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S_A_V [24]
3 years ago
6

How would you classify the following events?

Chemistry
1 answer:
ICE Princess25 [194]3 years ago
8 0

Answer:  Event 1 is an example of a physical change and Event 2 is an example of a chemical change.

Explanation: Physical change is one in which there is no change in chemical composition of the substance. There is only a change in phase change.

Chemical change is a change in which there is a change in chemical composition and there might or might not be a phase change.

On Boiling, the water molecules remain bonded in the same form and only covert from liquid to gaseous form, thus is a physical change.

On Rusting of iron nail, the iron changes to iron oxide by combining with oxygen, there is a rearrangement of atoms and thus is a chemical change.

Fe+O_2+H_2O\rightarrow Fe_2O_3.xH_2O

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How does weathering change rocks and minerals
PilotLPTM [1.2K]
It changes rocks and minerals by water, ice, acids, salt, and changes in the temperature. Once the rock has been broken down a process named erosion happens, it transports bits of rocks and minerals away
3 0
3 years ago
Which statements describe band theory? Check all that apply.
Papessa [141]

Answer:

b c

Explanation:

right

8 0
3 years ago
The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
3 years ago
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kirill [66]
Well,





I guess the answer would be....


























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6 0
3 years ago
Read 2 more answers
The decomposition of carbon disulfide to carbon monosulfide and sulfur is first order with k=2.8 ×10^-7 at 1000°C .What is the h
RoseWind [281]

Answer:

2.5×10⁶ s

Explanation:

From the question given above, the following data were obtained:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

The half-life of a first order reaction is given by:

Half-life (t½) = 0.693 / Rate constant (K)

t½ = 0.693 / K

With the above formula, we can obtain the half-life of the reaction as follow:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

t½ = 0.693 / K

t½ = 0.693 / 2.8×10¯⁷

t½ = 2.5×10⁶ s

Therefore, the half-life of the reaction is 2.5×10⁶ s

6 0
2 years ago
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