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Dvinal [7]
3 years ago
7

If a small weather balloon with a volume of 58.0 L at a pressure of 1.00 atmosphere was deployed at the edge of Typhoon Odessa,

what was the volume of the balloon when it reached the center
Chemistry
1 answer:
belka [17]3 years ago
5 0

Answer:

final volume is 60.42 L

Explanation:

given data

initial volume v1 = 58 L

pressure P1 = 1 atm

we take P2 = 40 mbar

solution

we get here final volume

so p2 = 1 - 40 × 10^{-3}  = 0.96 atm

so

p1 × v1 = p2 × v2    .........1

so put here value

58 × 1 = v2 × 0.96

solve it we get

v2 = 60.42 L

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A student ran the following reaction in the laboratory at 671 K: 2NH3(g) N2(g) + 3H2(g) When she introduced 7.33×10-2 moles of N
vaieri [72.5K]

Answer:

Kc = 8.05x10⁻³

Explanation:

This is the equilibrium:

                 2NH₃(g)   ⇄     N₂(g)     +     3H₂(g)

Initially       0.0733

React         0.0733α          α/2                3/2α

Eq     0.0733 - 0.0733α    α/2                0.103

We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.

Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.

3/2α = 0.103

α = 0.103 . 2/3 ⇒ 0.0686

So, concentration in equilibrium are

NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682

N₂ = 0.0686/2 = 0.0343

So this moles, are in a volume of 1L, so they are molar concentrations.

Let's make Kc expression:

Kc= [N₂] . [H₂]³ / [NH₃]²

Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³

3 0
3 years ago
A 25.0-mL sample containing Cu2+ gave an instrument signal of 25.2 units (corrected for a blank). When exactly 0.500 mL of 0.027
irga5000 [103]

Answer:

The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.

Explanation:

The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

C = \frac{n}{V} \\0.0275 = \frac{n}{0.0005} \\

n = 1.375x10⁻⁵ mol

The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):

1.375x10⁻⁵ mol _________ 19.9 units

        x              _________  25.2 units

x = 1.741x10⁻⁵mol

Finally, we can calculate the Cu²⁺ concentration :

C = 1.741x10⁻⁵mol / 0.025 L

C = 6.964x10⁻⁴ M

7 0
3 years ago
Increasing the temperature, as noted in the graph, increases the rate of this chemical reaction. What is the relationship betwee
Soloha48 [4]

Answer: (C) Although the average kinetic energy of the colliding substances increases, this has no influence on activation energy.

Explanation:

After increasing the temperature of the reaction , the rate of the chemical reaction increases due to increase in the average kinetic energy of the particles. At increased temperature high proportions of particles can react making the reaction faster.

8 0
3 years ago
Why is hydrogen a light gas and iron is a heavy solid
lbvjy [14]
<span>Heavier atoms make denser materials</span>
8 0
3 years ago
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