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Morgarella [4.7K]
3 years ago
11

Be sure to answer all parts. Consider the reaction N2(g) + 3H2(g) → 2NH3(g)ΔH o rxn = −92.6 kJ/mol If 3.0 moles of N2 react with

9.0 moles of H2 to form NH3, calculate the work done (in joules) against a pressure of 1.0 atm at 25°C. w = 15,000 J What is ΔU for this reaction? Assume the reaction goes to completion. ΔU = -240 kJ

Chemistry
1 answer:
Alex787 [66]3 years ago
6 0

Answer:

Work done in Joules = -14865.432J

ΔU = -262.94KJ

Explanation:

The steps are as shown in the attachment

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What is the name of the phase where stars live out most of their lives?
lisov135 [29]
The correct option is B. 
Stars live out most of their lives at MAIN SEQUENCE. Stars generally are divided into three major stages, these are:
1. Pro stars and pre-main sequence star
2. Main sequence  and giant star
3. Variable stars
Major stages in the life of a star can last for millions of years.

8 0
3 years ago
What pressure will be exerted by 0.675 moles of a gas at 25*C if it is in a 0.750-L container?
Westkost [7]
PV=nRT

P=nRT/V

P=[(0.650mol)(0.08206)(298K)]/(0.750L)=21.2atm



6 0
3 years ago
After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of
SIZIF [17.4K]

Answer:

b) The dehydrated sample absorbed moisture after heating

Explanation:

a) Strong initial heating caused some of the hydrate sample to splatter out.

This will result in a higher percent of water than the real one, because you assume in the calculation that the splattered sample was only water (which in not true).

b) The dehydrated sample absorbed moisture after heating.

Usually inorganic salts may absorbed moisture from the atmosphere so this will explain the 13% difference between calculated water percent the real content of water in the hydrate.

c) The amount of the hydrate sample used was too small.

It will create some errors but they do not create a difference of 13% difference as stated in the problem.

d) The crucible was not heated to constant mass before use.

Here the error is small.

e) Excess heating caused the dehydrated sample to decompose.

Usually the inorganic compounds are stable in the temperature range of this kind of experiments. If you have an organic compound which retain water molecules you may decompose the sample forming volatile compounds which will leave crucible so the error will be quite high.

6 0
3 years ago
How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield
ahrayia [7]
<span>Answer: 100 ml
</span>

<span>Explanation:


1) Convert 1.38 g of Fe₂S₃ into number of moles, n


</span>i) Formula: n = mass in grass / molar mass
<span>
ii) molar mass of </span><span>Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
</span>

iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of <span>Fe₂S₃
</span>

<span>2) Use the percent yield to calculate the theoretical amount:
</span>

<span>65% = 0.65 = actual yield/ theoretical yield =>


</span>theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol <span>Fe₂S₃</span><span>

3) Chemical equation:
</span>

<span> 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)


4) Stoichiometrical mole ratios:
</span>

<span>3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl


5) Proportionality:


</span>2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
<span>
=> x = 0.020 mol FeCl₃


6) convert 0.020 mol to volume
</span>

<span>i) Molarity formula: M = n / V
</span>

<span>ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
</span>

3 0
3 years ago
Read 2 more answers
is this information about magnesium true symbol..mg number..12 atomic mass..24.305 number of protons 12.... PLZ HELP :)
maxonik [38]
Yes, that is all correct! 

The atomic number is the same as the number of protons (12).
Magnesium = Mg
Atomic mass is 24. 305.
3 0
2 years ago
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