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2 years ago

Helppp plss

Chemistry

1 answer:

wlad13 [49]2 years ago

7 0

Answer:

Earth's atmosphere contains a huge pool of nitrogen gas (N2). But this nitrogen is “unavailable” to plants, because the gaseous form cannot be used directly by plants without undergoing a transformation. To be used by plants, the N2 must be transformed through a process called nitrogen fixation.

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How many valance electrons does group #11 have?

galina1969 [7]

It would have 11 valance electrons.

Example/Explanation:

Say we are talking about groups 10. Group 10 would have 10 valance electrons because of the atom's electronic arrangement in the periodic table.

6 0

3 years ago

During photosynthesis, plant cells use sunlight as an energy source; animal cells cannot do this. What substance do Animal cells

8090 [49]

Answer:

Chlorophyll

Explanation:

Photosynthesis is the process through which plant cells use carbon dioxide & water to create oxygen & energy rich organic compounds. They do so by converting sunlight ie light energy into chemical energy. This is possible due to presence of chlorophyll in them. Animals can't do so, as they dont have chlorophyll.

6 0

2 years ago

The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe

Irina-Kira [14]

Answer:

$t=147.24\ min$

Explanation:

Given that:

Half life = 30 min

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30}\ min^{-1}$

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration $[A_0]$ = 7.50 mg

Final concentration $[A_t]$ = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

$0.25=7.50e^{-0.0231\times t}$

$750e^{-0.0231t}=25$

$x=\frac{\ln \left(30\right)}{0.0231}$

$t=147.24\ min$

6 0

3 years ago

Please help!!! will give brainliest!!! 30 points! (need to put roman numeral in answer)

mariarad [96]

1. Nickel (II) Bromide

2. Iron (II) Oxide

3. Iron (III) Oxide

4. Tin (IV) Chloride

5. Lead (IV) tetrachloride

6. Tin (II) Bromide

7. Chromium (III) Phosphide

8. Iron (II) Fluoride

9. Gold (III) Chloride

I hope this helps. I'm more than 100% sure that all the answers except for number 7 are correct. I knew all of them off the top of my head except for this one. I hope the other answer has the correct answer for that one. Good luck and have a great day.

3 0

3 years ago

What volume of a 0.232 M hydrobromic acid solution is required to neutralize 10.8 mL of a 0.162 M potassium hydroxide solution?

Naily [24]

Answer:

156 mL hydrobromic acid

Explanation:

3 0

3 years ago