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anygoal [31]
3 years ago
14

When a teacher dilutes 50 mL of 2.0 M NaOH to 0.50 M, what volume of NaOH results?

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

Answer:

= 200 mL

Explanation:

Using the dilution formula;

M1V1 = M2V2 ;

Where, M1 is the concentration before dilution, V2 is the volume before dilution, while M2 is the concentration after dilution and V2 is the volume after dilution.

M1 = 2.0 M

V1 = 50 mL

M2 = 0.50 M

V2 = ?

V2 = M1V1/M2

     = ( 2.0 × 50 )/ 0.5

     = 200 mL

Therefore, the volume after dilution will be, 200 mL

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Explanation:

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As it has 6 valency of electron it must be in the 16 group of the table that comprises the 6 valency and as it is located in the 3rd row it must be sulfur that also has an atomic mass between selenium and oxygen.

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3 years ago
As the pendulum moves from point 2 to point 3, what happens to its mechanical energy?potential energy is converted to kinetic en
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Answer: Potential energy is converted to kinetic energy and back again.

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5 0
2 years ago
The decomposition of N2O5 in solution in carbon tetrachloride proceeds via the reaction 2 N2O5(soln) → 4 NO2(soln) + O2(soln) Th
Fantom [35]

<u>Answer:</u> The amount remained after 151 seconds are 0.041 moles

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,  

k = rate constant  = 4.82\times 10^{-3}s^{-1}

t = time taken for decay process = 151 sec

[A_o] = initial amount of the reactant = 0.085 moles

[A] = amount left after decay process =  ?

Putting values in above equation, we get:

4.82\times 10^{-3}=\frac{2.303}{151}\log\frac{0.085}{[A]}

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Hence, the amount remained after 151 seconds are 0.041 moles

7 0
3 years ago
Using the systematic approach for equilibrium problems, calculate the pH of 0.05 M HOCl. Ka= 3.0*10-8 Group of answer choices 3.
SVEN [57.7K]

Answer:

The pH is equal to 4.41

Explanation:

Since HClO is a weak acid, its dissociation in aqueous medium is:

                HClO   ⇄   ClO-  +  H+

start:          0.05            0         0

change       -x               +x       +x

balance     0.05-x         x         x

As it is a weak acid it dissociates very little, in its ClO- and H + ions, so the change is negative, where x is a degree of dissociation.

the acidity constant when equilibrium is reached is equal to:

Ka=\frac{[ClO-]*[H+]}{[HClO]}=\frac{x*x}{0.05-x}=3x10^{-8}

The 0.05-x fraction can be approximated to 0.05, because the ionized fraction (x) is very small, therefore we have:

3x10^{-8}=\frac{x^{2} }{0.05}

clearing the x and calculating its value we have:

x=3.87x10^{-5}=[H+]=[ClO-]

the pH can be calculated by:

pH=-log[H+]=-log[3.87x10^{-5}]=4.41

7 0
3 years ago
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