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Anit [1.1K]
3 years ago
10

Which of these is most similar to the Aztec political system?

Chemistry
1 answer:
Mnenie [13.5K]3 years ago
5 0
B the app and then it was just a was the last time you got it to a friend or we
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N sub 2 +3H sub 2 rightwards arrow 2NH sub If 6 liters of hydrogen gas are used, how many liters of nitrogen gas will be needed
astraxan [27]

Answer:

2L of nitrogen gas will be needed

Explanation:

Based on the following reaction:

N₂ + 3H₂ → 2NH₃

<em>1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.</em>

<em />

If 6L of hydrogen (In a gas, the volume is directly proportional to the moles, Avogadro's law) react, the volume of nitrogen gas required will be:

6L H₂ * (1mol N₂ / 3 moles H₂) =

<h3>2L of nitrogen gas will be needed</h3>
7 0
2 years ago
Atoms of the same element can have different forms; for example, carbon-12 and carbon-14.
cricket20 [7]
A carbon-12 atom has 6 protons (6P) and 6 neutrons (6N). But some types of carbon have more than six neutrons. We call forms of elements that have a different number of neutrons, isotopes. For example, carbon-14 is a radioactive isotope of carbon that has six protons and eight neutrons in its nucleus.
Hope that helps
5 0
3 years ago
An archer pulls her bowstring back 0.370 m by exerting a force that increases uniformly from zero to 264 N. (a) What is the equi
Arte-miy333 [17]

Answer:

713.51 N/m

Explanation:

Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.

From hook's law,

F = ke ...........................Equation 1

Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.

Make k the subject of the equation,

k = F/e ............................ Equation 2

Given: F = 264 N, e = 0.37 m.

Substitute into equation 2

k = 264/0.37

k = 713.51 N/m

Hence the spring constant of the bow  = 713.51 N/m

3 0
2 years ago
A scientist discovered a microscopic unicellular organism with no nucleus which of the following correctly describes the organis
boyakko [2]
The correct option is this: THE ORGANISM IS A PROKARYOTES.
There are basically two types of cells, prokaryotic and eukaryotic cells. The prokaryotic cells are primitive cells which contain only a few materials which are not well organised. This type of cells is usually found in microscopic organisms. The cells lack organised nucleus and cell organelles which have membranes.<span />
7 0
3 years ago
Read 2 more answers
Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79
BARSIC [14]

Answer :  The mole fraction and partial pressure of CH_4,CO_2 and He gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of CH_4 = 1.79 mole

Moles of CO_2 = 1.20 mole

Moles of He = 3.71 mole

Now we have to calculate the mole fraction of CH_4,CO_2 and He gases.

\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267

and,

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179

and,

\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554

Thus, the mole fraction of CH_4,CO_2 and He gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of CH_4,CO_2 and He gases.

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 5.78 atm

X_i = mole fraction of gas

p_{CH_4}=X_{CH_4}\times p_T

p_{CH_4}=0.267\times 5.78atm=1.54atm

and,

p_{CO_2}=X_{CO_2}\times p_T

p_{CO_2}=0.179\times 5.78atm=1.03atm

and,

p_{He}=X_{He}\times p_T

p_{He}=0.554\times 5.78atm=3.20atm

Thus, the partial pressure of CH_4,CO_2 and He gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0
3 years ago
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