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3 years ago

How does air above a heated surface move?

1 answer:

6 0

<span> When air is warmed up, its molecules move faster and faster and as a result they move further from each other. They still have the same mass, but they now occupy a larger volume. This means that its density is smaller.
The opposite when air is cooled off. The molecules slow down, get closer together, occupy a smaller volume and therefore its density is bigger.

When air is warmed up, it goes up. Once it's up there, is cools off and goes back down. Near the heated surface the air gets warmed up again, goes up, cools down, goes back down, and again and again.
that is called convection cells
</span>

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What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm

Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{freezing}}{T_f}$

where,

$\Delta S$ = change in entropy

$\Delta H_{fus}$ = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

$\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol$

$T_f$ = freezing point temperature = $-97.6^oC=273+(-97.6)=175.4K$

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

$\Delta S=\frac{\Delta H_{freezing}}{T_m}$

$\Delta S=\frac{-3170J/mol}{175.4K}$

$\Delta S=-18.07J/mol.K$

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

4 0

3 years ago

A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume

pshichka [43]

Answer:

mole fraction of N_2 O = 0.330

mole of fraction SF_4 = 0.669

PRESSURE OF N_2 O = 39127.053 Pa

pressure of SF_4 = 792126.36

Total pressure = 118253.413 Pa

Explanation:

Given data:

volume of tank 8 L

Weight of dinitrogen difluoride gas 5.53 g

weight of sulphur hexafluoride gas 17.3 g

Amount of $N_2 O = \frac{5.53}{14*2 + 16} = 0.1256 mol$

amount of $SF_4 = \frac{17.3}{32.1 + 19*4} = 0.254 mol$

mole fraction of $N_2 O = \frac{0.1256}{0.1256 + 0.254} = 0.330$

mole of fraction $SF_4 = \frac{0.254}{0.1256 + 0.254} = 0.669$

PV = nRT

P of N_2 O $= \frac{0.1256 *8.31 (273 + 26.9}{0.008} = 39127.053 Pa$

mole of SF_4 $=\frac{0.254 *8.31*(273+26.9)}{.008} = 79126.36 Pa$

Total pressure = 39127.053 + 79126.36 = 118253.413 Pa

6 0

3 years ago

What relationship exists between an enzyme and a catalyst?

shepuryov [24]

A catalyst is a chemical that increases the rate of a chemical reaction without itself being changed by the reaction. The fact that they aren't changed by participating in a reaction distinguishes catalysts from substrates, which are the reactants on which catalysts work. Enzymes catalyze biochemical reactions.

5 0

3 years ago

Read 2 more answers

A pharmaceutical company wants to test the efficiency of its new drug production techniques so they run 3 shifts of production f

spayn [35]

The percentage yield of the new production technique is 82.8%

<h3>What is the percentage yield?</h3>

Production is the procedure by which finished products are obtained form the raw materials. The production process involves the passing of raw materials through a certain procedure that involves the use of certain machines and equipment to give us the required products.

We are told in the question that there are three shifts;

Shift 1 produces 4562 grams

Shift 2 produces 5783 grams

Shift 3 produces 5247 grams

Average production from the three shifts = 4562 grams + 5783 grams + 5247 grams/3 = 5197 grams

The theoretical average yield is = 7000 grams + 7000 grams + 7000 grams/3 = 7000 grams

Now the percentage yield = actual yield/ theoretical yield * 100/1

percentage yield = 5197 grams/7000 grams * 100/1

percentage yield = 82.8%

Learn more about percentage yield :brainly.com/question/27492865

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3 0

2 years ago

A mixture of two or more kinds of molecules, evenly dispersed would be a(n)

Dvinal [7]

Azeotropic mixture. I think

5 0

3 years ago