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3 years ago

6

How are a sodium atom and a sodium ion alike?

1 answer:

grigory [225]3 years ago

3 0

They are not allike as na atom has 11 elctrons and Na ion has 10 electons. <span>They both have the same nucleus (same number of protons and neutrons). The only difference is that the ion has one less electron than the atom (in this case).</span>

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Answer:

Particle energy

Explanation:

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3 years ago

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Niels Bohr’s model of the atom helped to explain

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Answer:

a theory for the hydrogen atom based on quantum theory that energy is transferred only in certain well defined quantities. Electrons should move around the nucleus but only in prescribed orbits.

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3 years ago

A helium balloon has a volume of 735 mL when it is at ground level. The balloon is

bagirrra123 [75]

Answer:

The answer to your question is P1 = 1.4 atm

Explanation:

Data

Volume 1 = V1 = 735 ml

Pressure 1 = P1 = ?

Volume 2 = V2 = 1286 ml

Pressure 2 = P2 = 0.8 atm

Process

To solve this problem use Boyle's law.

P1V1 = P2V2

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3 years ago

Balancing chemical equation:

tino4ka555 [31]

2Na + MgF2 → 2NaF + Mg

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3 years ago

Calculate either [H3O+] or [OH−] for each of the solutions at 25 °C. Solution A: [OH−]=1.55×10−7 M Solution A: [H3O+]= M Solutio

Ahat [919]

Answer: A. $[H_3O^+]=0.64\times 10^{-7}M$

B. $[OH^-]=0.11\times 10^{-5}M$

C. $[OH^-]=\frac{10^{-14}}{0.000775}=1.3\times 10^{-11}M$

Thus solution B is basic in nature.

Explanation:

pH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

pH for acidic solutions is less than 7, for basic solutions it is more than 7 and for neutral solutions it is equal to 7.

$pH=-\log [H^+]$

$pOH=-log[OH^-]$

$pH+pOH=14$

or $[H^+][OH^-]=10^{-14}$

A. $[OH^-]=1.55\times 10^{-7}M$

$[H_3O^+]=\frac{10^{-14}}{1.55\times 10^{-7}}=0.64\times 10^{-7}M$

$pH=-log[H_3O^+]=-log[0.64\times 10^{-7}]=7$

B. $[H_3O^+]=9.43\times 10^{-9}M$

$[OH^-]=\frac{10^{-14}}{9.43\times 10^{-9}}=0.11\times 10^{-5}M$

$pH=-log[H_3O^+]=-log[9.43\times 10^{-9}]=8$

C. $[H_3O^+]=0.000775M$

$[OH^-]=\frac{10^{-14}}{0.000775}=1.3\times 10^{-11}M$

$pH=-log[H_3O^+]=-log[0.000775]=3$

Thus solution B is basic.

4 0

4 years ago