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2 years ago

(0904) How many atoms are there in 56.2 grams of Krypton, Kr?

Chemistry

1 answer:

Luba_88 [7]2 years ago

8 0

Answer:

56.2÷6.02×10^23

=9.34×10^23

Explanation:

Divide the given mass of the atom by the mass of an Atom (the avogadro's constant) to find the number of atoms in the given mass.

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Law Incorporation [45]

At the half equivalence point [HA] = [A-] and pH = pKa

<span>if Ka is 5.2e-5 then pKa = pH = 4.28</span>

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3 years ago

2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

No. of moles of CO = 0.800 mol

No. of moles of $H_2$ = 2.40 mol

Volume = 8.00 L

Concentration = $\frac{Moles}{Volume\ in\ L}$

Concentration of CO = $\frac{0.800}{8.00} = 0.100\ mol/L$

Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

$Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90$

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3 years ago

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Fiesta28 [93]

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2 years ago

Read 2 more answers

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anyanavicka [17]

Answer:

use visual studio code and put in this print{3:2}-1

Explanation:

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2 years ago

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