All categories

3 years ago

Phosphorus has an atomic mass of 31 and an atomic number of 15, so the number of neutrons must be _______________.

1 answer:

6 0

Remember this:

atomic mass= atomic number + number of neutrons

The atomic number (same thing as number of protons) in this case is 15. The atomic mass is 31.

So, we subtract 15 from 31 to find the number of neutrons.

31-15=16

16 neutrons!!!

You might be interested in

Mario uses a hot plate to heat a beaker of 50mL of water. He used a thermometer to measure the temperature of the water. The wat

Alex777 [14]

Answer:

The boiling point decreases as the volume decreases.

Explanation:

The Temperature - Volume law otherwise called as Charles law is applied, which says that the volume of the given gas at constant pressure is directly proportional to the temperature measured in Kelvin. As the volume increases, the temperature also increases, if the volume decreases, then the temperature also decreases.

As per the Charles law, here the volume is decreased from 50 ml to 25 ml so the boiling point also decreases.

8 0

2 years ago

Read 2 more answers

Can anyone solve this?

Sonja [21]

Answer:

3P2O5:

P:6 O:15

5CO2

C:5 O:10

3C6H12O6

C:18 H:36 O:18

4C6H12

C:24 H:48

3Mg3(PO4)2

Mg:9 P:2 O: 8

4(NH4)2S

N:8 H:32 S:4

4 0

2 years ago

How do you find a theoretical mass? Is there a difference between theoretical mass and theoretical yield?

swat32

Theoretical yield is the quantity of a product obtained from the complete conversion of the limiting reactant in a chemical reaction. It is the amount of product resulting from a perfect chemical reaction and thus not the same as the amount you'll actually get from a reaction.

3 0

3 years ago

As matter changes, what happens to the energy?

OlgaM077 [116]

Energy, Temperature, and Changes of State

Matter either loses or absorbs energy when it changes from one state to another. For example, when matter changes from a liquid to a solid, it loses energy. The opposite happens when matter changes from a solid to a liquid.

4 0

2 years ago

A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the

Irina18 [472]

Answer:

Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

$\displaystyle c = \frac{n}{V}$ ,

where

$c$ is the concentration of the solute,

$n$ is the number of moles of the solute, and

$V$ is the volume of the solution.

$V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}$ .

Acetic (ethanoic) acid:

$\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}$ .

Hydrochloric acid HCl:

$\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}$ .

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be $5.00\times 10^{-4}\;\text{M}$ .

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be $-x\;\text{M}$ . $x > 0$ .

$\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}$ .

$\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}$ .