Ask question

Ask question

All categories

2 years ago

8

If a sample of pure calcium (Ca) contains 4.69 x 1022 atoms of iodine, how many moles of calcium are in the sample?. A) 0.0779 m

ol. . B) 0.130 mol. C) 7.78 mol. D) 12.8 mol

1 answer:

timama [110]2 years ago

5 0

If a sample of pure calcium 9Ca) contains 4.69 x 1022 atoms of iodine, there are 0.0779 moles of calcium in the sample. The correct answer will be A.

You might be interested in

What is the easiest way to determine the reactivity of an element

makkiz [27]

I know the answer to your question but if you want me to answer this question first, I need your help with my question "How do I solve number 12? Quick please I need help ASAP!!!"

4 0

2 years ago

0.17 M KCHO2 Find the pH

SashulF [63]

I believe the pH is 7. If you would like me to explain comment.

6 0

3 years ago

Carrying capacity is _____.

Alinara [238K]

<span>Carrying capacity is the number of organisms an ecosystem can support. It is the maximum size of a population that can survive in the ecosystem. If the animals reach the carrying capacity, the population may crash. As the consequence, the number of animals will decrease due to predators or diseases.</span>

5 0

3 years ago

Read 2 more answers

Separation of mixtures and purification of chemical substances

Mnenie [13.5K]

Answer:

I don't understand what to do with this

4 0

3 years ago

Nitrogen dioxide is a red-brown gas responsible for the brown color of smog. A container of nitrogen dioxide that is at low pres

____ [38]

Answer:

Initially $1.51\times 10^{-2} moles$ of nitrogen dioxide were in the container .

Explanation:

Volume of the container at low pressure and at room temperature = $V_1=3.4 L$

Number of moles in the container = $n_1$

After more addition of nitrogen gas at the same pressure and temperature.

Volume of the container after addition = $V_2=5.11 L$

Number of moles in the container after addition= $n_2=2.28\times 10^{-2} mol$

Applying Avogadro's law:

$\frac{Volume}{Moles}=constant$ (at constant pressure and temperature)

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

$n_1=\frac{V_1\times n_2}{V_2}=\frac{3.4 L\times 2.28\times 10^{-2} mol}{5.11 L}$

$n_1=1.51\times 10^{-2} mol$

Initially $1.51\times 10^{-2} moles$ of nitrogen dioxide were in the container .

8 0

3 years ago