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Radda [10]
3 years ago
13

A weather balloon is inflated to a volume of 2.2 x 10^3 L with 37.4 g of helium. What is the density of helium in grams per lite

r?
Chemistry
1 answer:
Ede4ka [16]3 years ago
8 0
Density = mass/volume
substituting the given values :
We get :
D = 37.4 / 2.2 x 10^3
D = 0.017 grams/liter
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General Chemistry fourth edition by McQuarrie, Rock, and Gallogly. University Science Books presented by Macmillan Learning.
Helen [10]

Answer:

3.07 Cal/g

Explanation:

Step 1: Calculate the heat absorbed by the calorimeter

We will use the following expression.

Q = C × ΔT

where,

  • Q: heat absorbed
  • C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)
  • ΔT: temperature change (2.29 °C)

Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ

According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.

Step 2: Convert 86.1 kJ to Cal

We will use the conversion factor 1 Cal = 4.186 kJ.

86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal

Step 3: Calculate the number of Cal per gram of candy

20.6 Cal/6.70 g = 3.07 Cal/g

3 0
3 years ago
(please, I need some help)
olga55 [171]
c) the salt solubility decreases with temperature.

Salts usually dissolve in water at a given temperature. When water cannot dissolve anymore salt at that same temperature, it is known as a saturation point. With most substances the solubility increases with increase in temperature. Same is the case for a salt like potassium nitrate. With increase in temperature the ability of it to dissolve in water increases. And so with decrease in temperature, the solubility decreases.
7 0
3 years ago
How many liters of Cl2 gas will you have if you are using 63 g of Na?
ELEN [110]

Answer:

You will have 19.9L of Cl2

Explanation:

We can solve this question using:

PV = nRT; V = nRT/P

<em>Where V is the volume of the gas</em>

<em>n the moles of Cl2</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is 273.15K assuming STP conditions</em>

<em>P is 1atm at STP</em>

The moles of 63g of Cl2 gas are -molar mass: 70.906g/mol:

63g * (1mol / 70.906g) = 0.8885 moles

Replacing:

V = 0.8885mol*0.082atmL/molK*273.15K/1atm

V = You will have 19.9L of Cl2

6 0
3 years ago
Why air is classified as a mixture​
Vika [28.1K]

Answer:

See explanation

Explanation:

Air contains a mixture of several molecules and compounds such as oxygen and carbon dioxide.

5 0
2 years ago
Complete the statement<br>qxy- bxy+cxy= xy( )​
andrey2020 [161]

Answer:

xy (-b+c+q) is the answer to this

3 0
3 years ago
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