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ryzh [129]
3 years ago
8

Isobutylene is a hydrocarbon used in the manufacture of synthetic rubber. when 0.847 g of isobutylene was subjected to combustio

n analysis, the gain in mass of the co2 absorber was 2.657 g and that of the h2o absorber was 1.089 g. what is the empirical formula of isobutylene?
Chemistry
1 answer:
Firlakuza [10]3 years ago
6 0
Find the mass of C in the 2.657 g CO2:
(2.657 g CO2) / (44.01 g/mol) = 0.06037 mol CO2
Since each mole of CO2 also has 1 mole of C, this is equivalent to 0.06037 mol C.

Find the mass of H in the 1.089 g H2O:
(1.089 g H2O) / (18.02 g/mol) = 0.06043 mol H2O
Since 1 mol H2O has 2 mol H, this is equivalent to (0.06043)*2 = 0.1209 mol H.

Taking the ratio of H to C: 0.1209 / 0.06037 = 2.002 ~ 2
Therefore, the empirical formula of isobutylene is CH2.
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A 2.241-g sample of nickel reacts with oxygen to form 2.852 g of the metal oxide. Calculate the empirical formula of the oxide?
astraxan [27]

Answer:

NiO

Explanation:

Firstly, we need to know the mass of oxygen used up. This is the difference between the mass of the nickel and the mass of the oxide. This is : 2.852 - 2.241 = 0.611g

Now we convert these masses to mole by dividing by their atomic masses. The atomic mass of oxygen is 16 while that of nickel is 59

O = 0.611/16 = 0.0381875

Ni = 2.241/59 = 0.037983050847458

The next thing is to divide each by the smaller number of moles, which is that of the nickel.

O = 0381875 /0.037983050847458= 1

Ni= 0.037983050847458/0.037983050847458= 1

The empirical formula is thus NiO

7 0
3 years ago
Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4. Calculate the pH of a solution containing a caffeine concentration of 41
RSB [31]

Answer:

pH → 7.47

Explanation:

Caffeine is a sort of amine, which is a weak base. Then, this pH should be higher than 7.

Caffeine + H₂O  ⇄  Caffeine⁺  +  OH⁻      Kb

1 mol of caffeine in water can give hydroxides and protonated caffeine.

We convert the concentration from mg/L to M

415 mg = 0.415 g

0.415 g / 194.19 g/mol = 2.14×10⁻³ mol

[Caffeine] = 2.14×10⁻³  M

Let's calculate pH. As we don't have Kb, we can obtain it from pKb.

- log Kb = pKb → 10^-pKb = Kb

10⁻¹⁰'⁴ = 3.98×10⁻¹¹

We go to equilibrium:

Caffeine + H₂O  ⇄  Caffeine⁺  +  OH⁻      Kb

Initially we have 2.14×10⁻³ moles of caffeine, so, after the equilibrium we may have (2.14×10⁻³ - x)

X will be the amount of protonated caffeine and OH⁻

     Caffeine     +    H₂O  ⇄  Caffeine⁺  +  OH⁻      Kb

   (2.14×10⁻³ - x)                         x                x

We make the expression for Kb:

3.98×10⁻¹¹ = x² / (2.14×10⁻³ - x)

We can missed the -x in denominator, because Kb it's a very small value.

So: 3.98×10⁻¹¹ = x² / 2.14×10⁻³

√(3.98×10⁻¹¹ . 2.14×10⁻³) = x → 2.92×10⁻⁷

That's the [OH⁻].  - log [OH⁻] = pOH

- log 2.92×10⁻⁷ = 6.53 → pOH

14 - pOH = pH → 14 - 6.53 = 7.47

4 0
3 years ago
Saturated fatty acids and unsaturated fatty acids differ in.
ratelena [41]

Answer:

the number off double bonds in a fatty acid chain

4 0
1 year ago
50 POINTS FOR THE CORRECT ANSWER SOMEONE HELP ME ON THIS QUESTION PLEASE !!!!!!
oee [108]

Answer:

it is positive and it reacts

7 0
3 years ago
Read 2 more answers
How many grams are in 6.00 moles of NaCl?
Anastasy [175]

Is it 350.4, I assume?

5 0
3 years ago
Read 2 more answers
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