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deff fn [24]
3 years ago
12

How would you calculate the molar mass of 2AgCl?

Chemistry
1 answer:
Rashid [163]3 years ago
3 0
2 × (atomic mass of Ag) + (atomic mass of Cl (
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8. 00 g of ethane gas, c2h6, is burned in oxygen. What volume of carbon dioxide gas is produced at 1. 00 atm and 25. 0°c?.
Sonbull [250]

Answer:

Explanation:1 g CH

4

​

,C

2

​

H

6

​

C

3

​

H

8

​

C

4

​

H

10

​

T = 350 K, P = 1 atm

PV=nRT=

M

w×R×T

​

1×V=

58

1

​

×0.0821×350

V=0.495 L

V=495cm

3

8 0
1 year ago
4. How is a lead storage battery recharged?
alekssr [168]
The answer is c Yep Allll day
6 0
2 years ago
Read 2 more answers
Plsplsplsplsplsplsplsplsplspls help
MA_775_DIABLO [31]

Answer:

this is difficult but simple to answer

Explanation:

all atoms move in 1 direction no more than 2

6 0
3 years ago
Create the Equation: What is the Percent Yield of Ammonia (NH3) if 11.8 g is recovered in a reaction with 7.02 x 10^23 molecules
insens350 [35]

Answer:

Explanation:

The first thing that you need to do here is to calculate the theoretical yield of the reaction, i.e. what you get if the reaction has a

100

%

yield.

The balanced chemical equation

N

2

(

g

)

+

3

H

2

(

g

)

→

2

NH

3

(

g

)

tells you that every

1

mole of nitrogen gas that takes part in the reaction will consume

3

moles of hydrogen gas and produce

1

mole of ammonia.

In your case, you know that

1

mole of nitrogen gas reacts with

1

mole of hydrogen gas. Since you don't have enough hydrogen gas to ensure that all the moles of nitrogen gas can react

what you need

3 moles H (sub 2)

>

what you have

1 mole H (sub2)

you can say that hydrogen gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of nitrogen gas will get the chance to take part in the reaction.

So, the reaction will consume

1

mole of hydrogen gas and produce

1

mole H

2

⋅

2 moles NH

3

3

moles H

2

=

0.667 moles NH

3

at

100

%

yield. This represents the reaction's theoretical yield.

Now, you know that the reaction produced

0.50

moles of ammonia. This represents the reaction's actual yield.

In order to find the percent yield, you need to figure out how many moles of ammonia are actually produced for every

100

moles of ammonia that could theoretically be produced.

You know that

0.667

moles will produce

0.50

moles, so you can say that

100

moles NH

3

.

in theory

⋅

0.50 moles NH

3

.

actual

0.667

moles NH

3

.

in theory

=

75 moles NH

3

.

actual

Therefore, you can say that the reaction has a percent yield equal to

% yield = 75%

−−−−−−−−−−−−−

or 75 moles NH sub3

I'll leave the answer rounded to two sig figs.

5 0
3 years ago
How many grams of NH3 can be produced from 12.0g of H2?
RSB [31]

Answer:

Balanced reaction:

3 H2 (g)  + N2 (g)  → 2 NH3 (g)

Use stoichiometry to convert g of H2 to g of NH3.  The process would be:

g H2 → mol H2 → mol NH3 → g NH3

12.0 g H2 x (1 mol H2 / 2.02 g H2) x (2 mol NH3 / 3 mol H2) x (17.03 g NH3 / 1 mol NH3) = 67.4 g NH3

Explanation: See above

Hope this helps, friend.

8 0
2 years ago
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