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aleksandrvk [35]
3 years ago
10

how many grams of solid silver nitrate would you need to prepare 200.0 mL of a 0.150 M AgNO3 solution

Chemistry
1 answer:
aleksklad [387]3 years ago
4 0
0.150 M AgNO3 = x mol / 0.200 Liters
x mol = 0.03 mol AgNO3
0.03 mol AgNO3 (169.9g AgNO3 / 1 mol AgNO3) We are converting moles to grams here with stoichiometry.

Final answer = 5.097 grams, but if you want it in terms of sig figs then it is 5.09 grams.
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Explanation:

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3. People and instruments produce different ranges of sound temperatures is this true or false
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True I believe since every person has a different temperature and sound waves.
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Solid Mg(OH)2 is mixed with water and the following reversible reaction occurs:________.
hram777 [196]

Answer:

Mg^2+ and OH- are the chemical species present at the equilibrium. Mg(OH)2 will not affect the equilibrium.

Explanation:

Step 1: data given

Reactants are Solid Mg(OH)2 and H2O(l)

Kc1 = 1.8 * 10^-11

Step 2: The balanced equation

Mg(OH)2(s)  ⇄ Mg2+(aq) + 2OH-(aq)

Step 3: Define the equilibrium constant Kc

Kc = [OH-]²[Mg^2+]

Pure solids and liquids do not have any effect or influence on the equilibrium in the reaction. So they are not included in the equilibrium constant expression.

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7 0
3 years ago
1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium
lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

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How many moles are in 3.46 g of chromium?
Bezzdna [24]

Answer:

0.06654345229738384 moles of chromium.

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