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Firlakuza [10]
3 years ago
6

Which sub-layer thins out into space, where there is no air

Chemistry
2 answers:
Tomtit [17]3 years ago
7 0

Answer:

The exosphere

Explanation:

In Earth atmosphere, the closer to the ground is the troposphere, then the stratosphere, then the mesosphere, in which the highest clouds and radar can reach. Then in the thermosphere, where the auroras and the satellites are form, the air glow starts to disappear until we reach the space in the last layer called exosphere.

KengaRu [80]3 years ago
3 0
<span>The "exosphere" is the most distant and tenuous "layer" of our atmosphere.</span>
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If you trace back the history of a carbon atom in your little finger through all of cosmic history, where did this atom most lik
Kamila [148]

Answer:

It was fused from 3 helium nuclei in the core of a red giant star long before the Sun existed

Explanation:

8 0
2 years ago
Bob, Jill, Kim, and Steve measure an object's length, density, mass, and brightness, respectively. Which student must derive a u
ValentinkaMS [17]

Answer:

The answer is Bob

Explanation:

The science exist to kinds of measurements base units and derive units of measurement.

Base units are the ones that measure: length, mass, time, electric current, temperature, luminous intensity and amount of substance.

Derivative units of measurement are obtain when we combine the base units.

Know from your question

Bob measures length is a base unit

Jill measures density is a derivative unit, is the combination of mass and volumen

Kim measures mass is a base unit

Steve measures brightness is a base unit.

8 0
3 years ago
Select all of the BENEFITS of a PARALLEL circuit. If one bulb burns out the rest of the bulbs will stay lit. The bulbs do not ge
artcher [175]

Answer:

The first two options are correct

Explanation:

The first two options are part of the benefits of a parallel connection of bulbs in a circuit. Here, the voltage of each connecting bulb is the same as the voltage of the bulb in the circuit hence all the bulbs have the same voltage running through them. Thus, when one bulb is removed/burns out, it does not affect the remaining bulbs (those ones will remain lit). Also, the addition of bulb(s) does not cause the remaining bulbs in the circuit to get dimmer (since they will all have the same voltage).

4 0
3 years ago
How many joules of heat are required to raise the temperature of 174g of gold from 22°C to 85°C? The specific heat of gold is 0.
Slav-nsk [51]
Remembering the equation Q=MCdeltaT where
q=is the amount of heat energy
M=mass
C=specific heat
deltaT= change in temperature

Therefore, using the equation we can substitute values and solve for q.
Q= (15 grams) (0.129 J/(gx°C))(85-22)
Q=(15) ((0.129 J/(gx°C)) (63)
Q=121.9 Joules

The energy needed to raise the temperature of 15 grams of gold from 22 degrees Celsius to 85 degrees Celsius is then 121.9 Joules or 122 Joules (if rounded up).

7 0
3 years ago
What is the percent yield of a reaction in which 51.5 g of tungsten(VI) oxide (WO3) reacts with excess hydrogen gas to produce m
Rus_ich [418]

Answer:

The percent yield of a reaction is 48.05%.

Explanation:

WO_3+3H_2\rightarrow W+3H_2O

Volume of water obtained from the reaction , V= 5.76 mL

Mass of water = m = Experimental yield of water

Density of water = d = 1.00 g/mL

M=d\times V = 1.00 g/mL\times 5.76 mL=5.76 g

Theoretical yield of water : T

Moles of tungsten(VI) oxide = \frac{51.5 g}{232 g/mol}=0.2220 mol

According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2220 moles of tungsten(VI) oxide will give:

\frac{3}{1}\times 0.2220 mol=0.6660 mol

Mass of 0.6660 moles of water:

0.666 mol × 18 g/mol = 11.988 g

Theoretical yield of water : T = 11.988 g

To calculate the percentage yield of reaction , we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

=\frac{m}{T}\times 100=\frac{5.76 g}{11.988 g}\times 100=48.05\%

The percent yield of a reaction is 48.05%.

7 0
3 years ago
Read 2 more answers
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