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3 years ago

Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO

Chemistry

1 answer:

tekilochka [14]3 years ago

7 0

Answer: There are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms.

So, the number of atoms present in 0.67 moles are as follows.

$0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms$

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

$4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms$

Thus, we can conclude that there are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

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How many moles of helium are needed to fill a balloon to a volume of 5.3 L at 22 ℃ and 632 mmHg?

son4ous [18]

Answer:

0.18 moles

Explanation:

Applying,

PV = nRT................... Equation 1

Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = temperature.

make n the subject of the equation

n = PV/RT............... Equation 2

Given: V = 5.3 L, T = 22 °C = (22+272) K = 295 K, P = 632 mmHg = (0.00131579×632) = 0.8316 atm, R = 0.083 L.atm/K.mol

Substitute these values into equation 2

n = (0.8316×5.3)/(0.083×295)

n = 0.18 moles

6 0

3 years ago

Dalton’s Law CalculationA mixture of H₂, N₂ and Ar gases is present in a steel cylinder. The total pressure within the cylinder

Zolol [24]

Answer:

A) The partial presssure of CO₂ is 167 mm Hg

B) The partial presssure of N₂ is 354 mm Hg

C) The partial presssure of Ar is 235 mm Hg

D) The partial presssure of H₂ is 86 mm Hg

Explanation:

Dalton's law of partial pressures is basically expressed by the following statement:

The total pressure of a mixture is equal to the sum of the partial pressures of its components.

So initially we have:

$P_{T}$ = total presure of the system (675 mm Hg).

$P_{N_2}$ = partial pressure of N₂ (354 mm Hg).

$P_{Ar}$ = partial pressure of Ar (235 mm Hg).

Using Dalton's law we can find the partial pressure of H₂:

$P_{T}=P_{N_2}+P_{Ar}+P_{H_2}$

675 mm Hg=354 mm Hg + 235 mm Hg + $P_{H_2}$

$P_{H_2}$ = 675 mm Hg - 354 mm Hg - 235 mm Hg

$P_{H_2}$ =86 mm Hg

If CO₂ gas is added to the mixture, at constant temperature, and the volume is the same, the difference between the new total pressure and the previous total pressure is equal to the partial pressure of CO₂.

$P_{T}=P_{N_2}+P_{Ar}+P_{H_2}+P_{CO_2}$

842 mm Hg= 354 mm Hg + 235 mm Hg + 86 mm Hg + $P_{CO_2}$

$P_{CO_2}$ = 842 mm Hg - 354 mm Hg - 235 mm Hg - 86 mm Hg

$P_{CO_2}$ = 167 mm Hg

4 0

3 years ago

A serving of Cheez-Its releases 1.30 x 10^4 kcal (1 kcal = 4.18 kJ) when digested by your body. If this same amount of energy we

kondor19780726 [428]

Answer:

The final temperature is:- 7428571463.57 °C

Explanation:

The expression for the calculation of heat is shown below as:-

$Q=m\times C\times \Delta T$

Where,

$Q$ is the heat absorbed/released

m is the mass

C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

Mass of water = 1.75 mg = 0.00175 g ( 1 g = 0.001 mg)

Specific heat of water = 4.18 J/g°C

Initial temperature = 35 °C

Final temperature = x °C

$\Delta T=(x-35)\ ^0C/tex]
Q = [tex]1.3\times 10^4$ kcal

Also, 1 kcal = 4.18 kJ = $4.18\times 10^3$ J

So, Q = $1.3\times 10^4\times 4.18\times 10^3$ J = 54340000 J

So,

$54340000=0.00175\times 4.18\times (x-35)$

$0.00175\times \:4.18\left(x-35\right)=54340000$

$x-35=\frac{54340000}{0.007315}$

$x=7428571463.57$

Thus, the final temperature is:- 7428571463.57 °C

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Answer:

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