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3 years ago

A solution of pH 5 is diluted 100 times. Find the pH of the resulting solution.

2 answers:

3 0

First, we calculate of the concentration of the H+ ions in the solution from the pH given. Then, calculate the new concentration after dilution. Calculation are as follows:

pH = -log[H+]
5 = -log[H+]
[H+] = 1 x 10^-5 M

M1V1 = M2V2
1 x 10^-5 M (V1) = M2(100V1)
M2 = 1 x 10^-7

pH = -log[1 x 10^-7]
pH = 7

Paraphin [41]3 years ago

3 0

Answer:

The pH of the resulting solution is 7. This means that the solution is neutral.

Explanation:

To reach the pH of the resulting solution some calculations are required. They are simple calculations and you can use a calculator to help. Well, first you have to find out the concentration of hydrogen atoms that the solution had before it was diluted, for that you will use the formula pH = -log [H +]. Substituting the values in the formula, we have:

pH = -log [H +]

5= 5 = -log [H +]

[H +] = 1 x 10 ^ -5 M

Once you find this value, you will find the H + atom concentration after dilution using the formula C1V1 = C2V2. Substituting the values in the formula we will have:

C1V1 = C2V2

1 x 10 ^ -5 M (V1) = M2 (100V1)

M2 = 1x10 ^ -7

You can now reuse the first formula to finally find the pH of this solution after all dilution. The calculation will be:

pH = -log [1 x 10 ^ -7]

pH = 7

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For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow

lora16 [44]

Answer:

Rate constant of the reaction is $3.3\times 10^{-3} M^{-2} s^{-1}$ .

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

$R=k[A]^a[B]^b[C]^c$

Rate(R) of the reaction in trail 1 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.30 M$

$R=9.0\times 10^{-5} M/s$

$9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c$ ...[1]

Rate(R) of the reaction in trail 2 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.90 M$

$R=2.7\times 10^{-4} M/s$

$2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c$ ...[2]

Rate(R) of the reaction in trail 3 ,when :

$[A]=0.60 M,[B]=0.30 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c$ ...[3]

Rate(R) of the reaction in trail 4 ,when :

$[A]=0.60 M,[B]=0.60 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c$ ...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

$R=k[A]^2[B]^0[C]^1$

Rate constant of the reaction = k

$9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1$

$k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}$

$k=3.3\times 10^{-3} M^{-2} s^{-1}$

7 0

3 years ago

How many grams of Cl2 are consumed to produce 12.0 g of KCl?

Alla [95]

The reaction is:

Cl2 + 2 KBr --> 2 KCl + Br2

Moles of KCl is

n = m /M = 12 /74 = 0.16 mol

As, twice the moles of KCl is producing from 1 mol of chlorine

mole of Cl2 = 0.16 /2 = 0.08 mol

Mass of Cl2

m /70 = 0.08 = 5.6 g

Hence, 5.6 g mol Cl2 consumed to produce KCl

7 0

3 years ago

A lead mass is heated and placed in a foam cup calorimeter containing 50.0 ml of water at 18.0°c. the water reaches a temperatur

garri49 [273]

Heat gained or loss in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:

Heat = mC(T2-T1)

When two objects are in contact, it should be that the heat lost is equal to what is gained by the other. So, the heat released by the lead is equal to the heat that is absorbed by the water.

Heat = mC(T2-T1) = 50.0 mL (1.00 g/mL) (4.18 J/g °C) (20 °C - 18 °C) = 418 J

8 0

3 years ago

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Given that sodium chloride is 39.0% sodium by mass, how many grams of sodium chloride are needed to have 100.mg of Na present? G

seraphim [82]

Answer: The mass of sodium chloride solution present is 0.256 grams.

Explanation:

We are given:

39.0 % of sodium in sodium chloride solution

This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution

Mass of sodium given = 100 mg = 0.1 g (Conversion factor: 1 g = 1000 mg)

Applying unitary method:

If 39 grams of sodium metal is present in 100 grams of sodium chloride solution

So, if 0.1 grams of sodium metal will be present in = $\frac{100}{39}\times 0.1=0.256g$ of sodium chloride solution.

Hence, the mass of sodium chloride solution present is 0.256 grams.

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