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2 years ago

2.92 A 50.0-g silver object and a 50.0-g gold object are both added

Chemistry

1 answer:

Trava [24]2 years ago

3 0

Answer:

82.9 mL

Explanation:

1. Volume of silver

$\begin{array}{rcl}\text{Density}&=& \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho&=& \dfrac{m}{V}\\\\V &=& \dfrac{m}{\rho}\\\\& = & \dfrac{\text{50.0 g}}{\text{10.49 g$\cdot$mL}^{-1}}\\\\& = & \text{4.766 mL}\\\end{array}\\\text{The volume of the silver is $\large \boxed{\textbf{4.766 mL}}$}$

2. Volume of gold

$\begin{array}{rcl}V& = & \dfrac{\text{50.0 g}}{\text{19.30 g$\cdot$mL}^{-1}}\\\\& = & \text{2.591 mL}\\\end{array}\\\text{The volume of the gold is $\large \boxed{\textbf{2.591 mL}}$}$

3. Total volume of silver and gold

V = 4.766 mL + 2.591 mL = 7.36 mL

4 New reading of water level

V = 75.5 mL + 7.36 mL = 82.9 mL

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A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?

Tanzania [10]

<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume = 0.50 cm × 10.0 cm × 2.0 cm

= 10 cm³

Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

= 19.3 g/cm³

Thus, the density of the gold bar is 19.3 g/cm³

3 0

2 years ago

Industrial production of nitric acid, which is used in many products including fertilizers and explosives, approaches 10 billion

mylen [45]

Answer: $9.361\times 10^{4} kJ$

Explanation:

The balanced chemical equation :

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$ $\Delta H^0_{rxn}=-902.0kJ$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{7.056\times 10^3g}{17g/mol}=415.1moles$

According to stoichiometry:

4 moles of $NH_3$ produces = 902.0 kJ of energy

415.1 moles of $NH_3$ produces = $\frac{902.0}{4}\times 415.1=9.361\times 10^{4} kJ$ of energy

Thus the change in enthalpy is $9.361\times 10^{4} kJ$

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2 years ago

A proton transfer reaction can occur when an aldehyde is placed in strong base, such as an alkoxide ion, producing an alcohol an

Pani-rosa [81]

Hi, you have not provided structure of the aldehyde and alkoxide ion.

Therefore i'll show a mechanism corresponding to the proton transfer by considering a simple example.

Explanation: For an example, let's consider that proton transfer is taking place between a simple aldehyde e.g. acetaldehyde and a simple alkoxide base e.g. methoxide.

The hydrogen atom attached to the carbon atom adjacent to aldehyde group are most acidic. Hence they are removed by alkoxide preferably.

After removal of proton from aldehyde, a carbanion is generated. As it is a conjugated carbanion therefore the negative charge on carbon atom can conjugate through the carbonyl group to form an enolate which is another canonical form of the carbanion.

All the structures are shown below.

7 0

2 years ago

A 15.0 mL sample of 0.013 M HNO3 is titrated with 0.017 M CH$NH2 which he Kb=3.9 X 10-10. Determine the pH at these points: At t

kramer

<u>Answer:</u> The pH of the solution in the beginning is 1.89 and the pH of the solution after the addition of base is

<u>Explanation:</u>

<u>For 1:</u> At the beginning

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

Nitric acid is a monoprotic acid and it dissociates 1 mole of hydrogen ions. So, the concentration of hydrogen ions is 0.013 M

$[H^+]=0.013M$

Putting values in above equation, we get:

$pH=-\log(0.013)\\\\pH=1.89$

<u>For 2:</u>

To calculate the number of moles, we use the equation:

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

<u>For nitric acid:</u>

Molarity of nitric acid solution = 0.013 M

Volume of solution = 15 mL

Putting values in above equation, we get:

$0.013M=\frac{\text{Moles of }HNO_3\times 1000}{15}\\\\\text{Moles of }HNO_3=1.95\times 10^{-4}mol$

<u>For methylamine:</u>

Molarity of methylamine solution = 0.017 M

Volume of solution = 10 mL

Putting values in above equation, we get:

$0.017M=\frac{\text{Moles of }CH_3NH_2\times 1000}{10}\\\\\text{Moles of }CH_3NH_2=1.7\times 10^{-4}mol$

The chemical equation for the reaction of nitric acid and methylamine follows:

$HNO_3+CH_3NH_2\rightarrow CH_3NH_3^++NO_3^-$

As, the mole ratio of nitric acid and methyl amine is 1 : 1. So, the limiting reagent will be the reactant whose number of moles are less, which is methyl amine.

By Stoichiometry of the reaction:

1 mole of methyl amine produces 1 mole of $CH_3NH_3^+$

So, $1.7\times 10^{-4}mol$ of methyl amine will produce = $\frac{1}{1}\times 1.7\times 10^{-4}=1.7\times 10^{-4}\text{ moles of }CH_3NH_3^+$

To calculate the $pK_b$ of base, we use the equation:

$pK_b=-\log(K_b)$

where,

$K_b$ = base dissociation constant = $3.9\times 10^{-10}$

Putting values in above equation, we get:

$pK_b=-\log(3.9\time 10^{-10})\\\\pK_b=9.41$

To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[salt]}{[base]})$

$pOH=pK_b+\log(\frac{[CH_3NH_3^+]}{[CH_3NH_2]})$

We are given:

$pK_b=9.41$

$[CH_3NH_3^+]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M$

$[CH_3NH_2]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M$

Putting values in above equation, we get:

$pOH=9.41+\log(\frac{6.8\times 10^{-6}}{6.8\times 10^{-6}})\\\\pOH=9.41$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-9.41=4.59$

Hence, the pH of the solution is 4.59

4 0

2 years ago

A tank containing both hf and hbr gases developed a leak. The ratio of the rate of effusion of hf to the rate of effusion of hbr

Pavlova-9 [17]

Answer:

2.01

Explanation:

The effusion is the passage of the molecules by a small hole by a difference of pressure. By Graham's Law, the rate of the effusion is inversely proportional to the square of the molar mass of the compound. Thus,

rateHF/rateHBr = √MHBr /√MHF

MHBr = 81 g/mol

MHF = 20 g/mol

rateHF/rateHBr = √81/√20

rateHF/rateHBr = 2.01

4 0

2 years ago