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mixas84 [53]
3 years ago
13

In the reaction N2(g) + O2(g) 2NO(g), how will increasing pressure on the system affect the equilibrium?

Chemistry
2 answers:
marissa [1.9K]3 years ago
7 0

Nothing will happen.

Since there are equal moles of gas on both sides of the equation, increasing the pressure will have no effect.

Leno4ka [110]3 years ago
4 0
The pressure increase does not affect the equilibrium shift reaction.
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Density of a piece of wood that has a measurement of 24 cm3 and 768 g
Brut [27]

Answer:

<h3>The answer is 32 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 768 g

volume = 24 cm³

We have

density =  \frac{768}{24}  \\

We have the final answer as

<h3>32 g/cm³</h3>

Hope this helps you

4 0
3 years ago
Read 2 more answers
If excess ammonium sulfate reacts with 20.0g of calcium hydroxide how many grams of ammonia are produced
bagirrra123 [75]
The reaction between ammonium sulfate and calcium hydroxide is given below. 
            (NH₄)₂SO₄ + Ca(OH)₂ --> 2NH₃ + CaSO₄ + 2H₂O
From the balance equation, we can conclude that every 74 g of calcium sulfate reacted with enough amount of ammonium sulfate will yield 34 grams of ammonia. From the given amount,
               (20 g calcium sulfate) x (34 grams ammonia / 74 g calcium sulfate)
                               = <em>9.19 g ammonia</em>
3 0
3 years ago
The reactant concentration in a second-order reaction was 0.560 m after 115 s and 3.30×10−2 m after 735 s . what is the rate con
Aloiza [94]
Use the formula for second order reaction:

\frac{1}{C} = \frac{1}{C_0} + kt

C = concentration at time t 
C0 =  initial conc.
k = rate constant
t = time

1st equation :   \frac{1}{0.56} = \frac{1}{C_0} + 115k

2nd Equation: \frac{1}{0.033} = \frac{1}{C_0} + 735k

Find \frac{1}{C_0} from 1st equation and put it in 2nd equation:

\frac{1}{0.033} = \frac{1}{0.56} - 115k + 735k

\frac{1}{0.033} - \frac{1}{0.56} = 620k

k = 0.046
3 0
3 years ago
Calculate pH value in a solution that has a<br> concentration of 0.04 mol / dm^3 of H30+ ions?
Sergio039 [100]

Answer:288229262427020202625369201010910000000000000

Explanation: bcuz I said

4 0
3 years ago
How much water would I need to add to 700 mL of a 2.7 M KCl solution to make a 1.0 M solution?
geniusboy [140]

Answer:

1190\ \text{mL}

Explanation:

M_1 = Initial Concentration of KCl = 2.7 M

V_1 = Volume of KCl = 1 M

M_2 = Final concentration of KCl = 1 M

V_2 = Amount of water

We have the relation

M_1V_1=M_2V_2\\\Rightarrow V_2=\dfrac{M_1V_1}{M_2}\\\Rightarrow V_2=\dfrac{2.7\times 700}{1}\\\Rightarrow V_2=1890\ \text{mL}

The amount of water that is to be added is 1890-700=1190\ \text{mL}.

3 0
2 years ago
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