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tino4ka555 [31]
3 years ago
10

How many GRAMS are in 3.78 mol of NaOH? A. 150 g/mol B. 151.12 g/mol C. 151.1 g/mol

Chemistry
1 answer:
Rzqust [24]3 years ago
3 0

Answer:

C.151.1 g/mol

Cause there are 40g in a mole of NaOH, and 40*3.78 is 151.1

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A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibriu
tatyana61 [14]

Answer:

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

Explanation:

The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.

                  2NO(g)         +    2H2(g)    ⇒        N2(g)      +         2H2O(g)

i  mol            0.10                   0.050                                             0.10

c mol            -0.038                -0.038                +0019                +0.038                                                

e mol            0.062                 0.012                  00.019               0.057

Since the volume of the vessel is 1.0 L, the concentrations in molarity are:

[NO] =   0.062 M

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

5 0
3 years ago
10 points. Please help.
ratelena [41]

Answer:

-191.7°C

Explanation:

P . V = n . R . T

That's the Ideal Gases Law. It can be useful to solve the question.

We replace data:

2.5 atm . 8 L = 3 mol . 0.082 L.atm/mol.K . T°

(2.5 atm . 8 L) / (3 mol . 0.082 L.atm/mol.K) = T°

T° = 81.3 K

We convert T° from K to C°

81.3K - 273 = -191.7°C

6 0
3 years ago
Read 2 more answers
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