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seraphim [82]
3 years ago
5

If we added heat (Q) to the, and we knew this resulted in an increase of internal energy (U) of + 15, what would the change in w

ork (W) have to be in order for the equation to balance.
Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
6 0

The question is incomplete, here is the complete question:

If we added 10 J of heat (Q) to the system, and we knew this resulted in an increase of internal energy (U) of +15 J, what would the change in work have to be in order for the equation to balance.

<u>Answer:</u> The work done for compression is +5 J

<u>Explanation:</u>

According to first law of thermodynamics:

dU=q+W

where,

q = heat released or absorbed = +10 J  

When heat is added, it is getting absorbed and sign will be positive and when heat is removed, it is getting released and sign will be negative

dU = internal energy = +15 J

W = work done

Putting values in above equation, we get:

15=10+W\\\\W=+5J

<u>Sign convention of work: </u>

Work done for expansion process is taken as negative and work done for compression is taken as positive.

Hence, the work done for compression is +5 J

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Calculate the value of the equilibrium constant, Kc , for the equilibrium shown below, if 0.124 moles of NO, 0.0240 mole of H2,
sdas [7]

the reaction is

2NO(g) + 2H2(g) <—> N2(g) + 2H2O (g)

Kc = [N2] [ H2O]^2 / [NO]^2 [ H2]^2

Given

moles of NO = 0.124 therefore [NO] = moles /volume = 0.124 /2 = 0.062

moles of H2 = 0.0240 , therefore [H2] = moles / volume = 0.0240 / 2 = 0.012

moles of N2 = 0.0380 , therefore [N2] = moles / volume = 0.0380 / 2 = 0.019

moles of H2O  = 0.0276 , therefore [H2O] = moles / volume = 0.0276 / 2 = 0.0138

Kc = (0.019) ( 0.0138)^2 / (0.062)^2 ( 0.012)^2 = 6.54



4 0
3 years ago
Calculate the mass percent (m/m) of a solution prepared by dissolving 45.09 g of NaCl in 174.9 g of H2O.
inysia [295]

Answer:

20.3 %  NaCl

Explanation:

Given data:

Mass of solute = 45.09 g

Mass of solvent = 174.9 g

Mass percent of solution = ?

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Mass of solution = 45.09 g + 174.9 g

Mass of solution = 220 g

The solute in 220 g is 45.09 g

220 g = 2.22 × 45.09

In 100 g solution amount of solute:

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5 0
3 years ago
Balance the following Equation:
pashok25 [27]

Answer:

HCl

Explanation:

Given data:

Mass of Zn = 50 g

Mass of HCl = 50 g

Limiting reactant = ?

Solution:

Chemical equation:

Zn + 2HCl      →     ZnCl₂ + H₂

Number of moles of Zn:

Number of moles = mass / molar mass

Number of moles = 50 g/ 65.38 g/mol

Number of moles = 0.76 mol

Number of moles of HCl:

Number of moles = mass / molar mass

Number of moles = 50 g/ 36.5 g/mol

Number of moles = 1.4 mol

Now we will compare the moles of Reactant with product.

                 Zn         :          ZnCl₂

                  1           :             1

                 0.76     :           0.76

                Zn         :             H₂

                  1           :             1

                 0.76     :           0.76

               HCl         :          ZnCl₂

                  2           :             1

                 1.4         :           1/2×1.4 = 0.7

                HCl         :             H₂

                  2           :             1

                 1.4         :           1/2×1.4 = 0.7

Less number of moles of product are formed by HCl it will act limiting reactant.

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