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Dennis_Churaev [7]
2 years ago
7

Calculate the mass, in grams, of 2.74 l of co gas measured at 33°c and 945 mmhg.

Chemistry
1 answer:
nalin [4]2 years ago
4 0
<span>T = 33 + 273 =306 K moles CO = pV/RT = 1.24 x 2.74 / 0.08206 x 306=0.135 mass CO = 0.135 mol x 28.01 g/mol=3.78 g</span>
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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
puteri [66]

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}  

6 0
2 years ago
A piece of iron (C=0.449 J/g°C) and a piece of gold (C=0.128 J/g°C) have identical masses. If the iron has an initial temperatur
spin [16.1K]

Answer:

The correct answer is B. Since the two metals have the same mass, but the specific heat capacity of iron is much greater than that of gold, the final temperature of the two metals will be closer to 498 K than to 298 K

Explanation:

Iron is hotter and gold is colder, therefore, according to laws of thermodynamics, iron will lose heat to gold until they are at the same temperature.

The specific heat capacity of iron(0.449) is over three times that of gold(0.128). Since masses are equal, this means that each time iron's temperature drops by one degree, the energy released it releases makes gold's temperature increase by more than 3 degrees. So gold's temperature will be climbing much faster than iron's is falling. Meaning they will meet closer to the initial temperature of iron than that of gold

4 0
2 years ago
How many grams of N2 gas are present in 1.13 L of gas at 2.09 atm and 291 K?
solong [7]

Answer:

Mass= 2.77g

Explanation:

Applying

P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28

PV=nRT

NB

Moles(n) = m/M

PV=m/M×RT

m= PVM/RT

Substitute and Simplify

m= (2.09×1.13×28)/(0.082×291)

m= 2.77g

6 0
2 years ago
If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm?
Llana [10]
Since the temperature is constant, therefore, this problem can be solved based on Boyle's law.
Boyle's law states that: " At constant temperature, the pressure of a certain mass of gas is inversely proportional to its pressure".

This can be written as:
P1V1 = P2V2
where:
P1 is the initial pressure = 1 atm
V1 is the initial volume = 3.6 liters
P2 is the final pressure = 2.5 atm
V2 is the final volume that we need to calculate

Substitute with the givens in the above mentioned equation to get the final volume as follows:
P2V1 = P2V2
1(3.6) = 2.5V2
3.6 = 2.5V2
V2 = 3.6 / 2.5 = 1.44 liters
8 0
2 years ago
Read 2 more answers
How much energy is required to melt a 500. gram block of iron? The heat of vaporization is 6090 J/g and the heat of fusion is 24
galben [10]

The mass of iron block is 500 g. The amount of energy required to melt the iron block needs to be calculated. Melting means conversion of solid to liquid thus, heat of fusion is used which is 247 J/g.

From heat of fusion, 247 J of energy is released by melting 1 g of iron block. Thus, the amount of heat released by melting 500 g of iron rod will be:

H= 247 J/g× 500 g=1.23×10^{5}

Hence, option B is correct.

3 0
2 years ago
Read 2 more answers
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