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White raven [17]
3 years ago
15

1. Balance the reaction: Ag + H2S + O2 Ag2S + H2O

Chemistry
1 answer:
Murrr4er [49]3 years ago
8 0
It’s 0 because if u add them and divide 0 it’s 0
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Starting with 250 mL of a 0.250 M solution of HBr; a) Calculate the initial pH of the solution.b) Calculate the pH after adding
maxonik [38]

Answer:

a.

pH =  0.602

b.

pH = 1.5

c.

pH = 7

d.

pH = 12.1

Explanation:

a ) To calculate the pH, use the following equation:

pH = -log [H+]

Hbr is a strong acid, so the [H+] concentration can be calculated as follow:

[HBr] = 0.250 M

As acid Hbr:

Hbr = H+ + Br-

As strong acid HBr dissociates at all, so

[HBr] = [H+] = 0.250 M

So the pH:

<u>pH = -log [0.250 M] = 0.602</u>

<u></u>

<u>b) Calculate the pH after adding 250 mL of 0.125M NaOH</u>

<u></u>

<u>I</u>n this point, the reactions starts:

<u></u>

HBr + NaOH = H2O + NaBr

- First, we gonna find the mol of each reactant:

HBr:

mol = [M] × L

mol = 0.250 M × 0.250 L

mol = 0.0625 mol HBr

NaOH:

mol = [M] × L

mol = 0.125 M × 0.250L

mol = 0.03125 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.0625 mol HBr – 0.03125 mol HBr = 0.03125 mol HBr

These are the moles free in the solution, and we going to use them to calculate the pH:

pH = -log [ H]

pH = - log [ 0.03125 ] = 1.5

<u>c) Calculate the pH after adding 500 mL of 0.125M NaOH</u>

NaOH:

mol = [M] × L

mol = 0.125 M × 0.500L

mol = 0.0625 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.0625 mol HBr – 0.0625 mol HBr = 0 HBr

pH = 7

<u>d) Calculate the pH after adding 600 mL of 0.125M NaOH.</u>

<u>NaOH:</u>

mol = [M] × L

mol = 0.125 M × 0.600L

mol = 0.075 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.075 mol NaOH – 0.0625 mol NaOH = 00125 NaOH

These are the moles free in the solution, and we going to use them to calculate the pH:

pOH = -log [ OH]

pOH = - log [ 0.0125 ] = 1.90

pH + pOH = 14

pH = 14- pOH = 14 – 1.90 = 12.1

4 0
3 years ago
Peepeepoopoo? peepeepoopoo?
Keith_Richards [23]

Answer:

I just need points

Explanation:

4 0
3 years ago
Read 2 more answers
Calculate the wavelength of a photon of light having a frequency of 3.50 x 1015 Hz.
Aliun [14]

Answer:

Wavelength, \lambda=8.57\times 10^{-8}\ m

Explanation:

Given that,

Frequency, f=3.5\times 10^{15}\ Hz

We need to find the wavelength of a photon of light. The relation between frequency and wavelength is as follows :

c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{3.5\times 10^{15}}\\\\\lambda=8.57\times 10^{-8}\ m

So, the wavelength of the light is 8.57\times 10^{-8}\ m.

4 0
3 years ago
To what final concentration of NH3 must a solution be adjusted to just dissolve 0.060 mol of NiC2O4 (Ksp = 4×10−10) in 1.0 L of
ololo11 [35]
Given:
0.060 mol of NiC2O4
Ksp = 4 x 10⁻¹⁰
1.0 L of solution
Kf of Ni(NH3)6 2⁺ = 1.2 x 10⁹
<span>NiC2O4 + 6NH3 ⇋ Ni(NH3)6 2+ + 2O4 2- </span>
<span>NiC2O4 ⇋ Ni 2+ + C2O4 2- ...Ksp </span>
<span>Ni2+ + 6NH3 ⇋ Ni(NH3)6 2+...Kf </span>

Ksp * Kf = (4 x 10⁻¹⁰) * (1.2 x 10⁹) = 0.48

K = 0.48 = [Ni(NH3)6 2+][C2O4 2-] / [NH3]⁶<span> 
</span>0.48 = (0.060)² / [NH3]⁶<span> ... (dissolved C2O4 2- = 0.060M) 
</span><span>[NH3]</span>⁶<span> = (0.060)</span>²<span> / 0.48 = </span>0.0036 / 0.48 = 0.0075
NH3 = ⁶√0.0075 

NH3 = 0.44 M

3 0
3 years ago
What is the density(in g/L) of a gas with a molar mass of 32.49g/mol at 2.569atm and 458K?
garri49 [273]

Answer:

2.22~g/L

Explanation:

According to the ideal gas law equation:

pV = nRT

Let's express the number of moles in terms of mass and molar mass:

pV = \frac{m}{M} RT

Let's divide both sides by volume:

p = \frac{m}{MV} RT

Notice that:

\frac{m}{V} = d

So the equation becomes:

p = \frac{dRT}{M}

Expressing density:

d = \frac{pM}{RT}

Substituting the given values:

d = \frac{2.569~atm\cdot 32.49~g/mol}{0.08206~\frac{L~atm}{mol~K}\cdot 458~K} = 2.22~g/L

7 0
4 years ago
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