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3 years ago

In which sample are the particles arranged in a regular geometric pattern?

Chemistry

1 answer:

andrew11 [14]3 years ago

7 0

There is no picture hunny. Please add a picture please boo.

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Inside an atom, which of these occupies the maximum volume? a) the electrons b) the protons c) the neutrons d) empty space (vacu

icang [17]

Answer: Electrons

Explanation:The electron itself is small but it takes space as much as an atom by circling around the nucleus.

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2 years ago

Soap is made of molecules with atoms that share electrons equally at one end and atoms that have a slight charge at the other en

kicyunya [14]

Which could be soluble in soap?

Answer: Out of all the options presented above the one that represents which substance is soluble in soap is answer choice C) both because soap is part polar and part nonpolar.

I hope it helps, Regards.

6 0

2 years ago

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

$x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L$

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0

3 years ago

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Lera25 [3.4K]

Answer:

NH3

Explanation:

2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)

So for two moles of NH3 we need one mole of CO2. So let's count moles for each reagent.

n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol

n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol

From equation we have to divide n(NH3) by 2 because we need two equivalent per one CO2. That will be 3984.145. So the limiting agent is NH3 because it's not enough of it to react with all CO2

4 0

3 years ago

This is hard, isn't it? please help me though. PLEASE

Liono4ka [1.6K]

I think it might be A. i am not totally sure though

6 0

2 years ago