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Pani-rosa [81]
3 years ago
9

The density of air under ordinary conditions at 25°C is 1.19 g/L. How many kilograms of air are in a room that measures 10.0 ft

× 11.0 ft and has an 10.0 ft ceiling? 1 in = 2.54 cm (exactly); 1 L = 103 cm3.
Chemistry
2 answers:
Misha Larkins [42]3 years ago
5 0

Answer: There are 37 kg of air in the room.

Explanation:

To calculate the volume of cuboid (room), we use the equation:

V=lbh

where,

V = volume of cuboid

l = length of room = 11 ft

b = breadth of room =  10 ft

h = height of room= 10 ft

Putting values in above equation, we get:

V=10\times 11\times 10=1100ft^3=1100\times 28.3L=31130L  (Conversion factor: 1ft^3=28.3L

To calculate mass of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

We are given:

Density of air = 1.19 g/L

Volume of air = volume of room =  31130 L

Putting values in above equation, we get:

1.19g/L=\frac{\text{Mass of air}}{31130L}\\\\\text{Mass of air}=37000g=37.0kg    (1kg=1000g)

Hence, the mass of air is 37 kg.

Lunna [17]3 years ago
3 0

Answer: The mass of air present in the room is 37.068 kg

Explanation :  Given,

Length of the room = 10.0 ft

Breadth of the room = 11.0 ft

Height of the room = 10.0 ft

To calculate the volume of the room by using the formula of volume of cuboid, we use the equation:

V=lbh

where,

V = volume of the room

l = length of the room

b = breadth of the room

h = height of of the room

Putting values in above equation, we get:

V=10.0ft\times 11.0ft\times 10.0ft=1100ft^3=31148.53L

Conversion used : 1ft^3=28.3168L

Now we have to calculate the mass of air in the room.

Density=\frac{Mass}{Volume}

1.19g/L=\frac{Mass}{31148.53L}

Mass=37066.7507g=37.068kg

Conversion used : (1 kg = 1000 g)

Therefore, the mass of air present in the room is 37.068 kg

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Answer:

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Explanation:

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4(12.011g/mol) + 10(1.008 g/mol) = 58.124 g/mol C₄H₁₀

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--------------------------  x  ----------------------  x  ---------------------  
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Answer:

28.01g

Explanation:

Given the weight of one mole of Cabon as 12.01g and that of oxygen as 16.00g.

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A 256 mL sample of HCl gas is in a flask where it exerts a force (pressure) of 67.5 mmHg. What is the pressure of the gas if it
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Answer:

The pressure in the new flask would be 128\; \rm mmHg if the \rm HCl here acts like an ideal gas.  

Explanation:

Assume that the \rm HCl sample here acts like an ideal gas. By Boyle's Law, the pressure P of the gas should be inversely proportional to its volume V.

For example, let the initial volume and pressure of the sample be V_1 and P_1. The new volume V_2 and pressure P_2 of this sample shall satisfy the equation: P_1 \cdot V_1 =P_2 \cdot V_2.

In this question,

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The goal is to find the new pressure of this gas, P_2.

Assume that this sample is indeed an ideal gas. Then the equation P_1 \cdot V_1 =P_2 \cdot V_2 should still hold. Rearrange the equation to separate the unknown, P_2. Note: make sure that the units for V_1 and V_2 are the same before evaluating. That way, the unit of

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Explanation:

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