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3 years ago

Given the equation representing a reaction at equilibrium:

Chemistry

1 answer:

Margaret [11]3 years ago

4 0

Your answer would be 3) The equilibrium shifts to the right, and the concentration of N2 (g) decreases.

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when 3.18 g of copper (||) oxide were carefully heated in a stream of dry hydrogen, 2.54 g of copper and 0.72 g of water were fo

gavmur [86]

Answer:

the number of moles of atom is 0.91584

8 0

3 years ago

Can somebody please give me an answer for this question that doesn't consist of building or construction?

zavuch27 [327]

Answer:

to build a house you need land and people

Explanation:

4 0

3 years ago

You are burning wood to heat water for your industrial process. What is the mass of wood required to raise the temperature of 10

natta225 [31]

Answer:

18,8kg of wood

Explanation:

The energy you need to to raise the temperature of 1000 kg of water from 25.0 to 100.0 °C is:

q = C×m×ΔT

Where: q is heat, C is specific heat of water (4,184J/g°C), m is mass in grams (1000x10³g), and ΔT is 100,0°C - 25,0°C = 75,0°C

Replacing:

q = 4,184J/g°C×1000x10³g×75,0°C

q = 3,14x10⁸ J of heat are required

Now, if the heating value of dry wood is 16,72 MJ/kg = 16,72x10⁶ J/kg, mass of wood required is:

3,14x10⁸J × (1kg / 16,72x10⁶ J) = 18,8 kg of wood are required

I hope it helps!

5 0

3 years ago

Why isn't the mass of the electron included in the mass of an atom on the periodic table?

marta [7]

Its A because............

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4 years ago

Calculate the number of grams of solute needed to make each of the following solutions:

AleksAgata [21]

Explanation:

(w/w) % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.

$w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

1) 100 g of 0.500% (w/w) NaI

Mass of solution = 100 g

Mass of solute = x

Required w/w % of solution = 0.500%

$0.500\%=\frac{x}{100 g}\times 100$

$x=\frac{0.500\times 100 g}{100}=0.500 g$

0.500 grams of solute needed to make 100 g of 0.500% (w/w) NaI.

2) 250 g of 0.500% (w/w) NaBr

Mass of solution = 250 g

Mass of solute = x

Required w/w % of solution = 0.500%

$0.500\%=\frac{x}{250 g}\times 100$

$x=\frac{0.500\times 250 g}{100}=1.25 g$

1.25 grams of solute needed to make 250 g of 0.500% (w/w) NaBr

3) 500 g of 1.25% (w/w) glucose

Mass of solution = 500 g

Mass of solute = x

Required w/w % of solution = 1.25%

$1.25\%=\frac{x}{500 g}\times 100$

$x=\frac{1.25\times 500 g}{100}=6.25 g$

6.25 grams of solute needed to make 500 g of 1.25% (w/w) (glucose)

4) 750 g of 2.00% (w/w) sulfuric acid.

Mass of solution = 750 g

Mass of solute = x

Required w/w % of solution = 2.00%

$2.00\%=\frac{x}{750 g}\times 100$

$x=\frac{2.00\times 750 g}{100}=15.0 g$

15.0 grams of solute needed to make 750 g of 2.00% (w/w) sulfuric acid.

3 0

3 years ago