How many Zn atoms are there in 4.65 moles of Zn?

1 answer:

8 0

So your answer would pretty much be 2.80 x 10^24. The picture is just the explanation and how you would get that answer.

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A 8L sample of gas at 250 K is cooled to 75 K. What is the volume of the gas after it is cooled?

Stells [14]

$by \: Charles'\: law: \: \\ \frac{v_{1}}{t_{1}} = \frac{v_{2}}{t_{2}} \\$

Where v is the volume(in L) and t is the temperature(in °K)

$\frac{8}{250} = \frac{v_{2}}{75} \\\\4/5 =v_2/3 \\\\\boxed{v_{2} = \frac{12}{5}}\\\\\\\huge{=2.5 l}$

7 0

2 years ago

What is the energy released in this β − β − nuclear reaction 40 19 K → 40 20 C a + 0 − 1 e 19 40 K → 20 40 C a + − 1 0 e ? (The

Effectus [21]

<u>Answer:</u> The energy released in the given nuclear reaction is 1.3106 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

$_{19}^{40}\textrm{K}\rightarrow _{20}^{40}\textrm{Ca}+_{-1}^{0}\textrm{e}$

We are given:

Mass of $_{19}^{40}\textrm{K}$ = 39.963998 u

Mass of $_{20}^{40}\textrm{Ca}$ = 39.962591 u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(39.963998-39.962591)=0.001407u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.001407u)\times c^2$

$E=(0.001407u)\times (931.5MeV)$ (Conversion factor: $1u=931.5MeV/c^2$ )

$E=1.3106MeV$

Hence, the energy released in the given nuclear reaction is 1.3106 MeV.

6 0

3 years ago

A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. Wha

Ainat [17]

We have that the molecular weight (3sf) of the compound (g/mol)

$m=44.15g/mol$

From the question we are told

A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. What is the molecular weight (3sf) of the compound (g/mol).

Generally the equation for the Rouault's law is mathematically given as

P=P_0 N

$22.67=23.8*\frac{\frac{12.5}{18}}{\frac{125}{18}+\frac{15}{m}}\\\\\6.95+\frac{15}{m}=7.29\\\\\frac{15}{m}=7.29-6.95\\\\m=\frac{15}{0.34}\\\\m=44.11g/mol$