During which event does the moon appear closer? Select one: a. Apogee b. Penumbra c. Perigee d. Umbra

A sample of a gas at 25°C has a volume of 150 mL when its pressure is 0.947 atm. What will the temperature of the gas be at a pr

sertanlavr [38]

Answer: The temperature of the gas at a pressure of 0.987 atm and volume of 144mL is $25.16^0C$

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 0.947 atm

$P_2$ = final pressure of gas = 0.987 atm

$V_1$ = initial volume of gas = 150 ml

$V_2$ = final volume of gas = 144 ml

$T_1$ = initial temperature of gas = $25^0C=(25+273.15)K=298.15K$

$T_2$ = final temperature of gas = ?

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

$\frac{0.947\times 150}{298.15}=\frac{0.987\times 144}{T_2}$

$T_2=298.31K=(298.31-273.15)^0C=25.16^0C$

The temperature of the gas at a pressure of 0.987 atm and volume of 144mL is $25.16^0C$

4 0

3 years ago

What are alveoli and what do they do?

zlopas [31]

Alveoli are tiny sacs within our lungs that allow oxygen and carbon dioxide to move between the lungs and bloodstream

6 0

3 years ago

Read 2 more answers

What mass of aluminum is needed to produce 0.500 mole of aluminum chloride?

3241004551 [841]

Answer: " 13.5 g Al " ;

→ that is: "13.5 grams of aluminum."

____________________________

Explanation:

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Note: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:

The reactants: "aluminum (Al) " ; and "chlorine (Cl) " ; and:

The product: "aluminum choloride (AlCl₃) " .

____________________________

The "balanced chemical equation" is:

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2 Al + 3 Cl₂ → 2 AlCl₃ ;

_____________________________

Note: The molecular weight of "aluminum (Al)" is: " 26.98 g /mol " .

____________________________

So: We call solve using a technique known as: "dimensional analysis" :

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0.500 mol AlCl₃ * $(\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?$

____________________________

Note: The units of "mol AlCl₃" cancel out to "1' ; and:

The units of "mol Al" cancel out to "1" ; and we are left with:

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" $\frac{(0.500 * 2 * 26.98)}{2}$ g Al ["grams of aluminum"] ;

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Note: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:

→ (0.500 * 26.98) g Al ;

= 13.49 g Al ;

→ Round to 3 (Three) significant figures;

→ Since: "0.500" has 3 (Three) significant figures:

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= 13.5 g Al ; that is: "13.5 grams of aluminum."

____________________________

Hope this is helpful!

Best wishes to you in your academic pursuits—and within the "Brainly" community!

____________________________

3 0

3 years ago

15 POINTS PLEASE HELP What volume of water must be added to 35mL of 2.6m KCl to reduce its concentration to 1.2m? Please explain

BartSMP [9]

First, find the volume the solution needs to be diluted to in order to have the desired molarity:
You have to use the equation M₁V₁=M₂V₂ when ever dealing with dilutions.

M₁=the starting concentration of the solution (in this case 2.6M)
V₁=the starting volume of the solution (in this case 0.035L)
M₂=the concentration we want to dilute to (in this case 1.2M)
V₂=the volume of solution needed for the dilution (not given)

Explaining the reasoning behind the above equation:
MV=moles of solute (in this case KCl) because molarity is the moles of solute per Liter of solution so by multiplying the molarity by the volume you are left with the moles of solute. The moles of solute is a constant since by adding solvent (in this case water) the amount of solute does not change. That means that M₁V₁=moles of solute=M₂V₂ and that relationship will always be true in any dilution.

Solving for the above equation:
V₂=M₁V₁/M₂
V₂=(2.6M×0.035L)/1.2M
V₂=0.0758 L
That means that the solution needs to be diluted to 75.8mL to have a final concentration of 1.2M.

Second, Finding the amount of water needed to be added:
Since we know that the volume of the solution was originally 35mL and needed to be diluted to 75.8mL to reach the desired molarity, to find the amount of solvent needed to be added all you do is V₂-V₁ since the difference in the starting volume and final volume is equal to the volume of solvent added.
75.8mL-35mL=40.8mL
40.8mL of water needs to be added

I hope this helps. Let me know if anything is unclear.
Good luck on your quiz!

5 0

3 years ago

The activation energy of a reaction is 56.9 kj/mol and the frequency factor is 1.5×1011/s. Part a calculate the rate constant of

Vanyuwa [196]

We will use Arrehenius equation

lnK = lnA -( Ea / RT)

R = gas constant = 8.314 J / mol K

T = temperature = 25 C = 298 K

A = frequency factor

ln A = ln (1.5×10 ^11) = 25.73

Ea = activation energy = 56.9 kj/mol = 56900 J / mol

lnK = 25.73 - (56900 / 8.314 X 298) = 2.76

Taking antilog

K = 15.8

5 0

3 years ago