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bearhunter [10]
3 years ago
13

What type of front in which warm air mass is cut off from the ground by cool air beneath it?

Chemistry
1 answer:
vodomira [7]3 years ago
7 0

Answer:

Occluded Front

Explanation:

"Occluded Front Forms when a warm air mass gets caught between two cold air masses. The warm air mass rises as the cool air masses push and meet in the middle. The temperature drops as the warm air mass is occluded, or “cut off,” from the ground and pushed upward."

  - www.eduplace.com › science › hmxs › pdf

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When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit
pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

The process:

\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq).

How many moles of this process?

Relative atomic mass from a modern periodic table:

  • K: 39.098;
  • N: 14.007;
  • O: 15.999.

Molar mass of \text{KNO}_3:

M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of the process = Number of moles of \text{KNO}_3 dissolved:

\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}.

What's the enthalpy change of this process?

Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ} for 0.212162\;\text{mol}. By convention, the enthalpy change \Delta H measures the energy change for each mole of a process.

\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}.

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

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3 years ago
How would I complete and balance the equation <br> Mg + Br2 &gt;
Katyanochek1 [597]
Hope this helps you.

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During the phase change from solid ice to liquid water, which bonds/forces are being weakened?
vampirchik [111]
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Calculate the molarity 20ml of 3.5 m kci in to a final volume of 100 ml​
satela [25.4K]

Answer:

Molarity =  0.7 M

Explanation:

Given data:

Volume of KCl = 20 mL ( 0.02 L)

Molarity = 3.5 M

Final volume = 100 mL (0.1 L)

Molarity in 100 mL = ?

Solution:

Molarity = number of moles of solute / volume in litter.

First of all we will determine the number of moles of KCl available.

Number of moles = molarity × volume in litter

Number of moles = 3.5 M × 0.02 L

Number of moles = 0.07 mol

Molarity in 100 mL.

Molarity = number of moles / volume in litter

Molarity =  0.07 mol /0.1 L

Molarity =  0.7 M

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Would an opah fish be better suited to live in cold water or warm water? explain why
Kaylis [27]

Don't blame me if you get it wrong i'm a dumb but yet successful student

Warm blood gives deep-sea fish a boost, according to Wegner. The opah's muscles and nervous system likely function faster than an equivalent fish with cold blood. ... This fish, the southern opah, lives in colder waters than the northern opah, so it would be harder to keep warm, Wegner said — but even more beneficial.

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