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Naddik [55]
3 years ago
15

The rate constant of the elementary reaction CH3OCH3(g) CH4(g) +CH2O(g) is k = 8.33×10-6 s-1 at 427°C, and the reaction has an a

ctivation energy of 245 kJ mol-1. (a) Compute the rate constant of the reaction at a temperature of 545°C. s-1 (b) At a temperature of 427°C, 8.32×104 s is required for half of the CH3OCH3 originally present to be consumed. How long will it take to consume half of the reactant if an identical experiment is performed at 545°C?
Chemistry
1 answer:
zepelin [54]3 years ago
7 0

Answer:

(a) The rate constant is 3.61×10^-3 s^-1

(b) 7.12×10^4 s

Explanation:

(a) Log (K2/K1) = Ea/2.303R × [1/T1 - 1/T2]

K1 = 8.33×10^-6 s^-1

Ea = 245 kJ = 245,000 J

R = 8.314 J/mol.K

T1 = 427°C = 427+273 = 700 K

T2 = 545°C = 546+273 = 818 K

Log (K2/8.33×10^-6) = 245,000/2.303 × [1/700 - 1/818]

Log (K2/8.33×10^-6) = 2.637

K2/8.33×10^-6 = 10^2.637

K2 = 8.33×10^-6 × 433.51 = 3.61×10^-3 s^-1

(b) The relationship between temperature and the time required for reactants to be consumed is inverse

t2 = T1t1/T2

T1 = 427 °C = 700 K

t1 = 8.32×10^4 s

T2 = 545 °C = 818 K

t2 = 700×8.32×10^4/818 = 7.12×10^4 s

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Answer:

Explanation:

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4 0
3 years ago
Read 2 more answers
To measure the amount of iron in a certain type of iron ore, an analytical chemist dissolves a sample in strong acid and titrate
Fiesta28 [93]

Answer:

\% Fe=11.4\%

Explanation:

Hello!

In this case, for the described chemical reaction, we find it is:

8H^+(aq)+5Fe^{2+}(aq)+MnO_4^-(aq)\rightarrow 5Fe^{3+}(aq)+Mn^{2+}(aq)+4H_2O(l)

Because it says that the iron is dissolved in a strong acid which provides addition hydrogen ions to the reaction media. Thus, for the questions attached on the figure we find:

- This a REDOX reaction because we see iron is being oxidized from 2+ to 3+ and manganese reduced from +7 to +2.

- Since it is a redox reaction and the oxidized species is that undergoing an oxidation number increase, we evidence iron goes from +2 to +3, which means iron is the oxidized species.

- In this case, for the used 59.2 mL (0.0592 L) of the 0.2000 M solution of potassium permanganate, we can compute the consumed grams of iron via stoichiometry including the 5:1 mole ratio between them in the chemical reaction:

m_{Fe^{2+}}=0.0592L*0.2000\frac{molMnO_4^-}{L}*\frac{5molFe^{2+}}{1molMnO_4^-}  *\frac{55.85gFe^{2+}}{1molFe^{2+}} \\\\m_{Fe^{2+}}=3.31gFe^{2+}

It means that the percent of iron in that sample is:

\% Fe=\frac{3.31g}{29.00g} *100\%\\\\\% Fe=11.4\%

Best regards.

5 0
3 years ago
Three compounds containing potassium and oxygen are compared. Analysis shows that for each 1.00 g of O, the compounds have 1.22
kicyunya [14]
<span>The law of proportion states that elements combine in whole number ratios. The gram readings for K are multiples of each other, both in grams and moles.
Let us compare the ratios:
</span>2.44 grams/1.22 grams = 2
<span>4.89 grams/2.44 grams = 2</span>

<span>Therefore, Potassium always combines with Oxygen in a ratio of 2  is to 1.</span>
4 0
4 years ago
2. Answer the following questions about a sample of calcium phosphate:
weqwewe [10]

Answer:

a) <u>310.18 g/mol</u>

<u>b) 4.352 moles Ca3(PO4)2</u>

<u>c) 2.6 * 10^24 molecules</u>

<u>d) 5.24 * 10^24 P atoms</u>

<u>e)13.056 moles Ca</u>

<u>f)</u>10825.3 grams

Explanation:

Step 1: Data given

Atomic mass of Ca = 40.08 g/mol

Atomic mass of P = 30.97 g/mol

Atomic mass of O = 16.0 g/mol

Step 2: Calculate molecular weight of Ca3(PO4)2

Molecular weight of Ca3(PO4)2 = 3*atomic mass of Ca + 2* atomic mass of P and 8* atomic mass of O

Molecular weight of Ca3(PO4)2 = 3*40.08 + 2*30.97 + 8*16.0  =<u> 310.18 g/mol</u>

Step 3: Calculate moles of Ca3(PO4)2 in 1350 grams

Moles Ca3(PO4)2 = mass Ca3(PO4)2 /molar mass

Moles Ca3(PO4)2 = 1350 grams / 310.18 g/mol

Moles Ca3(PO4)2 = <u>4.352 moles</u>

Step 4: Calculate molecules in 1350 grams

Molecules = moles * number of Avogadro

Molecules = 4.352 moles * 6.02 * 10^23

Molecules = <u>2.6 *10^24 molecules</u>

<u />

Step 5: Calculate moles Phosphorus

For 1 mol Ca3(PO4)2 we need 2 moles P

For 4.352 moles Ca3(PO4)2 we have 2*4.352 = 8.704 moles

Step 6: Calculate P atoms

Atoms P = 8.704 moles * 6.02*10^23

Atoms P =<u> 5.24 * 10^24 P atoms</u>

<u />

Step 7: Calculate moles Calcium in 1350 grams

For 1 mol Ca3(PO4)2 we have 3 moles Ca

For 4.352 moles we have 3*4.352 = <u>13.056 moles Ca</u>

<u />

<u />

<u>Step 8:</u> Calculate mass of 2.1 * 10^25 molecules of Ca3(PO4)2

Moles Ca3(PO4)2 = 2.1 * 10^25 / 6.02 * 10^23

Moles Ca3(PO4)2 = 34.9 moles

Mass Ca3(PO4)2 = 34.9 moles * 310.18 g/mol

Mass Ca3(PO4)2 = 10825.3 grams

4 0
3 years ago
A foundry produces circular utility access hatches (manhole covers). Currently, 120 covers are produced in a 10-hour shift. If l
SOVA2 [1]

Answer:

14.4 covers/hr.

Explanation:

First, we need to identify how much covers are produced in 1 hour, which will be total produced divided by the total time:

covers/hr = 120/10

covers/hr = 12

If the production will be increased by 20%, it means that new production will be the initial one plus 20% (0.20) of it:

12 + 0.20*12 =  14.4 covers/hr.

6 0
3 years ago
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