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vovikov84 [41]
3 years ago
14

For the following reaction,

Chemistry
1 answer:
jeka943 years ago
5 0

Answer:

The answer to your question is:

1.- CO

2.- 0.414 moles of CO2

Explanation:

Data

                               2CO   + O2      ⇒    2CO2

CO = 0.414 moles

O2 = 0.418

 

Process

theoretical ratio   CO/O2 = 2/1 = 1

experimental ratio  CO/O2 = 0.414/0.418 = 0.99

Then the limiting reactant is CO

2.-

                    2 moles of CO ---------------  2 moles of CO2

                    0.414 moles of CO ---------  x

                   x = (0.414 x 2) / 2

                   x = 0.414 moles of CO2

                   

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A 143.1 g sample of a compound contains 53.4 g of carbon, 16.9 g of hydrogen, 43 g of nitrogen, and some amount of oxygen. what
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I'm fairly sure this is it 20.82%
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Calculate the quantity of energy required to change 8.22 mol of liquid water to steam at 100oc. the molar heat of vaporization o
Ainat [17]
Since  the water  is  at  100 degrees  then   it take  40.6 kj/mol   to  change 1  mole of  water at 100 degrees into  steam at  100  degrees
  the  moles  of water  8.22mol
since  one  mole  take 40.6kj/mol   8.22mol  will  be  =
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4 0
3 years ago
Lauryl alcohol is a nonelectrolyte obtained from coconut oil and is used to make detergents. A solution of 6.80 g of lauryl alco
natka813 [3]

Answer:

The approximate molar mass of lauryl alcohol is 174.08 g/m

Explanation:

An excersise to apply the colligative property of Freezing-point depression.

This is the formula: ΔT = Kf . m

First of all, think the T° of fusion of benzene → 5.5°C

ΔT = T° pure solvent - T° fusion solution

Kf for benzene: 5.12 °C/m

5.5°C - 4.5°C = 5.12 °C /m  . m

1°C /  5.12 m /°C = m

0.195 m = molality

This moles of lauryl alcohol, solute, are in 1 kg of benzene, solvent.

I have to find out in 0.2 kg.

1 kg sv ____ 0.195 moles solute

0.2 kg sv ____ (0.195 . 0.2)/1 = 0.039 moles solute

The mass for these moles is 6.80 g, so if I want to know the molar mass, I have to divide mass / moles

6.80 g/ 0.039 moles = 174.08 g/m

5 0
3 years ago
How many calories of heat were added to 347.9 g of water to raise its temperature from 25oC to 55oC?
saw5 [17]

Answer:

10437calories

Explanation:

The following data were obtained from the question given:

M = 347.9g

C = 4.2J/g°C

T1 = 25°C

T2 = 55°C

ΔT = 55 — 25 = 30°C

Q =?

Q = MCΔT

Q = 347.9 x 4.2 x 30

Q = 43835.4J

Converting this to calories, we obtained the following:

4.2J = 1 calorie

43835.4J = 43835.4/ 4.2 = 10437calories

5 0
3 years ago
Helppppppp! why should i use spr??
Viktor [21]

SPR can be used in order to do real time monitoring and to evaluate the advancement of compounds.

<h3>Why should we use SPR?</h3>

SPR provide information to evaluate whether or not compounds should advance to the next stage of investigation. It also provides real-time monitoring.

So we can conclude that SPR can be used in order to do real time monitoring and to evaluate the advancement of compounds.

Learn more about monitoring here: brainly.com/question/13163394

#SPJ1

6 0
2 years ago
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