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aliya0001 [1]
3 years ago
6

A virus has a mass of ×9.010−12mg and an oil tanker has a mass of ×3.0107kg . Use this information to answer the questions below

. Be sure your answers have the correct number of significant digits.What is the mass of one mole of viruses in grams?
Chemistry
1 answer:
puteri [66]3 years ago
6 0

Answer: Mass of one mole of viruses in grams is 54\times 10^{8}  

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

Given : One virus has mass of = 9.0\times 10^{-12}mg=9.0\times 10^{-15}g     1mg=10^{-3}g

One mole of virus 6.023\times 10^{23} has mass of = \frac{9.0\times 10^{-15}}{1}\times 6.023\times 10^{23}=54\times 10^{8}g  

Thus mass of one mole of viruses in grams is 54\times 10^{8}  

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When most liquids freeze, explain what happens to the motion and the space between the atoms.
KiRa [710]

Answer:

atoms relative motion slow down and begin to vibrate in place

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3 years ago
In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce ga
Reptile [31]

Answer:

The rate at which ammonia is being produced is 0.41 kg/sec.

Explanation:

N_2+3H_2\rightarrow 2NH_3 Haber reaction

Volume of dinitrogen consumed in a second = 505 L

Temperature at which reaction is carried out,T= 172°C = 445.15 K

Pressure at which reaction is carried out, P = 0.88 atm

Let the moles of dinitrogen be n.

Using an Ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.88 atm\times 505 L}{0.0821 atm l/mol K\times 445.15 K}=12.1597 mol

According to reaction , 1 mol of ditnitrogen gas produces 2 moles of ammonia.

Then 12.1597 mol of dinitrogen will produce :

\frac{2}{1}\times 12.1597 mol=24.3194 mol of ammonia

Mass of 24.3194 moles of ammonia =24.3194 mol × 17 g/mol

=413.43 g=0.41343 kg ≈ 0.41 kg

505 L of dinitrogen are consumed in 1 second to produce 0.41 kg of ammonia in 1 second. So the rate at which ammonia is being produced is 0.41 kg/sec.

6 0
3 years ago
When 3.0 kg of water is warmed from 10 °C to 80 °C, how much heat energy is needed?
Naddik [55]

Answer:

THE HEAT NEEDED TO CHANGE 3KG OF WATER FROM 10 C TO 80 C IS 877.8kJ OR 877,800 J.

Explanation:

Mass = 3.0 kg = 3 * 1000 = 3000 g

Initial temperature = 10 C

Final temperature = 80 C

Change in temperature = 80 - 10 = 70 C

Specific heat of water = 4.18 J/g C

Heat needed = unknown

Heat is the amount of energy in joules needed to change a gram of water by 1 C.

Heat = mass * specific heat * change in temperature

Heat = 3000 g * 4.18 J/g C * 70 C

Heat = 877 800 Joules

Heat = 877.8 kJ.

The heat needed to change 3 kg mass of water from 10 C to 80 C is 877,800 J or 877.8 kJ.

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A longitudinal wave passing through a medium causes areas of:
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The answer is B. Compressions and rarefactions.

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2 years ago
What happens to the liquid in a thermometer when it is moved from cold water to boiling water?
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Nothing it just changes temp as it is protected by glass layers
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