All categories

3 years ago

What are the products of the balanced equation for the combustion of C8H17OH ?

Chemistry

1 answer:

erastova [34]3 years ago

3 0

C8H17OH + 12O2 —> 8CO2 + 9H2O

You might be interested in

What are the different characteristics that can be used to identify an unknown mineral

brilliants [131]

These are just a couple ways to identify unknown minerals,touch taste smell

4 0

4 years ago

Read 2 more answers

PLS HELP ME ON MY FINAL MULTIPLE CHOICE

dexar [7]

Answer:

The force would decrease

5 0

3 years ago

Read 2 more answers

This is true for the body being attracted as well. Bodies of water that are smaller than oceans, seas, and large lakes are _____

zhannawk [14.2K]

Answer:

Explanation:

4 0

3 years ago

How many atoms are there in 3.559*10^-6 mol of krypton?

algol [13]

Answer:

The answer is

<h2> $2.143 \times {10}^{18} \: \: atoms$ </h2>

Explanation:

To find the number of atoms given the number of moles we use the formula

N = n × L

where

N is the number of entities

n is the number of moles

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question

$n = 3.559 \times {10}^{ - 6} \: mol$

Substitute the values into the above formula and solve

That's

<h3> $N = 3.559 \times {10}^{ - 6} \times 6.02 \times {10}^{23} \\ = 2.1425 \times {10}^{18}$ </h3>

We have the final answer as

<h3> $N = 2.143 \times {10}^{18} \: \: atoms$ </h3>

Hope this helps you

3 0

4 years ago

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mo

aleksandrvk [35]

Explanation:

Relation between temperature and activation energy according to Arrhenius equation is as follows.

k = $A exp^{\frac{-E_{a}}{RT}}$

where, k = rate constant

A = pre-exponential factor

$E_{a}$ = activation energy

R = gas constant

T = temperature in kelvin

Also,

$ln (\frac{k_{2}}{k_{1}}) = (\frac{-E_{a}}{R}) \times (\frac{1}{T_{2}} - \frac{1}{T_{2}})$

$T_{1}$ = $244^{o}C$ = (244 + 273) K = 517.15 K

$T_{2}$ = $324^{o}C$ = (597.15 + 273) K = 597.15 K

$k_{1}$ = 6.7 $M^{-1} s^{-1}$ , $k_{2}$ = ?

R = 8.314 J/mol K

$E_{a}$ = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows.

$ln (\frac{k_{2}}{6.7}) = (\frac{-71000}{8.314} \times (\frac{1}{597.15} - \frac{1}{517.15})$

= 2.2123

$\frac{k_{2}}{6.7} = exp(2.2123)$

= 9.1364

$k_{2} = 61 M^{-1}s^{-1}$

Thus, we can conclude that rate constant of this reaction is $61 M^{-1}s^{-1}$ .

3 0

3 years ago