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strojnjashka [21]
3 years ago
10

Within the atmosphere as the altitude increases?

Chemistry
2 answers:
Troyanec [42]3 years ago
6 0
What is the question?
marta [7]3 years ago
3 0

Answer:

Within the atmosphere, as altitude increases,

A.air pressure stays the same.

B.air becomes heavier.

C.air pressure decreases.  The answer.

D.air pressure increases.

Explanation:

That's the question. AND Pressure with Height: pressure decreases with increasing altitude. The pressure at any level in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels. Hope that helps. This means the answer is C.

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All of the following statements concerning voltaic cells are true EXCEPT
MissTica

Answer:

oxidation occurs at the cathode.

Explanation:

In a voltaic cell electrons move from anode to cathode. At the anode, species give up electrons. This is an oxidation reaction depicted by the oxidation half equation. At the cathode, species accept electrons and become reduced. This is depicted by the reduction half equation. In summary; in a Voltaic cell, oxidation occurs at the anode while reduction occurs at the cathode.

3 0
3 years ago
What is extensibility?​
Katena32 [7]

Answer:

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5 0
3 years ago
a container holds 7.4 moles of gas hydrogen gas makes up 25% of the total moles in the container if the pressure is 1.24 atm wha
Tema [17]
This question is based on Dalton's Law of Partial Pressure which states that "the total pressure of a system of gas is equal to the sum of the pressure of each individual gas (partial pressure).

Now, Partial Pressure of a gas = (mole fraction) × (total pressure)

⇒ Partial Pressure of Hydrogen = \frac{1}{4}   ×   \frac{1.24}{1}
                  
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6 0
3 years ago
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Likurg_2 [28]
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3 0
2 years ago
A certain substance X has a normal freezing point of -10.1 degree C and a molal freezing point depression constant K_f = 5.32 de
kirill115 [55]

Answer: -15.4^00C

Explanation:

T^0_f-T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_f = boiling point of solution = ?

T^o_f = boiling point of solvent (X) = -10.1^oC

k_f = freezing point constant  = 5.32^oC/kgmol

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte like urea)

= mass of solute (urea) = 29.82 g

= mass of solvent (X) = 500.0 g

M_2 = molar mass of solute (urea) = 60 g/mol

Now put all the given values in the above formula, we get:

(-10.1-(T_f))^oC=1\times (5.32^oC/m)\times \frac{(29.82g)\times 1000}{60\times (500.0g)}

T_f=-15.4^0C

Therefore, the freezing point of  solution is -15.4^0C

7 0
2 years ago
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