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3 years ago

Ice bomb

Chemistry

1 answer:

Ahat [919]3 years ago

5 0

Answer:

Formation of Gas

Explanation:

The solid Carbon Dioxide warms it sublimates to gas. With limited room for the gas to expand, the pressure in the bottle increases.

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When a forest burns and all that's left is some ash,

dalvyx [7]

Answer:

C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + H₂O

Explanation:

When a forest burns and all that's left is some ash, most of the mass of the trees go to the atmosphere, in the form of carbon dioxide.

In a total combustion process of organic matter,<em> the two molecules produced are carbon dioxide and water. </em>(CO₂ and H₂O)

The equation for the combustion of glucose is:

C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + H₂O

8 0

3 years ago

Read 2 more answers

What is the correct formula that would result from the combination of the two ionic species?

DerKrebs [107]

Answer:

2K+ + NO-2 -> K2NO

Explanation:

Message me for extra explanation.

snap- parkguy786

5 0

3 years ago

A self-contained living thing

sleet_krkn [62]

The correct answer Organism

3 0

3 years ago

Under certain conditions the rate of this reaction is zero order in dinitrogen monoxide with a rate constant of ·0.0047Ms−1: 2N2

leva [86]

Answer:

250 mmol are left after 26.60 secs ≅ 26.6 seconds

Explanation:

zero order kinetics formula is:

[A] = [A₀] - kt

where [A] = amount left; [A₀] = amount remaining; k = rate constant; t = time

concentration [A₀] = mole/volume = 0.5 mole/2.0L = 0.25

[A] = 0.25 mole/2.0L = 0.125 M

k = 0.0047 Ms⁻¹

t = {[A]-[A₀]}/-k = (0.125 - 0.25)/(-0.0047) = 26.60 seconds = 26.6 seconds

the amount is reduced by half after 26.6 seconds

6 0

3 years ago

Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

Temka [501]

The question is incomplete , complete question is:

Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

(a) n = 2 to n = 4

(b) n = 2 to n = 1

(d) n = 4 to n = 3

Answer:

Hence the order of the transition will be : d < a < c < b

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

Energy of n = 1 in an hydrogen atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = 2 in an hydrogen atom:

$E_2=-13.6\times \frac{1^2}{2^2}eV=-3.40eV$

Energy of n = 3 in an hydrogen atom:

$E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV$

Energy of n = 4 in an hydrogen atom:

$E_4=-13.6\times \frac{1^2}{4^2}eV=-0.85 eV$

Energy of n = 5 in an hydrogen atom:

$E_5=-13.6\times \frac{1^2}{5^2}eV=-0.544 eV$

a) n = 2 to n = 4 (absorption)

$\Delta E_1= E_4-E_2=-0.85eV-(-3.40eV)=2.55 eV$

b) n = 2 to n = 1 (emission)

$\Delta E_2= E_1-E_2=-13.6 eV-(-3.40eV)=-10.2 eV$

Negative sign indicates that emission will take place.

c) n = 2 to n = 5 (absorption)

$\Delta E_3= E_5-E_2=-0.544 eV-(-3.40eV)=2.856 eV$

d) n = 4 to n = 3 (emission)

$\Delta E_4= E_3-E_4=-1.51 eV-(-0.85 eV)=-0.66 eV$

Negative sign indicates that emission will take place.

According to Planck's equation, higher the frequency of the wave higher will be the energy:

$E=h\nu$

h = Planck's constant

$\nu$ frequency of the wave

So, the increasing order of magnitude of the energy difference :

$E_4$

And so will be the increasing order of the frequency of the of the photon absorbed or emitted. Hence the order of the transition will be :

: d < a < c < b

7 0

4 years ago