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sashaice [31]
3 years ago
8

How many liters are in a .00813M solution that contains 1.55 g of KBr?

Chemistry
1 answer:
Sloan [31]3 years ago
5 0

Answer:

                      1.602 L (or) 1602 mL

Explanation:

             Molarity is the amount of solute dissolved per unit volume of solution. It is expressed as,

                         Molarity  =  Moles / Volume of Solution    ----- (1)

Rearranging above equation for volume,

                         Volume of solution  =  Moles / Molarity    -------(2)

Data Given;

                  Molarity  =  0.00813 mol.L⁻¹

                  Mass  =  1.55 g

First calculate Moles for given mass as,

                   Moles  =  Mass / M.mass

                   Moles  =  1.55 g / 119.002 g.mol⁻¹

                   Moles  =  0.0130 mol

Now, putting value of Moles and Molarity in eq. 2,

                         Volume of solution  =  0.0130 mol / 0.00813 mol.L⁻¹

                         Volume of solution  = 1.60 L

or,

                         Volume of solution  =  1602 mL

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3 years ago
What is the concentration (M) of CH3OH in a solution prepared by dissolving 34.4 g of CH3OH in sufficient water to give exactly
bonufazy [111]

Answer:

4.67M

Explanation:

The concentration of methanol (CH3OH) can be calculated using the following:

Molarity (M) = number of moles(n)/volume(v)

However, mole is not given. It can be obtained by using:

Mole = mass / molar mass

Where; mass = 34.4g

Molar mass (MM) of CH3OH is:

= 12 + 1(3) + 16 + 1

= 12 + 3 + 17

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mole = 34.4/32

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The volume needs to be converted to L by dividing by 1000

230mL = 230/1000

= 0.230L

Molarity = mol/volume

Molarity = 1.075/0.230

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The concentration of CH3OH in solution is 4.67M

5 0
3 years ago
How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them? (answ
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Answer:

7.12 mm

Explanation:

From coulomb's law,

F = kqq'/r².................... Equation 1

Where F = force, k = proportionality constant, q and q' = The two point charges, r = distance between the two charges.

Make r the subject of the equation,

r = √(kqq'/F).......................... Equation 2

Given: q = q' = 75.0 nC = 75×10⁻⁹ C, F = 1.00 N

Constant: k = 9.0×10⁹ Nm²/C².

Substitute into equation 2

r = √[ (75×10⁻⁹ )²9.0×10⁹/1]

r = 75×10⁻⁹.√(9.0×10⁹)

r = (75×10⁻⁹)(9.49×10⁴)

r = 711.75×10⁻⁵

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Hence the distance between the point charge = 7.12 mm

3 0
3 years ago
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An atom gains an electron from another atom. Hence, option B is the correct answer.

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