All categories

4 years ago

A sample of carbon dioxide gas (CO2) contains 6 x 1022 molecules. How many moles of carbon dioxide does this represent?

Chemistry

1 answer:

Agata [3.3K]4 years ago

5 0

Answer:

Explanation:

1 ST METHOD

GIVEN DATA:

number of molecules=6.022×10²²molecules

Avogodro's number Na=6.022×10²²

TO FIND:

number of moles

SOLUTION:

As we knom that number of moles=number of molecules/Na

number of moles=6.022×10²²/6.022×10²²

number of moles=1 moles

2ND METHOD:

AS WE KNOW THAT A MOLE OF A SUBSTANCE CONTAINS 6.022×10²² PARTICLES OF THAT SUBATANCE SO

1MOLE OF CO2=6.022×10²²MOLECULES OF CO2

You might be interested in

1.As you move from left to right across a period,the number of valence electrons

Vitek1552 [10]

1. The number of valence electrons increases as you go from left to right across a period. This is because the number of electrons increases, so another electron will be added to the outer shell.

2. Group 6A elements will have 6 valence elecrons, while Group 2A elements only have 2, therefore Group 6A elements have more valence electrons that Group 2A elements.

3. Fluorine has a smaller atomic size than the other halogens (Cl, Br, I), so its valence electrons are nearer to its nucleus. This means that the attractive forces are stronger, so when another electron (from another atom) draws near the F atom, it is more likely that the electron will be pulled toward the nucleus and react with the F atom.

4 0

3 years ago

Read 2 more answers

A chemist wants to find Kc for the following reaction at 751 K: 2NH3(g) + 3 I2 (g) LaTeX: \Longleftrightarrow ⟺ 6HI(g) + N2(g) K

charle [14.2K]

Answer: The equilibrium constant for the total reaction is $4.09\times 10^{-6}$

Explanation:

We are given:

$K_{c_1}=0.282\\\\K_{c_2}=41$

We are given two intermediate equations:

Equation 1: $N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g);K_{c_1}=0.282$

The expression of $K_{c_1}$ for the above equation is:

$K_{c_1}=\frac{[NH_3]^2}{[N_2][H_2]^3}$

$0.282=\frac{[NH_3]^2}{[N_2][H_2]^3}$ .......(1)

Equation 2: $H_2(g)+I_2(g)\rightleftharpoons 2HI(g);K_{c_2}=41$

The expression of $K_{c_2}$ for the above equation is:

$K_{c_2}=\frac{[HI]^2}{[H_2][I_2]}$

$41=\frac{[HI]^2}{[H_2][I_2]}$ ......(2)

Cubing both the sides of equation 2, because we need 3 moles of HI in the main expression if equilibrium constant.

$(41)^3=\frac{[HI]^6}{[H_2]^3[I_2]^3}$

Now, dividing expression 1 by expression 2, we get:

$\frac{K_{c_1}}{K_{c_2}}=\left(\frac{\frac{[NH_3]^2}{[N_2][H_2]^3}}{\frac{[HI]^6}{[H_2]^3[l_2]^3}}\right)\\\\\\\frac{0.282}{68921}=\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}$