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liubo4ka [24]
3 years ago
12

What will happen to the pH of a solution when the [H+] is increased?

Chemistry
1 answer:
matrenka [14]3 years ago
8 0

Answer:

decreases

Explanation:

the more conc of H ions increases the more it becomes acidic so the pH goes down

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used in the manufacturing of glass, detergents chemicals and other industrial products.

4 0
3 years ago
The reaction between SO2 and O2 to give SO3 was studied to determine the rate law for the reaction. The following results were o
Tom [10]

Answer : (b) The rate law expression for the reaction is:

\text{Rate}=k[SO_2]^2[O_2]

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The general reaction is:

A+B\rightarrow C+D

The general rate law expression for the reaction is:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

R = rate  law

k = rate constant

[A] and [B] = concentration of A and B reactant

Now we have to determine the rate law for the given reaction.

The balanced equations will be:

2SO_2+O_2\rightarrow 2SO_3

In this reaction, SO_2 and O_2 are the reactants.

The rate law expression for the reaction is:

\text{Rate}=k[SO_2]^2[O_2]^1

or,

\text{Rate}=k[SO_2]^2[O_2]

8 0
3 years ago
Please help this is due in 30 minutes !!!
tamaranim1 [39]

Answer: so the answer would likely be

Explanation:

5 0
3 years ago
Please someone help I’m really confused
NARA [144]

Answer:10-3

Explanation:

7 0
3 years ago
For the reaction: 4PH3(g) → P4(g) + 6H2(g) about 0.065 mol/s of PH3 is consumed in a 5.0 L flask. What are the rates of producti
hoa [83]

Answer: The rates of production of P_4 is 3.25\times 10^{-3} mol/Ls and H_2 is 0.0195 mol/Ls.

Explanation:

4PH_3(g)\rightarrow P_4(g)+6H_2(g)

Rate with respect to reactants is shown by negative sign as the reactants are decreasing with time and Rate with respect to products is shown by positive sign as the products are increasing with time.

Rate of the reaction=-\frac{1}{4}\frac{[d[PH_3]}{dt}=\frac{[d[P_4]}{dt}=\frac{1}{6}\frac{[d[H_2]}{dt}

Rate of decomposition of PH_3=\frac{0.065 mol/s}{5.0L}=0.013mol/Ls

Rate of production of P_4=\frac{1}{4}\times {\text{rate of decomposition of}} PH_3=\frac{1}{4}\times 0.013=3.25\times 10^{-3}mol/Ls

Rate of production of  H_2=\frac{6}{4}\times {\text {rate of decomposition of}} PH_3=\frac{6}{4}\times 0.013=0.0195mol/Ls



7 0
4 years ago
Read 2 more answers
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