Which of these relationships is true of an uncharged atom?

Chemistry

1 answer:

marusya05 [52]2 years ago

7 0

Answer:

The correct option is C) The number of protons is equal to the number of electrons.

Explanation:

Atoms are made up of particles called protons, neutrons, and electrons. Protons and neutrons are in the center of the atom and form the nucleus, while electrons surround the nucleus. Electrons have a negative charge. The charge of the protons is positive and finally, the neutrons have no charge.

If the atom has no charge, this means that the total charge of the atomic nucleus, which is positive due to the presence of the protons, is equal to the negative charge of the electrons, so that it cancels out.

So, <u><em>the correct option is C) The number of protons is equal to the number of electrons.</em></u>

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Read 2 more answers

If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W

olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

$CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2$

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

$m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl$

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

$m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3$