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Snowcat [4.5K]
3 years ago
11

How many grams of NaCl are needed to make 0.800 liter of a 5.00 M solution?

Chemistry
1 answer:
ziro4ka [17]3 years ago
6 0

Answer:

             0.324 g is required to make 5.00 M solution of NaCl in 0.800 L.

Given data:

                 Molarity = 5.00 M

                 Formula Mass = 58.5 g/mol

                 Required volume = 0.800 L

To Find;

               Mass in gram = ?

Solution:

              Formula for calculating mass in gram is given as,

              Mass in gram = Molarity × Formula mass × Volume required / 1000 putting values

              Mass in gram = 5.00 M × 58.5 g/mol × 0.800 L / 1000

               Mass in gram = 0.234 g


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How many joules of heat is lost when a 64 g piece of copper cools from 375 oC to 26 oC? The specific heat of copper is 0.38 J/go
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Answer: 8500 J lost or -8500 J

Explanation:

q=cmt

t=375-26

t=349

q=0.38(64)(349)

q=8487.68 J

Answer must have 2 sig figs, which means it rounds to 8500 J.

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3 years ago
What is the percent composition of hydrogen in vinegar (CH3COOH)?​
barxatty [35]

Answer:

                      %H  = 6.72 %

Explanation:

                   Percent composition of an element is the total mass of that element divided by the molecular mass of compound (or molecular mass) of which it is present in.

So,

Percent composition of Hydrogen will be given as,

         %H  =  Total mass of H / Molecular Mass of Acetic Acid × 100

So,

Total Mass of H  =  1.01 × 4 = 4.04 g

Molecular Mass of Acetic acid  =  60.052 g/mol

Putting values in above formula,

         %H  =  4.04 g/mol ÷ 60.052 g/mol × 100

         %H  = 6.72 %

6 0
3 years ago
You throw a ball straight up in the air trying to throw it as high as possible. What do you think happens to the balls kinetic e
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Read 2 more answers
Determine the percent ionization of a 0.225 M solution of benzoic acid. Express your answer using two significant figures.
Assoli18 [71]

Answer:

1.68% is ionized

Explanation:

The Ka of benzoic acid, C₇H₆O₂, is 6.46x10⁻⁵, the equilibrium in water of this acid is:

C₇H₆O₂(aq) + H₂O(l) ⇄ C₇H₅O₂⁻(aq) + H₃O⁺(aq)

Ka = 6.46x10⁻⁵ = [C₇H₅O₂⁻] [H₃O⁺] / [C₇H₆O₂]

<em>Where [] are concentrations in equilibrium</em>

In equilibrium, some 0.225M of the acid will react producing both C₇H₅O₂⁻ and H₃O⁺, the equilibrium concentrations are:

[C₇H₆O₂] = 0.225-X

[C₇H₅O₂⁻] = X

[H₃O⁺] = X

Replacing:

6.46x10⁻⁵ = [X] [X] / [0.225-X]

1.4535x10⁻⁵ - 6.46x10⁻⁵X = X²

1.4535x10⁻⁵ - 6.46x10⁻⁵X - X² = 0

Solving for X:

X = -0.0038. False solution, there is no negative concentrations.

X = 0.00378M. Right solution.

That means percent ionization (100 times Amount of benzoic acid ionized  over the initial concentration of the acid) is:

0.00378M / 0.225M * 100 =

<h3>1.68% is ionized</h3>
8 0
3 years ago
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