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3 years ago

DESPERATE WILL GIVE BRAINLIST AND THANKS

Chemistry

1 answer:

Annette [7]3 years ago

8 0

Answer:

20 grams

Explanation:

40g= 300

x = 150

cross multiply, 40*150/300

6000/300= 20g

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EVIDENCE NOTEBOOK

BARSIC [14]

Answer:

Your body temperature would decrease

Explanation:

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3 years ago

N2 + 3H2 mc011-1.jpg 2NH3 What is the mole ratio of hydrogen to ammonia?

hjlf

1 N₂ + 3 H₂ = 2 NH₃

Mole ratio of hydrogen to ammonia :

3 moles H₂ : 2 moles NH₃ or 3 : 2

hope this helps!

3 0

3 years ago

Read 2 more answers

A bicyclist travels the first 800 m of a trip with a speed of 10 m/s, the next 500 m with an

bezimeni [28]

Answer:9.18 m/s

Explanation:

The average speed for the entire trip is found ......total distance/total time. Remember r*t=d, so divide both sides by t and get r = d/t.

So the cyclist went 800+500+1200 m = 2500 m for total distance.

10t =800 leads to t=80 sec

5t=500 leads to t=100 sec

13t = 1200 leads to t = 92.3 sec

total time is 272.3 sec

Average speed for the entire trip is 2500 / 272.3 = 9.18 m/s

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3 years ago

How does the sun transfer heat energy to various materials?

BARSIC [14]

I believe the answer is C.

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3 years ago

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The Haber Process synthesizes ammonia at elevated temperatures and pressures. Suppose you combine 1580 L of nitrogen gas and 351

ikadub [295]

Answer : The volume of reactant measured at STP left over is 409.9 L

Explanation :

First we have to calculate the moles of $N_2$ and $H_2$ by using ideal gas equation.

For $N_2$ :

$PV_{N_2}=n_{N_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $N_2$ gas = 1580 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 1580L=n_{N_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{N_2}=70.49mole$

For $H_2$ :

$PV_{H_2}=n_{H_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $H_2$ gas = 3510 L

n = number of moles $H_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 3510L=n_{H_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{H_2}=156.6mole$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the balanced reaction we conclude that

As, 3 mole of $H_2$ react with 1 mole of $N_2$

So, 156.6 moles of $H_2$ react with $\frac{156.6}{3}\times 1=52.2$ moles of $N_2$

From this we conclude that, $N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of $N_2$ reactant (unreacted gas).

Excess moles of $N_2$ reactant = 70.49 - 52.2 = 18.29 moles

Now we have to calculate the volume of reactant, measured at STP, is left over.

$PV=nRT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of gas = ?

n = number of moles of unreacted gas = 18.29 moles

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times V=18.29mole\times (0.0821L.atm/mol.K)\times 273K$

$V=409.9L$

Therefore, the volume of reactant measured at STP left over is 409.9 L

8 0

3 years ago