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Hunter-Best [27]
2 years ago
12

How is area and volume of cuboids calculated?​

Chemistry
1 answer:
julia-pushkina [17]2 years ago
8 0

Answer:

A = Base x Height, Volume = Length x Width x Height

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PLEASE HELP ME!!! ASAP
shtirl [24]

Answer:

Theoretical yield of the reaction = 34 g

Excess reactant is hydrogen

Limiting reactant is nitrogen

Explanation:

Given there is 100 g of nitrogen and 100 g of hydrogen

Number of moles of nitrogen = 100 ÷ 28 = 3·57

Number of moles of hydrogen = 100 ÷ 2 = 50

Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation

N2 + 3H2 → 2NH3

From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed

Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen

⇒ We require 10·71 moles of hydrogen

But we have 50 moles of hydrogen

∴ Limiting reactant is nitrogen and excess reactant is hydrogen

From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia

Molecular weight of ammonia = 17 g

∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g

5 0
3 years ago
Help me with this question please
Natasha_Volkova [10]

Answer:

I dont know sorry i will try my best htough

Explanation:

7 0
2 years ago
Calculate the standard molar enthalpy for the complete combustion of liquid ethanol (C2H5OH) using the standard enthalpies of fo
PSYCHO15rus [73]

Answer:

Explanation:

For the reaction

C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O

We can calculate the  standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:

ΔHºc =  2ΔHºf CO2 (g) + 3ΔHºfH2O(l)  - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )

( we were not give the water state but we know we are at standard conditions so it is in its liquid state )

The ΔHºfs can be found in appropiate reference or texts.

ΔHºc =  2ΔHºf CO2 (g)+ 3ΔHºfH2O(l)  - (  ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )

= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [(  -276.2 + 0 ) ] kJ

ΔHºc =  -1368.33 kJ

5 0
3 years ago
What can you do differently for
drek231 [11]

Answer:

I would say, what helps me is really paying attention in class and asking questions, also making sure you study for upcoming test's and quizzes and completely assingments on time

Explanation:

3 0
2 years ago
During nuclear fission, great amounts of energy are produced from____. A. particle accelerators B. tremendous amounts of mass C.
Phoenix [80]
During nuclear fission, great amounts of energy are produced from____.

Answer: Out of all the options presented above, the one that completes the statement and makes it true, is answer choice D) very small amounts of mass. Nuclear fission power does not produce carbon and also produces 12% of the world's power.

I hope it helps, Regards.
3 0
3 years ago
Read 2 more answers
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