A gaseous fluorinated methane compound has a mass of 8.00 g at 2.81 atm and 300. K in a 1.00 L container.

Imagine a 10 kg block moving with a velocity of 20/ms to the left calculate the kinetic energy of this block

tino4ka555 [31]

Answer:

KE = 1/2 mv^2=0.5x 10 X 20^2= 800

Explanation

4 0

3 years ago

5. A gas occupies 2000. Lat 100.0 K and exerts a pressure of 100.0 kPa. What volume will

Anna [14]

Answer:

4000 L

Explanation:

Step 1:

Data obtained from the question. This include the following:

Initial volume (V1) = 2000 L.

Initial temperature (T1) = 100 K.

Initial pressure (P1) = 100 kPa.

Final temperature (T2) = 400 K.

Final pressure (P2) = 200 kPa.

Final volume (V2) =..?

Step 2:

Determination of the new volume of the gas.

The new volume of the gas can be obtained by using the general gas equation as follow:

P1V1/T1 = P2V2/T2

100 x 2000/100 = 200 x V2/400

Cross multiply to express in linear form.

100 x 200 x V2 = 100 x 2000 x 400

Divide both side by 100 x 200

V2 = (100 x 2000 x 400)/(100 x 200)

V2 = 4000 L

Therefore, the new volume of the gas is 4000 L

5 0

3 years ago

The suns mass is very small compared to earth true or false

KonstantinChe [14]

I believe that is false

7 0

3 years ago

Read 2 more answers

Which elements have similar behavior? barium silicon aluminum strontium osmium beryllium

victus00 [196]

Answer:

Barium, strontium and beryllium, all belong to the group 2. Silicon belong to group 14. Aluminium belong to group 13 and Osmium belong to group 8. Hence, the elements which have similar behavior are Barium, strontium and beryllium.

Explanation:

hope this helps

6 0

3 years ago

A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h

statuscvo [17]

Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$ (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

Density = $\frac{mass}{volume}$

0.998 g/ml = $\frac{mass}{3.26 ml}$

mass = 3.25 g

Similarly, calculate the moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.25 g}{18.02 g/mol}$

= 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.183 mol = $\frac{mass}{12 g/mol}$

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.95 g}{16 g/mol}$

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

C H O

No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$ .

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$ .

6 0

3 years ago