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3 years ago

12. Electrons in the first energy level have the least amount of energy.

Chemistry

1 answer:

ludmilkaskok [199]3 years ago

6 0

Answer:

True

Explanation:

Electrons that are in the first energy level (energy level 1) are closest to the nucleus and will have the lowest energy. Electrons further away from the nucleus will have higher energy. An atom's electron shell can accommodate 2n2 electrons, where n is the energy level.

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I don't know how to do this, pls help!

Aliun [14]

Cations from smallest to largest
Li⁺ ,Na⁺, K⁺ (from Periodic Table, the bigger number of period, the bigger size, of atom, so the bigger size of cation)

1) LiF smaller cation then KF
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The lattice energy increases as cations get smaller, as shown by LiF and KF.
I think this one should be correct answer, because the compared substances have also the same anion, and we can compare cations in them.

2) The same cation Li , so wrong statement.

3)The same cation Na , so wrong statement.

4) NaCl smaller cation then KF
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3 years ago

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Why is green light cool

shepuryov [24]

Because

green insinuates a feeling of moving forward and gain confidence and comfort by the color

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3 years ago

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If 25 mL of a HCl solution of unknown concentration was neutralized with 10 mL of a 0.30 M NaOH solution, what was the original

QveST [7]

Answer:

0.12 M

Explanation:

Step 1: Write the balanced equation

NaOH + HCl ⇒ NaCl + H₂O

Step 2: Calculate the reacting moles of NaOH

10 mL of a 0.30 M NaOH solution react.

$0.010L \times \frac{0.30mol}{L} = 3.0 \times 10^{-3} mol$

Step 3: Calculate the reacting moles of HCl

The molar ratio of NaOH to HCl is 1:1. The reacting moles of HCl are 1/1 × 3.0 × 10⁻³ mol = 3.0 × 10⁻³ mol.

Step 4: Calculate the concentration of HCl

3.0 × 10⁻³ mol of HCl are in 25 mL of solution.

$M = \frac{3.0 \times 10^{-3} mol}{0.025L} = 0.12 M$

5 0

3 years ago

21.If the anion in an ionic compound is a nonmetal, you must change its ending

True [87]

Answer:

-ite --> Anionite

Explanation:

Feel free to give brainliest.

Have an awesome day!

7 0

3 years ago

For the reaction IO3–(aq) + 5 I–(aq) + 6 H+(aq) → 3 I2(aq) + 3 H2O(l) the rate of disappearance of I–(aq) at a particular time a

scoray [572]

Answer: The rate of appearance of $I_2(aq)$ is $1.44\times 10^{-3}mol/Ls$

Explanation:

Rate of a reaction is defined as the rate of change of concentration per unit time.

Thus for reaction:

$IO_3^-(aq)+5I^-(aq)+6H^+(aq)\rightarrow 3I_2(aq)+3H_2O(l)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{1d[I^-]}{5dt}=+\frac{d[I_2]}{3dt}$

Given: $-\frac{d[I^-]}{dt}]$ = $2.4\times 10^{-3}mol/Ls$

$+\frac{d[I_2]}{dt}=-\frac{3d[I^-]}{5dt}=-\frac{3}{5}\times 2.4\times 10^{-3}mol/Ls=1.44\times 10^{-3}mol/Ls$

The rate of appearance of $I_2(aq)$ is $1.44\times 10^{-3}mol/Ls$

6 0

3 years ago