Where do you find the reactants in a chemical equation

Chemistry

1 answer:

Alexxandr [17]2 years ago

5 0

On the left side of the equation is the reactants (starting side) and the right side (end side) is the products.

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A 75.0 g sample of dinitrogen monoxide is confined in a 3.q L vessel. What is the pressure( in atm) at 115 celsius

Nonamiya [84]

Data Given:
Pressure = P = ?

Volume = V = 3.0 L

Temperature = T = 115 °C + 273 = 388 K

Mass = m = 75.0 g

M.mass = M = 44 g/mol

Solution:
Let suppose the Gas is acting Ideally. Then According to Ideal Gas Equation,
P V = n R T
Solving for P,
P = n R T / V ------ (1)
Calculating Moles,
n = m / M

n = 75.0 g / 44 g.mol⁻¹

n = 1.704 mol

Putting Values in Eq. 1,

P = (1.704 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 388 K) ÷ 3.0 L

P = 18.08 atm

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in the event of a submarine disaster the___ can be used to rescue people from the depth up to 3500 feet.

gulaghasi [49]

DSVR because it only makes sense. Try it out and if it works give me a Brainly

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Read 2 more answers

Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500.

patriot [66]

Explanation:

Chemical reaction equation for the give decomposition of $NH_{3}$ is as follows:.

$2NH_{3}(g) \rightleftharpoons N_{2}(g) + 3H_{2}(g)$

And, initially only $NH_{3}$ is present.

The given data is as follows.

$P_{NH_{3}}$ = 2.3 atm at equilibrium

$P_{H_{2}} = 3 \times P_{N_{2}}$ = 0.69 atm

Therefore,

$P_{N_{2}} = \frac{0.69 atm}{3}$

= 0.23 aatm

So, $P_{NH_{3}}$ = 2.3 - 2(0.23)

= 1.84 atm

Now, expression for $K_{p}$ will be as follows.

$K_{p} = \frac{(P_{N_{2}})(P^{3}_{H_{2}})}{(P^{2}_{NH_{3}})}$

$K_{p} = \frac{(0.23) \times (0.69)^{3}}{(1.84)^{2}}$

= $\frac{0.23 \times 0.33}{3.39}$

= 0.0224

or, $K_{p} = 2.2 \times 10^{-2}$

Thus, we can conclude that the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture is $2.2 \times 10^{-2}$ .