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kow [346]
2 years ago
11

The chaparral is characterized as a _________________ biome with _________________ , dry summers and mild, rainy winters, making

it vulnerable to infrequent _________________ in late summer and fall. Most chaparral biomes occur in _________________ and the Mediterranean. please help fast due in 20 minutes
Chemistry
1 answer:
elena-14-01-66 [18.8K]2 years ago
8 0

Answer:

temperate, hot, degradation, California

Explanation:

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The partial pressure of N2 in the air is 593 mm Hg at 1 atm. What is the partial pressure of N2 in a bubble of air a scuba diver
Evgen [1.6K]

Answer: Partial pressure of N_{2} at a depth of 132 ft below sea level is 2964 mm Hg.

Explanation:

It is known that 1 atm = 760 mm Hg.

Also,   P_{N_{2}} = x_{N_{2}}P

where,    P_{N_{2}} = partial pressure of N_{2}

                 P = atmospheric pressure

            x_{N_{2}} = mole fraction of N_{2}

Putting the given values into the above formula as follows.

      P_{N_{2}} = x_{N_{2}}P

    593 mm Hg = x_{N_{2}} \times 760 mm Hg

       x_{N_{2}} = 0.780

Now, at a depth of 132 ft below the surface of the water where pressure is 5.0 atm. So, partial pressure of N_{2} is as follows.

         P_{N_{2}} = x_{N_{2}}P

                  = 0.78 \times 5 atm \times \frac{760 mm Hg}{1 atm}

                  = 2964 mm Hg

Therefore, we can conclude that partial pressure of N_{2} at a depth of 132 ft below sea level is 2964 mm Hg.

8 0
3 years ago
A compound has a molar mass of 100 g/mol and the percent composition (by mass) of 65.45% C, 5.45% H, and 29.09% 0. Determine the
coldgirl [10]

Answer:

B) C3H3O and C6H6O2

Explanation:

Given data:

Molar mass of compound = 100 g/mol

Percentage of hydrogen = 5.45%

Percentage of carbon = 65.45%

Percentage of oxygen = 29.09%

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of H = 5.45 / 1.01 = 5.4

Number of gram atoms of O = 29.09/ 16 = 1.8

Number of gram atoms of C = 65.45 / 12 = 5.5

Atomic ratio:

            C                      :      H            :         O

           5.5/1.8              :     5.4/1.8     :        1.8/1.8

            3                      :        3          :        1

C : H : O = 3 : 3 : 1

Empirical formula is C₃H₃O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = 12×3 + 1.01 ×3 + 16×1 = 55.03  

n = 100 / 5503

n = 2

Molecular formula = n (empirical formula)

Molecular formula = 2 (C₃H₃O)

Molecular formula = C₆H₆O₂

3 0
3 years ago
C5H12, pentane, is a liquid at room temperature.
zloy xaker [14]

Answer:

C)We cannot be sure unless we find out its boiling point.

Explanation:

I will like to clearly state that simply comparing two compounds will not tell us exactly which one will be a liquid, solid or gas at room temperature.

If I want to determine whether an unknown substance will be a liquid at room temperature, I will have to measure its boiling point. If the boiling point is above room temperature, and the melting point is below room temperature, it’s a liquid. If the boiling point of the unknown substance is below room temperature, it is a gas.

This confirms that we cannot conclude on the state of matter in which a compound exists unless we know something about its boiling point, not by inspecting the properties of neighbouring compounds in the same homologous series

4 0
3 years ago
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