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3 years ago

Electrons determine the of an atom

Chemistry

1 answer:

g100num [7]3 years ago

7 0

Answer:

if u know the number of electrons u also know the number of protons present in the atom.

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In what circumstance would the complete and net ionic equations for a reaction be identical?

Yakvenalex [24]

Answer: Complete ionic equations dissociate all aqueous solutions into ions. Net ionic equations show the change that occurs in chemical reactions and do not show spectator ions that are the same in reactants and products. (b) If no spectator ions were present then complete and net ionic equations would be identical.

Explanation:

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3 years ago

Help pls ASAP thank you!!

BigorU [14]

Answer:

Deer, Eagle, Raccoon, frog, turtle, fish, dragonfly.

Explanation:

They interact from the life cycle they eat each other...

5 0

3 years ago

What is the theory that can explain the model of Pangaea?

NISA [10]

The bones of the same animal found out continents far away from each other

4 0

3 years ago

In the following reaction HF(aq) + HPO42-(aq) = F-(aq) + H2PO4-(aq)

garik1379 [7]

Answer:

Explanation:

HF gives its H to HPO42 making it H2PO4, so HF is the acid and F- is the conj base

6 0

3 years ago

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

3 0

3 years ago