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Nina [5.8K]
3 years ago
13

How much time would it take to distribute one avagadro's number of wheat grains if 10^10 grains are A)1.9*10^2 years B)1.9*10^10

years C)1.9*10^8 years S)1.9*10^6 years
Chemistry
1 answer:
Makovka662 [10]3 years ago
7 0

Answer:

It takes  1.9 \times 10^{6} years to distribute all the grains.

Explanation:

In one second, 10^{10} grains of wheat are distributed.

We are supposed to find the time it would take to Distribute Avogadro number of grains.

1 avogadro number = 6.022 \times 10^{23}

Number of grains distributed in 1 day = 86400 \times 10^{10}

                                                             = 8.64 \times 10^{14}

Number of grains distributed in 1 year= 8.64 \times 10^{14} \times 365

                                 = 3.1536 \times 10^{17}

Time taken = \frac{Number of grains to be distributed}{grains distributed in 1 year}

            = \frac {6.022 \times 10^{23}}{3.1536\times 10^{17}}

            = 1.9 \times 10^{6}

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Answer:

22.44°C will be the final temperature of the water.

Explanation:

Heat lost by tin will be equal to heat gained by the water

-Q_1=Q_2

Mass of tin = m_1=18.5 g

Specific heat capacity of tin = c_1=0.21 J/g^oC

Initial temperature of the tin = T_1=97.38^oC

Final temperature = T_2=T

Q_1=m_1c_1\times (T-T_1)

Mass of water= m_2=75.7 g

Specific heat capacity of water= c_2=4.184 J/g^oC

Initial temperature of the water = T_3=21.52^oC

Final temperature of water = T_2=T

Q_2=m_2c_2\times (T-T_3)

-Q_1=Q_2

-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)

On substituting all values:

-(18.5 g\times 0.21 J/g^oC\times (T-97.38^oC))=75.7 g\times 4.184 J/g^oC\times (T-21.52 ^oC)

we get, T = 22.44°C

22.44°C will be the final temperature of the water.

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Every Sunday, Sarah prepares blue fruit punch to
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3 years ago
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A gas cylinder contains 2.5 atm of ne, 0.8 atm of co2, and 3.4 atm of ar. what is the total pressure in the cylinder? g
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To answer this problem, we must make assumptions for simplicity. The first assumption is that, the system only consist of these 3 gases. The second assumption is that, these gases behave ideally. Thus, from Dalton's Law of Partial Pressure, the total pressure is simply the sum of their individual partial pressures.

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Which of the following sentences describes oxygen at room temperature? (check all that apply)
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It has no definite volume and takes the shape of its container.

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3 years ago
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A tank at is filled with of chlorine pentafluoride gas and of sulfur hexafluoride gas. You can assume both gases behave as ideal
Ivan

Answer:

- Mole fraction of Chlorine Pentafluoride

= 0.265

- Partial Pressure of Chlorine Pentafluoride

= 16.05 kPa

- Mole fraction of Sulfur Hexafluoride

= 0.735

- Partial Pressure of Sulfur Hexafluoride

= 44.53 kPa

Total Pressure exerted by the gases = 60.58 kPa

Explanation:

First of, we calculate the number of moles of each gas present.

Number of moles = (Mass)/(Molar Mass)

For ClF₅

Mass = 4.28 g

Molar Mass = 130.445 g/mol

number of moles of Chlorine Pentafluoride

= (4.28/130.445) = 0.0328 moles

For SF₆

Mass = 13.3 g

Molar Mass = 146.06 g/mol

number of moles of Sulfur Hexafluoride

= (13.3/146.06) = 0.0911 moles

Total number of moles present = 0.0328 + 0.0911 = 0.1239 moles.

Using the ideal gas equation

PV = nRT

P = total pressure in the tank = ?

V = volume of the tank = 5.00 L = 0.005 m³

R = molar gas constant = 8.314 J/mol.K

T = temperature of the tank = 20.9°C = 294.05 K

n = total number of moles present = 0.1239 moles

P × 0.005 = (0.1239 × 8.314 × 294.05)

P = 60,580.45 Pa = 60.58 kPa.

- Mole fraction of a particular component of interest = (number of moles of the component of interest) ÷ (total number of moles)

- Partial Pressure of a particular component of interest = (mole fraction of that component of interest) × (total pressure)

This is Dalton's law of Partial Pressure.

- Mole fraction of Chlorine Pentafluoride

= (0.0328/0.1239) = 0.265

- Partial Pressure of Chlorine Pentafluoride

= 0.265 × 60.58 = 16.05 kPa

- Mole fraction of Sulfur Hexafluoride

= (0.0911/0.1239) = 0.735

- Partial Pressure of Sulfur Hexafluoride

= 0.735 × 60.58 = 44.53 kPa

Total Pressure exerted by the gases = 16.04 + 44.53 = 60.58 kPa

Hope this Helps!!!

3 0
3 years ago
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