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Savatey [412]
3 years ago
8

N the reaction Mg + 2HCl -> MgCl2 + H2 which element’s oxidation number does not change?

Chemistry
1 answer:
BartSMP [9]3 years ago
4 0

Answer:- Oxidation number of Cl does not change as it is -1 on both sides.

Explanations:- oxidation number of Mg on reactant side is 0 as it is in its elemental form(not combined with another element).

Oxidation number  of hydrogen in its compounds is +1, so if H is +1 in HCl the oxidation number of Cl is -1 as the sum has to be zero.

On product side, Mg oxidation number is +2 as the oxidation number of alkaline earth metals in their compounds is +2.

Two Cl are present in magnesium chloride, so if Mg is +2 then Cl is -1.

Oxidation number of H on product side is 0 as it is present in its elemental for, H_2 ,

So, it is only chlorine(Cl) whose oxidation number does not change for the given equation.


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Answer:

net ionic equation

\boxed{  CaCO_3{}_{(s)} + 2H^{+} {}_{(aq.)} \rightarrow Ca^{2+} {}_{(aq.)} + H_2O_{(l)} + CO_2{}_{(g)} }

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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
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Assoli18 [71]
Gamma is not affected by both fields because it isnt charged. 

Only alpha and beta ray will be affected because they are charged with positive and negative ions respectively.

In otherwords, <span>Gamma rays are high-energy photons of light, which have zero electric charge. So they are not affected by magnetic fields</span>

Here's a chart. .

If that doesnt help tell me :)

7 0
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morpeh [17]

Answer:

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