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yKpoI14uk [10]
3 years ago
6

What are the units of molar mass? A. L/g B. mol/g C. g/L D. g/mol SUBST

Chemistry
2 answers:
rosijanka [135]3 years ago
7 0

Answer:

D - g/mol

step by step method

den301095 [7]3 years ago
5 0
D
I think its the best answer
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As a cooking method, braising is valued for its A. ability to retain flavor. B. quickness and convenience. C. simplicity and ver
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Answer:

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Explanation:

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Cl2 + 2OH− → Cl− + ClO− + H2O
Nataly_w [17]
Cl2(g) -------> Cl-(aq) + ClO-(aq) 

2e- + Cl2(g) -------> 2Cl-(aq) [reduction] 

4OH-(aq) + Cl2(g) -----------> 2ClO-(aq) + 2H2O(l) + 2e- [oxidation] 
______________________________________... 
2OH-(aq) + Cl2(g) --------> Cl-(aq) + ClO-(aq) + H2O(l)
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How do catalysts increase the rate of a chemical reaction?
bogdanovich [222]
A, They Lower The Activation Energy.
7 0
3 years ago
Read 2 more answers
23.5 L of h2 is stored at a pressure of 58.7 Kpa what volume would the gas take up at stp
stepan [7]

Answer:-  13.6 L

Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.

Standard pressure is 1 atm that is 101.325 Kpa.

Boyle's law equation is:

P_1V_1=P_2V_2

From given information:-

P_1 = 58.7 Kpa

V_1 = 23.5 L

P_2 = 101.325 Kpa

V_2 = ?

Let's plug in the values and solve it for final volume.

58.7Kpa*23.5L=101.325Kpa*V_2

On rearranging the equation for V_2

V_2=\frac{58.7Kpa*23.5L}{101.325Kpa}

V_2 = 13.6 L

So, the volume of hydrogen gas at STP for the given information is 13.6 L.

3 0
3 years ago
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
2 years ago
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