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9966 [12]
3 years ago
13

The atmospheric gases in the thermosphere are heated _____. Pick the best one.

Chemistry
2 answers:
Trava [24]3 years ago
7 0

Answer:

Hi! It is acually directly by the sun.

Explanation:

lozanna [386]3 years ago
4 0

by radiation from the Earth's surface

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2. Which is not true of an electric current?
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B. Electric current requires a specific path to follow.
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2 years ago
How can substances in a compound and in a mixture be separated?
Ray Of Light [21]
<span>A chemical reaction is required to separate the substances in a compound. The components of a mixture can be separated based on their physical properties using techniques like filtration or distillation.</span>
8 0
3 years ago
A 2.026g sample of a hydrate of sodium carbonate (Na2CO3) is heated to remove water. After heating, the mass of the sample is 0.
OLEGan [10]

Answer:

62.98 % of the sample of hydrate is water

Explanation:

Step 1: Data given

Mass of the sample of a hydrate of sodium carbonate (Na2CO3) = 2.026 grams

After heating, the mass of the sample is 0.750 g

Molar mass H2O = 18.02 g/mol

Step 2: Calculate mass of water

Mass water = mass of hydrate - mass of sample after heating

Mass water = 2.026 grams - 0.750 grams

Mass water = 1.276 grams

Step 3: Calculate mass % percent of water

Mass % of water = (mass of water / total mass hydrate) * 100 %

Mass % of water = (1.276 grams / 2.026 grams) *100 %

Mass % of water = 62.98 %

62.98 % of the sample of hydrate is water

7 0
3 years ago
Read 2 more answers
A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
3 years ago
Which element gains two electrons to fill its outermost energy level?
laiz [17]
O magnesium Mg is the answer
4 0
2 years ago
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