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2 years ago

Heat is being removed from a substance, but its temperature does not change. What is

1 answer:

4 0

So its temperature will not rise, since kinetic energy of molecules remains the same. The quantity of heat absorbed or released when a substance changes its physical phase at constant temperature (e g. From solid to liquid at melting point or from liquid to gas at boiling point) is termed as its latent heat.

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A 3L sample of an ideal gas at 178 K has a pressure of 0.3 atm. Assuming that the volume is constant, what is the approximate pr

harina [27]

Answer: The approximate pressure of the gas after it is heated to 278 K is 0.468 atm.

Explanation:

Given: $T_{1}$ = 178 K, $P_{1}$ = 0.3 atm

$T_{2}$ = 278 K, $P_{2}$ = ?

According to Gay Lussac law, at constant volume the pressure of a gas is directly proportional to the temperature.

Formula used to calculate the pressure is as follows.

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\$

Substitute the values into above formula is as follows.

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\frac{0.3 atm}{178 K} = \frac{P_{2}}{278 K}\\P_{2} = 0.468 atm$

Thus, we can conclude that the approximate pressure of the gas after it is heated to 278 K is 0.468 atm.

3 0

3 years ago

Explain why salivary amylase would not digest protein

CaHeK987 [17]

Answer:This particular enzyme promotes the breakdown of starches into simpler sugars which can be absorbed by the body. Salivary amylase, like most other enzymes, is a protein. ... Consequentlysalivary amylase does not function once it enters the stomach.

Explanation:

6 0

3 years ago

What does your data indicate about the optimum ph level for this enzyme-catalyzed reaction?

mestny [16]

At an optimum pH of 7.0, there are more molecules per minute in all amounts of substrate thus this pH is ideal for maximum growth. 5. Enzymes function most efficiently at the temperature of a typical cell, which is 37 degrees Celsius. Increases or decreases in temperature can significantly lower the reaction rate.

3 0

3 years ago

Which of the following statements about salinity is true?

diamong [38]

Answer:

The answer would be B.

Explanation:

Ocean water near areas with low evaporation has higher salinity.

if im wrong please tell me .__.

4 0

3 years ago

Read 2 more answers

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago