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3 years ago

How many grams of nickel (+2) are required to replace all of the silver when 15.55 grams of AgNO3 are present?

Chemistry

1 answer:

Karolina [17]3 years ago

8 0

2AgNO3 + Ni2+ = Ni(NO3)2 + 2Ag+

From the reaction, it can be seen that AgNO3 and Ni2+ has following amount of substance relationshep:

n(AgNO3):n(Ni)=2:1

From the relationshep we can determinate requred moles of Ni2+:

n(AgNO3)=m/M= 15.5/169.87=0.09 moles

So, n (Ni)=n(AgNO3)/2=0.045 moles

Finaly needed mass of Ni2+ is:

m(Ni2+)=nxM=0,045x58.7=2.64g

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Suppose that you have 115 mL of a buffer that is 0.460 M in both benzoic acid ( C 6 H 5 COOH ) and its conjugate base ( C 6 H 5

77julia77 [94]

Answer: The maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL

Explanation:

We are given:

Concentration of buffer = 0.460 M

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})$ .....(1)

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of benzoic acid = 4.2

$[C_6H_5COOH]=0.460M$

$[C_6H_5COO^-]=0.460M$

pH = ?

Putting values in equation 1, we get:

$pH=4.2+\log(\frac{0.460}{0.460})\\\\pH=4.2$

When the pH of the buffer changes by 1 unit, the buffering capacity is said to be lost.

pH change for loosing buffer capacity = [4.2 - 1] = 3.2

Calculating the ratio of conjugate base and its acid by using equation 1:

$pK_a$ = 4.2

pH = 3.2

Putting values in equation 1, we get:

$3.2=4.2+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})\\\\\frac{[C_6H_5COO^-]}{[C_6H_5COOH]}=0.1$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ ......(2)

For benzoic acid and its conjugate base:

Molarity of benzoic acid and its conjugate base = 0.460 M

Volume of solution = 115 mL

Putting values in equation 2, we get:

$0.460=\frac{\text{Moles of benzoic acid and its conjugate base}\times 1000}{115mL}\\\\\text{Moles of benzoic acid and its conjugate base}=\frac{0.460\times 115}{1000}=0.053mol$

The chemical reaction for aniline and HCl follows the equation:

$C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-$

Let the moles of acid added to carry out the change is 'x' moles

Calculating the moles of acid added:

$\frac{[C_6H_5COO^-]-x}{[C_6H_5COOH]+x}=0.1\\\\\frac{0.053-x}{0.053+x}=0.1\\\\x=0.0434$

Calculating the volume of acid added by using equation 2, we get:

Moles of acid added = 0.0434 moles

Molarity of solution = 0.390 M

Putting values in equation 2, we get:

$0.390M=\frac{0.0434\times 1000}{\text{Volume of acid}}\\\\\text{Volume of acid}=\frac{0.0434\times 1000}{0.390}=111.3mL$

Hence, the maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL

7 0

3 years ago

During this reaction, P4 + 5O2 → P4O10, 1.5 moles of product were made in seconds. What is the rate of reaction? 0.011 g/min 210

Ivanshal [37]

The answer is: the rate of reaction is 850 g/min.

n(P₄O₁₀) = 1.5 mol; amount of product.

M(P₄O₁₀) = 283.9 g/mol; molar mass of phosphorus(V) oxide.

m(P₄O₁₀) = n(P₄O₁₀) · M(P₄O₁₀).

m(P₄O₁₀) = 1.5 mol · 283.9 g/mol.

m(P₄O₁₀) = 425.85 g in 30 seconds.

Make proportion: 425 g : 30 s = m(P₄O₁₀) : 60 s.

m(P₄O₁₀) = 850 g in 60 s.

7 0

3 years ago

Read 2 more answers

Nickel and gold combine to make white gold. White gold is considered a(n) of nickel and gold.

san4es73 [151]

White gold is considered an alloy of nickel and gold.
Alloy is the term which refers to the combination of metals, in which a new type of a metal is created. When you mix gold and some other white metal (such as nickel, palladium, or manganese), you get an alloy known as white gold.

8 0

3 years ago

Read 2 more answers

A softball is thrown with 145 Joules of kinetic energy. If the ball is moving at 20.0 m/s, what is the mass of the ball in kg?

Tasya [4]

Answer:

0.725 kg

Explanation:

Step 1: Given data

Kinetic energy of the softball (K): 145 J

Speed of the softball (v): 20.0 m/s

Mass of the softball (m): ?

Step 2: Calculate the mass of the softball

We will use the following expression.

K = 1/2 × m × v²

m = 2 K / v²

m = 2 × 145 J / (20.0 m/s)²

m = 0.725 kg

The mass of the softball is 0.725 kg.

3 0

3 years ago

What is easier to observe about an object,it’s physical properties or it’s chemical properties. Explain why.

Sladkaya [172]

Answer:

Physical Properties

Explanation:

You are able to see physical properties but are unable to see chemical properties.

7 0

3 years ago