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3 years ago

how many liters of 0.150 m hcl would be required to react completely with 8.00 grams of magnesium hydroxide

Chemistry

1 answer:

Artemon [7]3 years ago

4 0

Answer:

how many liters of 0.150 m hcl would be required to react completely with 8.00 grams of magnesium hydroxide

Explanation:

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Which participates in chemical reactions? electrons protons neutrons gamma rays

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I think the correct answer is electrons. All chemical reactions are dependent on the transfer of electrons. Other subatomic particles are located in the innermost portion of an atom and are bonded strongly with each other.

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3 years ago

Read 2 more answers

The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper

LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

$t_{1/2}=\frac{1}{k\times a_0}$

k = rate constant =?

$a_0$ = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

a) Initial concentration of XY = $a_o=0.100 M$

Rate constant of the reaction = k = $6.96\times 10^{-3} M^{-1} s^{-1}$

Half life of the reaction is:

$t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}$

$=1,436.78 s$

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

$\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}$

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

$\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = $6.20\times 10^{-2} M$

$\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become $6.20\times 10^{-2} M$ .

e) Initial concentration of $SO_2Cl_2= 0.050 M$

Final concentration after time t = x

t = 55.0 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

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3 years ago

What is the oxidation state of each element in the compound CaS04? Include +or - in your answer as appropriate

borishaifa [10]

Answer:therefore the oxidation state of sulphur is +6 in the compound CaSO4.

Explanation:

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3 years ago

For the separation of components of a mixture, we are following different methods. What is the condition in which the process of

soldier1979 [14.2K]

Answer:

The mixture 1/ must be homogeneous where there is one or more dissolved salts; 2/ should only contain one liquid component.

Explanation:

This method drives off the liquid components from the solid components. The process typically involves heating the mixture until no more liquid remains. Prior to using this method, the mixture should only contain one liquid component, unless it is not important to isolate the liquid components. This is because all liquid components will evaporate over time. This method is suitable to separate a soluble solid from a liquid.

Note : In many parts of the world, table salt is obtained from the evaporation of sea water. The heat for the process comes from the sun.

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Hope this answer can help you. Have a nice day!

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3 years ago

Explain the results for the tube in which 1.0 m naoh was added to benzoic acid. write an equation for this, using complete struc

nikklg [1K]

When sodium hydroxide (NaOH) is added to benzoic acid (C $C_{6}H_{5}COOH$ ), sodium benzoate and water ( $H_{2}O$ ) are formed because benzoic acid is weak acid and NaOH is strong base. The reaction is shown in diagram 1. To the tube of sodium benzoate, if 6 m HCl is added, sodium benzoate salt will react with HCl and produce benzoic acid and sodium chloride, NaCl. The reaction is shown in diagram 2.

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3 years ago