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3 years ago

At 333 k, which of the pairs of gases below would have the most nearly identical rates of effusion?

1 answer:

8 0

E. co and n2Effusion is the process where gas escapes through a hole. Gases with a lower molecular mass effuse more speedy than gases with a higher molecular mass. Relative rates of effusion is related to the molecular mass.
a) M(N₂)/M(O₂) = 28/32 = 0,875
b) M(N₂O)/M(NO₂) = 44/46 = 0,956
c) M(CO)/M(CO₂) = 28/44 = 0,636
d) M(NO₂)/M(N₂O₂) = 44/58= 0,758
e) M(CO)/M(N₂) = 28/28 = 1, CO and N₂ have iexact molecular masses and will effuse at nearly identical rates.

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A compound has been shown to contain 14.5 percent hydrogen and 85.5 percent carbon by mass.

Sedaia [141]

Answer:

Option d. Its empirical formula is CH₂.

Explanation:

The percent composition of the compound allow you to calculate the empirical formula of the compound but is not enough to calculate either the molar mass or the molecular formula. So, since now you can discard options b. and c.

Telling that it is a hydrocarbon (option e.) is true but very vague compared with finding the empirical formula. So, you can also discard the option e.

The fact that the product has a triple bond cannot be concluded from the percent composition, you should find the molecular formula to assert whether it contains or not a triple bond. So, you could discard option a., which lets you only with choice d.

Let us find the empirical formula to be certain that it is CH₂.

1. First, assume a basis of 100 g of compound:

H: 14.5% × 100 g = 14.0 g

C: 85.5% × 100 g = 85.5 g

2. Divide each element by its atomic mass to find number of moles:

H: 14.0 g / 1.008 g/mol = 14.38 mol

C: 85.5 g / 12.011 g/mol = 7.12 mol

3. Divide both amounts by the smallest number, to find the mole ratio:

H: 14.38 mol / 7.12 mol ≈ 2
C: 7.12 mol / 7.12 mol = 1.

Hence, the ratio is 2:1 and the empirical formula is CH₂.

7 0

3 years ago

Which is greater , 3 moles of sulfur or 2 moles of iron sulphide

QveST [7]

Answer:

well since 3 is greater than 2 it would be 3 moles of sulfur.

Explanation:

7 0

3 years ago

Read 2 more answers

Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks

charle [14.2K]

Answer:

pH = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH- Ksp = 3X10^-15

Zn+2 + 4OH- --> Zn(OH)4-2 Kf = 2X10^15

K = Ksp X Kf

= 3*2*10^-15 * 10^15

= 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

Zn(OH)₂ + 2OH⁻(aq) --> Zn(OH)₄²⁻(aq)

Initial: 0 0.3 0

Change: -2x +x

Equilibrium: 0.3 - 2x x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

= 0.3 -(2*0.089)

= 0.122

pOH = -log[OH⁻]

= -log 0.122

= 0.91

pH = 14-0.91

= 13.09

4 0

3 years ago

What would scientist’s next steps be if his/her data supported his/her hypothesis?

Lady bird [3.3K]

To communicate the results in an organized report

6 0

4 years ago

Read 2 more answers

A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what

Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. $[a]_{t}$

Calculations:

The second order rate equation is:

$1/[a]_{t} = kt +1/[a]$

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M

7 0

3 years ago