How does a chemist know how much is "right" in a chemical reaction?

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Dovator [93]

Answer:

5 × 10^-4 L

Explanation:

The equation of the reaction is;

2KClO3 = 2KCl + 3O2

Number of moles of KClO3 = 13.5g/122.5 g / mol = 0.11 moles

From the stoichiometry of the reaction;

2 moles of KClO3 yields 3 moles of O2

0.11 moles of KClO3 yields 0.11 × 3/2 = 0.165 moles of oxygen gas

From the ideal gas equation;

PV= nRT

P= 85.4 × 10^4 KPa

V=?

n= 0.165

R= 8.314 J K-1 mol-1

T= 40+273 = 313K

V= 0.165 ×8.134 × 313/85.4 × 10^4

V=429.4/85.4 × 10^4

V= 5 × 10^-4 L

3 0

3 years ago

NEED HELP PLEASE!!!!

VladimirAG [237]

When you first pull back on the pendulum, and when you pull it back really high the Potential Energy is high and the Kinetic Energy is low, But when up let go, and it gets right around the middle, that's when the Potential energy transfers to Kinetic, at that point the kinetic Energy is high and the potential Energy is low. But when it comes back up at the end. The same thing will happen, the Potential Energy is high, and the Kinetic Energy is low. Through all of that the Mechanical Energy stays the same.

I hope this helps. :)

4 0

3 years ago

Where was the first oil well drilled

zepelin [54]

Answer:

First oil well in the United States, built in 1859 by Edwin L. Drake, Titusville, Pennsylvania.

7 0

3 years ago

A 4.86-gram sample of calcium reacted completely with oxygen to form 6.80 grams of calcium oxide. This reaction is represented b

sesenic [268]

The number of mole of Ca reacted is:
4.86 g Ca/ (40.08 g/mol Ca)= 0.121 mol Ca

Because Ca reacted completely with oxygen and there is 2 mol Ca, there is 1 mol O2 reacted.

Total mass of oxygen that reacted is:
0.121 mol Ca* (1mol O2/ 2 mol Ca)* (32 g O2/ 1 mol O2)= 1.94 g O2 reacted.

Hope this would help~

8 0

3 years ago

An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?

umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

$K_b$ = molal boiling point elevation constant = 0.51°C/m

$m_{solute}$ = Given mass of solute (CsCl) = 8.00 g

$M_{solute}$ = Molar mass of solute (CsCl) = 168.4 g/mol

$W_{solvent}$ = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC$

Hence, the boiling point of solution is 100.53

6 0

3 years ago