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Law Incorporation [45]
3 years ago
7

What is the concentration of a potassium iodate solution after you complete the following porcedure? Pipette 10 mL of a 0.31 M p

otassium iodate stock solution to a 100-mL volumetric flask and then add water to the mark.
Chemistry
1 answer:
zubka84 [21]3 years ago
4 0

<u>Given:</u>

Initial concentration of potassium iodate (KIO3) M1 = 0.31 M

Initial volume of KIO3 (stock solution) V1 = 10 ml

Final volume of KIO3 V2 = 100 ml

<u>To determine:</u>

The final concentration of KIO3 i.e. M2

<u>Explanation:</u>

Use the relation-

M1V1 = M2V2

M2 = M1V1/V2 = 0.31 M * 10 ml/100 ml = 0.031 M

Ans: The concentration of KIO3 after dilution is 0.031 M

You might be interested in
Determine the molarity for each of the following solution solutions:
____ [38]

Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of H_2SO_4 solution is, 0.00525 mole/L

(c)The molarity of Al(NO_3)_3 solution is, 0.0612 mole/L

(d)The molarity of CuSO_4.5H_2O solution is, 7.61 mole/L

(e)The molarity of Br_2 solution is, 0.0565 mole/L

(f)The molarity of C_2H_5NO_2 solution is, 0.0113 mole/L

Explanation :

<u>(a) 1.457 mol of KCl in 1.500 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Solute is KCl.

\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L

The molarity of KCl solution is, 0.9713 mole/L

<u>(b) 0.515 gram of H_2SO_4, in 1.00 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Solute is H_2SO_4

Molar mass of H_2SO_4 = 98 g/mole

\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L

The molarity of H_2SO_4 solution is, 0.00525 mole/L

<u>(c) 20.54 g of Al(NO_3)_3 in 1575 mL of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Solute is Al(NO_3)_3

Molar mass of Al(NO_3)_3 = 213 g/mole

\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L

The molarity of Al(NO_3)_3 solution is, 0.0612 mole/L

<u>(d) 2.76 kg of CuSO_4.5H_2O in 1.45 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Solute is CuSO_4.5H_2O

Molar mass of CuSO_4.5H_2O = 250 g/mole

\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L

The molarity of CuSO_4.5H_2O solution is, 7.61 mole/L

<u>(e) 0.005653 mol of Br_2 in 10.00 ml of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

Solute is Br_2.

\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L

The molarity of Br_2 solution is, 0.0565 mole/L

<u>(f) 0.000889 g of glycine, C_2H_5NO_2, in 1.05 mL of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Solute is C_2H_5NO_2

Molar mass of C_2H_5NO_2 = 75 g/mole

\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L

The molarity of C_2H_5NO_2 solution is, 0.0113 mole/L

5 0
3 years ago
5. Chlorine gas was formed at the ... *
Marat540 [252]

Clorine gas was formed at the <em><u>positive</u></em><em><u> </u></em><em><u>electrode</u></em><em><u>.</u></em><em><u> </u></em>

5 0
3 years ago
A student submits the following work on the mass number, but he has made a few mistakes. Select the sentences that are incorrect
rewona [7]

Answer

 

Calculating the mass number for an atom requires that we know the atomic number and the number of protons in the atom’s nucleus. The mass number then gives us the average weight of atoms of a given element. However, as long as the number of protons equals the number of neutrons, the values balance out and we always obtain a whole number for the mass number.

Explanation:

those 3

7 0
3 years ago
How many grams of Na2SO4 can be produced from 423.67 g of NaCl?
san4es73 [151]

Mass of  Na2SO4= 514.18 grams

<h3>Further explanation</h3>

Given

423.67 g of NaCl

Required

mass of  Na2SO4

Solution

Reaction

2NaCl + H2SO4 → Na2SO4 + 2HCl

mol NaCl :

= 423.67 g : 58.5 g/mol

= 7.24

From the equation, mol Na2SO4 :

= 1/2 x mol NaCl

= 1/2 x 7.24

= 3.62

Mass  Na2SO4 :

= 3.62 mol x 142,04 g/mol

= 514.18 grams

4 0
3 years ago
Determine the mechanical energy of this object: a 1-kg ball sits on a 10-meter high cliff.
Bond [772]
Mechanical energy = mass x acceleration due to gravity x height 
                               =     1x9.8x10 = 98J
8 0
3 years ago
Read 2 more answers
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