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Anna [14]
3 years ago
8

. Write an equation for the combined gas law, using temperature in degrees Celsius. Explain why the Kelvin scale is convenient.

Chemistry
1 answer:
NemiM [27]3 years ago
8 0

Answer:

<em>P1V1</em><em>/</em><em>(</em><em>c1</em><em>+</em><em>2</em><em>3</em><em>7</em><em>)</em><em>=</em><em>P2V2</em><em>/</em><em>(</em><em>c2</em><em>+</em><em>2</em><em>3</em><em>7</em><em>)</em><em> </em><em>where </em><em>c</em><em>=</em><em>Celsius</em>

Explanation:

kelvin scale is alot covinent because of it absolute zero

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How many grams of NO can be produced if 204 g of NO2 is mixed with 58.1 g of H2O?
Goshia [24]

Answer:

44.4 grams of NO can be produced

Explanation:

Step 1: Data given

Mass of NO2 = 204 grams

Molar mass NO2 = 46.0 g/mol

Mass of H2O = 58.1 grams

Molar mass H2O = 18.02 g/mol

Step 2: The balanced equation

3NO2 + H2O→ 2HNO3 + NO

Step 3: Calculate moles NO2

Moles NO2 = 204 grams / 46.0 g/mol

Moles NO2 = 4.43 moles

Step 4: Calculate moles H2O

Moles H2O = 58.1 grams / 18.02 g/mol

Moles H2O = 3.22 moles

Step 5: Calculate limiting reactant

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

NO2 is the limiting reactant. It will completely be consumed (4.43 moles). H2O is in excess. there will react 4.43 /3 = 1.48 moles. There will remain 3.22 - 1.48 = 1.74 moles

Step 6: Calculate moles NO

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

For 4.43 moles NO2 we'll have 4.43/3 = 1.48 moles NO

Step 7: Calculate mass NO

Mass NO = 1.48 moles * 30.01 g/mol

Mass NO = 44.4 grams

44.4 grams of NO can be produced

3 0
3 years ago
There are many sources of air pollution in the united states and around the world. Some are natural sources, and other come from
KATRIN_1 [288]
B should be your best guess
4 0
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A squirrel running up a tree and sitting on a branch shows what kind of energy conversion
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Kinetic energy converting to gravitational potential energy
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2 years ago
Read 2 more answers
A 13.5 g sample of an aqueous solution of nitric acid contains an unknown amount of the acid. If 23.5 mL of 0.434 M barium hydro
zavuch27 [327]

The percent by mass of nitric acid in the mixture is 3.36 %

2 HNO3 + Ba(OH)2 --->Ba(NO3)2 + 2H2O

so 1 mole Ba(OH)2 neutralizes 2 mol HNO3

moles of Ba(OH)2 present = molarity of base* volume of base

= 0.229 M * 15.6 ml

= 3.5724 milli mols

mols of HNO3 neutralized = 2* mols of Ca(OH)2 used

= 2* 3.5724 milli mols = 7.1448 x10^-3 mols

molar mass of HNO3 = 63.01 g/mol

mass of HNO3 present = molar mass * mols of HNO3

= 63.01 g/mol * 7.1448 x10^-3 mols

=0.4502 g

mass of sample = 13.4 g

mass % = mass of HNO3/mass of sample * 100

= 0.4502/13.4 *100

= 3.36 %

Nitric acid is a colorless, fuming, and distinctly corrosive liquid that could be a not unusual laboratory reagent and an important commercial chemical for the manufacture of fertilizers and explosives.

Learn more about Nitric acid here: brainly.com/question/22698468

#SPJ4

3 0
1 year ago
A chemist wants to extract a solute from 100 mL of water using only 300 mL of ether. The partition coefficient between ether and
ExtremeBDS [4]

Answer:

a. 0,903

b. 0,985

c. 0,996

Explanation:

Partition coefficient between ether and water is:

K = \frac{CONCether}{CONCwater}

a. As k = 3,10, a single extraction with 300mL ether will give:

3,10 = \frac{(q/300mL)}{(1-q)/100mL}

9,30 - 9,30q = q

9,30 = 10,30q

<em>q = 0,903</em>

b. Three extractions of 100 mL will extract:

First extraction:

3,10 = \frac{(q/100mL)}{(1-q)/100mL}

3,10 - 3,10q = q

3,10 = 4,10q

<em>q = 0,756</em>

The second extraction will 0,756 times of the solute that is in water. This solute is 1-0,756=0,244

Second extraction will extract 0,244*0,756 = <em>0,184</em>

In the same way, third extraction will extraxt: (0,244-0,184)*0,756 = <em>0,045</em>

Total solute extracted:

<em>q = 0,756 + 0,184 + 0,045 = 0,985</em>

<em />

c. Six extractions with 50mL of ether extract:

First extraction:

3,10 = \frac{(q/50mL)}{(1-q)/100mL}

3,10 - 3,10q = 2q

3,10 = 5,10q

<em>0,608 = q</em>

Second extraction = (1-0,608)*0,608 = 0,234

Third extraction =(1-0,608-0,234)*0,608 = 0,096

Fourth extraction =(1-0,608-0,234-0,096)*0,608 = 0,038

Fifth extraction =(1-0,608-0,234-0,096-0,038)*0,608 = 0,015

Sixth extraction =(1-0,608-0,234-0,096-0,038-0,015)*0,608 = 0,005

q = 0,608+0,234+0,096+0,038+0,015+0,005 = <em>0,996</em>

<em />

I hope it helps!

3 0
3 years ago
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