The two ions are separated by a dist

strojnjashka [21]

Answer:

Force of attraction = 35.96 $\times 10^{27}$ N

Explanation:

Given: charge on anion = -2

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Solution: The force of attraction is calculated by using equation,

$F = \dfrac{k \times q_1 q_2}{ \r^2}$ ---(1)

where, q represents the charge and the subscripts 1 and 2 represents cation and anion.

k = $8.99 \times 10^9 \ Nm^{2}C^{-2}$

F = force of attraction

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Substituting all the values in the equation (1) the equation becomes

$F = \dfrac{8.99 \times 10^9 \times 2 \times 2}{ \left ( 10^-9 \right )^2 }$

Force of attraction = 35.96 $\times 10^{27}$ N

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Zielflug [23.3K]

Answer:

true

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3 years ago