Ask question

Ask question

All categories

3 years ago

5

What prevents most small meteoroids from hitting earth surface?

1 answer:

Ostrovityanka [42]3 years ago

4 0

Gravity is the solution!. We are lucky to have Jupiter in our solar system. Jupiter's gravity is strong enough to pull meteorites towards itself.

You might be interested in

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

Masja [62]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration equilibrium constant is $K_c = 14.39$

Explanation:

The chemical equation for this decomposition of ammonia is

$2 NH_3$ ↔ $N_2 + 3 H_2$

The initial concentration of ammonia is mathematically represented a

$[NH_3] = \frac{n_1}{V_1} = \frac{29}{75}$

$[NH_3] = 0.387 \ M$

The initial concentration of nitrogen gas is mathematically represented a

$[N_2] = \frac{n_2}{V_2}$

$[N_2] = 0.173 \ M$

So looking at the equation

Initially (Before reaction)

$NH_3 = 0.387 \ M$

$N_2 = 0 \ M$

$H_2 = 0 \ M$

During reaction(this is gotten from the reaction equation )

$NH_3 = -2 x$ (this implies that it losses two moles of concentration )

$N_2 = + x$ (this implies that it gains 1 moles)

$H_2 = +3 x$ (this implies that it gains 3 moles)

Note : x denotes concentration

At equilibrium

$NH_3 = 0.387 -2x$

$N_2 = x$

$H_2 = 3 x$

Now since

$[NH_3] = 0.387 \ M$

$x= 0.387 \ M$

$H_2 = 3 * 0.173$

$H_2 = 0.519 \ M$

$NH_3 = 0.387 -2(0.173)$

$NH_3 = 0.041 \ M$

Now the equilibrium constant is

$K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}$

substituting values

$K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}$

$K_c = 14.39$

3 0

2 years ago

What is the volume of ammonia produced at 243 K at a pressure of 1.38 atm by the unbalanced reaction on the left if 5740 moles o

Neporo4naja [7]

Answer:

49671 L is the produced volume of ammonia

Explanation:

We think the reaction of ammonia 's production:

N₂(g) + 3H₂(g) → 2NH₃ (g)

We have the moles of each reactant so let's determine the limiting reactant:

Ratio is 1:3. 1 mol of nitrogen reacts with 3 moles of H₂

Then, 1720 moles of N₂ will react with (1720 .3) /1 = 5160 moles of H₂

We have 5740 moles of hydrogen, so we have enough hydrogen. This is the excess reagent, so the limiting is the N₂

1 mol of N₂ can produce 2 moles of ammonia

Therefore 1720 moles of N₂ will produce (1720 . 2) /1 = 3440 moles of NH₃

We apply now, the Ideal Gases Law → P . V = n . R .T

V = (n . R . T) /P → V = (3440 mol . 0.082 L.atm/mol.K . 243K) / 1.38 atm

V = 49671 L

We confirm that the nitrogen was the limiting reactant

3 moles of H₂ need 1 mol of nitrogen to react

Therefore, 5740 moles of H₂ will react with (5740 . 1) /3 = 1913 moles of N₂

It was ok to say, that N₂ was the limiting reactant because we need 1913 moles in the reaction, and we only have 1720 moles

6 0

3 years ago

Many plants obtain glucose through the process of ___

ryzh [129]

Answer:

vb

Explanation:

b

8 0

3 years ago

Read 2 more answers

What is the chemical name of the compound represented by the formula Ni2O3?

faust18 [17]

Answer:

Nickel(III) oxide is the chemical name of the compound represented by the formula Ni2O3?

3 0

3 years ago

A gas balloon has a volume of 106.0 liters when the temperature is 25.0 °C and the pressure is 740.0 mm Hg. What will its volume

ipn [44]

Answer : The final volume of gas will be, 103.3 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 740.0 mmHg = 98.4 kPa

Conversion used : (1 mmHg = 0.133 kPa)

$P_2$ = final pressure of gas = 99.3 kPa

$V_1$ = initial volume of gas = 106.0 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $25.0^oC=273+25.0=298K$

$T_2$ = final temperature of gas = $20.0^oC=273+20.0=293K$

Now put all the given values in the above equation, we get:

$\frac{98.4kPa\times 106.0L}{298K}=\frac{99.3kPa\times V_2}{293K}$

$V_2=103.3L$

Therefore, the final volume of gas will be, 103.3 L

8 0

3 years ago