HELP ME ASAPPPP PLZZZZZ!!!!!

A sample of helium gas has a volume of 1.50 L at 159 K and 5.00 atm. When the gas is compressed to 0.200 L at 50.0 atm, the temp

Brums [2.3K]

Answer:

ghdtgfgrdvreeeegrwwggegteefewqrwefrwerrftrtdsfgyuytfgererdf

Explanation:

The given data is as follows.

$V_{1}$ = 1.50 L, $T_{1}$ = 159 K,

$P_{1}$ = 5.00 atm, $V_{2}$ = 0.2 L,

$T_{2}$ = ?, $P_{2}$ = 50.0 atm

And, according to ideal gas equation,

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

Hence, putting the given values into the above formula to calculate the value of final temperature as follows.

$\frac{5.0 atm \times 1.50 L}{159 K} = \frac{5 atm \times 0.2 L}{T_{2}}$

$T_{2}$ = 21.27 K

Thus, we can conclude that the final temperature is 21.27 K .

5 0

3 years ago

What is the number of significant figures in 12.0160 mL of water?

likoan [24]

Answer:

To calculate the number of significant figures in a number with decimal,

start at the first non-zero digit in the number and count the total number of digits till the end

hence, in this number:

we will have 6 significant figures

4 0

4 years ago

What are limitations of an egg without the shell but has a very thin layer holding it all together

dedylja [7]

Answer:

As a hen ages, the eggs that she lays get gradually larger. However, the calcium content deposited in the shell remains the same despite the size of the egg. So the eggshells become thinner as the hen ages.

Explanation:

6 0

4 years ago

5. The heat of fusion of lead is 25 J/g and its melting point is 601 K. How much heat is given off as 3.0 g

PIT_PIT [208]

Answer:

the heat given off is 75 J.

Explanation:

Given;

latent heat of fusion of lead, L= 25 J/g

mass of liquid lead, m = 3.0g

The heat given off is calculated as;

H = Lm

Where;

H is the quantity of heat given off

H = 25 x 3

H = 75 J.

Therefore, the heat given off is 75 J.

5 0

3 years ago

A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.50 grams of aluminum foil in a solut

andreyandreev [35.5K]

Answer & Explanation:

The replacement of aluminium (Al) by copper (Cu) can be represented by the following chemical equation:

3CuCl2 + 2Al = 2AlCl3 + 3Cu

We are given 0.50g of Al, and 0.75g of CuCl2

We calculate the maximum amount of copper by assuming one of the reactants (CuCl2 or Al) will be depleted.

Calculations:

The corresponding molecular masses are :

3CuCl2 = 3 (63.546+2*35.457) = 403.38 ...........(1)

2Al = 2*26.9815 = 53.963........................................(2)

2AlCl3 = 2 (26.9815+3*35.457) = 266.705 (we don't need to know this!)

3Cu = 3*63.546 = 190.638.......................................(3)

Assuming 0.50g of aluminium will be depleted with unlimited CuCl2, the amount of copper recuperated will be:

M1 = 0.50 *( (3)/(2) ) = 0.50*190.638/53.963 = 1.766 g

If 0.75g of CuCl2 were depleted with unlimited Al, the amount of copper recuperated will be

M2 = 0.75 * ( (3) / (1) ) = 0.75 * 190.638 / 403.38 = 0.354 g < M1

Therefore CuCl2 is the limiting reagent.

Since we can obtain the maximum amount of copper from the limiting reagent (CuCl2), so the required maximum amount is 0.35 g of copper.

8 0

3 years ago