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Chemistry! Help! Please!

1. Which of the following solutions would have a pH value greater than 7?
a. [OH-] = 2.4 × 10^-2 M
b. [H3O+] =1.53 × 10^-2 M
c. 0.0001 M HCl
d. [OH^-] = 4.4 × 10^-9 M

2. Consider the following equation for an equilibrium system: (attached)
Which concentration(s) would be included in the denominator of the equilibrium constant expression?
a. Pb(s), CO2(g), and SO2(g)
b. PbS(s), O2(g), and C(s)
c. O2(g), Pb(s), CO2(g), and SO2(g)
d. O2(g)

3. The oxidation number of the sulfur atom in the SO^2- 4 ion is
a. +2.
b. -2.
c. +6.
d. +4.

4. In the following reaction, which is the oxidizing agent? (attached - reaction starts with AgNO2)
a. AgNO2
b. Cl2
c. KOH
d. KCl

5. Complete the following nuclear equation (attached w/ choices)

6. Which two particles have the same mass but opposite charge?
a. a beta particle and a positron
b. a neutron and a proton
c. a proton and an electron
d. an alpha particle and a proton

Chemistry Help Please 1 Which of the following solutions would have a pH value greater than 7 a OH 24 102 M b H3O 153 102 M c 00001 M HCl d OH 44 109 M 2 Consid class=
Chemistry Help Please 1 Which of the following solutions would have a pH value greater than 7 a OH 24 102 M b H3O 153 102 M c 00001 M HCl d OH 44 109 M 2 Consid class=
Chemistry Help Please 1 Which of the following solutions would have a pH value greater than 7 a OH 24 102 M b H3O 153 102 M c 00001 M HCl d OH 44 109 M 2 Consid class=

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mathjohndough

Question 1

The correct answer is a.

Explanation

The relationship between [tex]pH[/tex] and [tex]pOH[/tex] is given by

[tex]pH + pOH = 14[/tex], We can use [tex]pH=-log[H^+][/tex] and [tex]pOH= -log[OH^-][/tex]. In a we are given the concentration of [OH] and so we use that to find the pOH, then from pOH we can find the pH.

[tex]pOH= -log [2.4 \times 10^-^2] = 1.62\\\\pH = 14 - 1.62= 12.38[/tex]

[tex] b.pH=-log(1.53 \times 10^-^2) = 1.82[/tex]

c. HCl is a strong acid so it dissociates to 0.0001 [tex]H^+[/tex] and 0.0001 [tex]Cl^-[/tex], hence

[tex]pH= -log[0.0001] = 4[/tex]

[tex] d.pOH= -log [4.4 \times 10^-9] = 8.36\\\\pH= 14- 8.36 = 5.64[/tex]

[tex] a.[OH^-] = 2.4 \times 10^-^2[/tex] has a pH 12.36 which is greater than 7

Question 2

The correct answer is d [tex]O_2(g)[/tex].

Explanation

This is because solids and liquids do not appear in the equilibrium constant expression. Since [tex]K= \frac{[products]}{[reactants]}[/tex].

For the equation

[tex]2PbS_(_s_) + 3O_2_(_g_) + C_(_s_) \implies 2Pb _(_s_) + CO_2_(_g_) + SO_2(_g_)[/tex]

[tex]K= \frac{[CO_2][SO_2]}{[O_2]}[/tex]

Question 3

The correct answer is c

In a molecule the oxidation number are assigned to get the sum of a neutral charge or ion. The overall charge of [tex]SO_4^2^-[/tex] is -2 and in all its compounds oxygen has -2 charge, hence

[tex]S+ 4O = -2\\S+4(-2)=-2\\\implies S= -2 +8 = +6[/tex]

Question 4

The correct answer is b. An oxidising agent must gain electrons in a reaction, thus it is the one that undergoes reduction.In order find out what is being reduced we write half reactions for molecules that change their state. [tex]Ag^+[/tex] and [tex]K^+[/tex] are spectactor ions in this reaction.

[tex]Cl_2 + 2e \implies 2Cl^-\\\\NO_2 +H_2O \implies NO_3 +2H^+ +2e\\\\H^+ + OH^- \implies H_2O\\[/tex]

In the chloride half reaction [tex]Cl_2[/tex] gains electrons to become [tex]Cl^-[/tex] ions.

Question 5

The last element D is correct. When a nucleus decays by beta emission it produces a daughter nucleus that has same mass number but different atomic number. Therefore beta-decay will have equation

[tex]\frac{187}{75} Re \implies \frac{187}{76} Os + \beta[/tex]

Question 6

The correct answer is a. A beta particle is an electron because it has a charge of -1 and has same mass as an electron, while a positron is  a particle with the same mass as an electron  but with a positive charge.

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