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3 years ago

Can you quess my Hogwarts house? Please explain why you think that and I will give brainliest

Chemistry

1 answer:

vladimir1956 [14]3 years ago

8 0

Answer:

I prob can bc I'm a bad b*tch

lol

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1. If you have 8.1 g CO₂ and 2.3 g H₂, which one is the limiting reactant. Show your work. 2. Calculate the theoretical yield of

ZanzabumX [31]

Answer:

2.95 g of CH₄

Explanation:

To start this, we determine the equation:

4H₂ + CO₂ → CH₄ + 2H₂O

4 moles of hydrogen react to 1 mol of carbon dioxide in order to produce 1 mol of methane and 2 moles of water.

To determine the limiting reactant, we need to know the moles of each reactant.

8.1 g . 1 mol/ 44g = 0.184 moles of carbon dioxide

2.3 g . 1mol / 2g = 1.15 moles of hydrogen

4 moles of hydrogen react to 1 mol of CO₂

Then, 1.15 moles may react to (1.15 . 1) /4 = 0.2875 moles

We only have 0.184 moles of CO₂, so this is the limiting reactant. Not enough CO₂ to complete the 0.2875 moles that are needed.

Ratio is 1:1. 1 mol of CO₂ produces 1 mol of methane

Then, 0.184 moles of CO₂ will produce 0.184 moles of CH₄

We convert moles to mass: 0.184 mol . 16 g /mol = 2.95 g

7 0

3 years ago

What is the concentration in mass percent of a solution prepared from 50.0g nacl and 150.0g of water?

otez555 [7]

M(NaCl)=50.0 g
m(H₂O)=150.0 g

m(solution)=m(NaCl)+m(H₂O)

w(NaCl)=100m(NaCl)/m(solution)=100m(NaCl)/{m(NaCl)+m(H₂O)}

w(NaCl)=100*50.0/(50.0+150.0)=25%

3 0

4 years ago

I need help on number 2&3

AVprozaik [17]

Go to google and look up how are air bags changing for number 2. and look up new airbag technologies in google and rewrite that in a few paragraphs for number 3.

6 0

3 years ago

Read 2 more answers

Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907

melamori03 [73]

Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent. $\frac{^0 C}{m}$

m = molality of solute (m or $\frac{mol}{Kg}$ )

Given : kf = 1.86 $\frac{^0 C*Kg}{mol}$

m = 0.907 $\frac{mol}{Kg}$ )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 $\frac{^0 C*Kg}{mol}$ * 0.907 $\frac{mol}{Kg}$ )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

5 0

3 years ago

Science chemistry Question:<br>

shutvik [7]

Are you doing a quiz?

6 0

3 years ago