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3 years ago

What are the smallest sub atomic structure

Chemistry

1 answer:

Anton [14]3 years ago

7 0

Answer:

Electrons

Explanation:

In an atom there would be three subatomic particles: Neutrons, electrons, protons. The smallest and lightest in terms of mass is electrons. This is because the nucleus is comprised of the protons and the neutrons, these have a greater mass than electrons as electrons has very little mass that can considered to be 0.

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What is required for an electron to move between energy levels?

Bingel [31]

It required a fixed finite amount these zones are known as energy levels

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3 years ago

Read 2 more answers

Calculate the dissociation constant of nh4oh(aq) if the degree of dissociation of 0.006 mol/kg solution is 0.053 and the activit

Anastasy [175]

The dissociation equation will be

NH4OH ---> NH4+ + OH-

Initial 0.006 0 0

Change -0.006 X 0.053 +0.006 X 0.053 -0.006 X 0.053

Equlibrium 0.006 -0.006 X 0.053 0.006 X 0.053 0.006 X 0.053

Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053

Ka = 1.78 X 10^-5

7 0

3 years ago

The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A

liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

$p^{OH}$ =14-56.17

=8.823

The volume of the $NH_{3}$ = 40.00 ml

convert into the liter= 0.040L

The value of the concentrated $NH_{3}$ =0.10 M

The volume of the $NH_{4}Cl$ = 50.00 ml

convert into the liter= 0.050L

The value of concentrated $NH_{4}Cl$ = 0.10 M

The volume of the $H_{2}So_{4}$ = 30 ml

convert into the liter= 0.030L

The value of concentrated $H_2So_4$ =0.05 M

Calculating total volume=(0.40+0.050+0.030)

=0.120 L

calculating the new concentrated value of $NH_3$ = $\frac{0.10\times 0.040}{0.120}= 0.33 \ M$

calculating the new concentrated value of $NH_4Cl$ = $\frac{0.050\times 0.10}{0.120}= 0.04166 \ M$ calculating the new concentrated value of $H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M$ when 1 mol $H_2So_4$ produced 2 mols $H^{+}$ so, 0.0125 in $H_2So_4$ produced:

$=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}$

create the ICE table:

$NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}$

I (m) 0.033(m) 0.025 0.04166

C -0.025 -0.025 + 0.025

E 8.3\times 10^{-3} 0 0.0667

now calculating pH:

when ph= 8.83:

$P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769$

5 0

3 years ago

One beaker contains 100 mL of pure water and second beaker contains 100 mL of seawater. The two beakers are left side by side on

gayaneshka [121]

Explanation:

When we add a non-volatile solute in a solvent then due to the impurity added to the solution there will occur an increase in the boiling point of the solution.

This increase in boiling point will be known as elevation in boiling point.

As one beaker contains seawater (water having NaCl) will have some impurity in it. So, more temperature is required by seawater to escape into the atmosphere.

Whereas another beaker has only pure water so it is able to easily escape into the atmosphere since, it contains no impurity.

Thus, we can conclude that level of pure water will decrease more due to non-volatile solute present in it as compared to seawater.

7 0

3 years ago

Please help~~ thank you so much

deff fn [24]

Answer:

c. 2 and 3

Explanation:

Ca(NO3)2 names calcium nitrate.

Ca metal, O, N - nonmetals, so Ca(NO3)2 is ionic compound.

It is solid and white.

4 0

3 years ago