Ask question

Ask question

All categories

4 years ago

9

An atom has 3 protons, two electrons and 1 neutrons. Describe this atom using as many vocabulary words

1 answer:

andre [41]4 years ago

4 0

Answer: This is a very unstable atom. It is a cation, meaning it is positively charged. This atom doesn't have an equal amount of neutrons to protons, so it is extremely unstable. It has a charge of +1 because it is missing a negatively charged particle, making a positive charge.

Explanation:

You might be interested in

What do many employers require of respiratory technician? Check all that apply.

meriva

Answer:

1 & 4

Explanation:

just got it right on edge

6 0

3 years ago

Typically, a solid chloride unknown (not the AgCl product) for analysis is dried in an oven to remove absorbed water before the

tatyana61 [14]

Answer:

%Cl would be too high if not dried.

Explanation:

Cl (chlorine) is a liquid. During the process of drying AgCl, the chlorine particles evaporate and chlorine is lost.

7 0

3 years ago

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv

Triss [41]

Answer:

1. $3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2. $5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

Explanation:

1.

The $pH$ value of Aspirin solution = 2.62

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

$[H^+]=10^{-2.62}=0.00240 M$

Moles of s asprin = $\frac{2.00 g}{180 g/mol}=0.01111 mol$

Volume of the solution = 0.600 L

The initial concentration of Aspirin = c = $\frac{0.01111 mol}{0.600 L}=0.0185 M$

$HAs\rightleftharpoons As^-+H^+$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_a=\frac{[As^-][H^+]}{[HAs]}$ :

$K_a=\frac{x\times x }{(c-x)}$

$=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}$

$K_a=3.57\times 10^{-4}$

$3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2.

The $pH$ value of ethylamine = 11.87

$pH+pOH=14$

$pOH=14-11.87=2.13$

$pOH=-\log[OH^-]$

$[OH^-]=10^{-2.13}=0.00741 M$

The initial concentration of ethylamine = c = 0.100 M

$C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}$ :

$K_b=\frac{x\times x}{(c-x)}$

$=\frac{0.00741\times 0.00741}{(0.100-0.00741)}$

$K_b=5.93\times 10^{-4}$

$5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

3 0

3 years ago

What elements were formed in Nuclear fusion?

Sonja [21]

Answer:

tritium and deuterium are combined and result in the formation of helium

6 0

3 years ago

Please help with 48!

natima [27]

A granite catches fire more quickly

4 0

4 years ago