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Ksivusya [100]
3 years ago
12

Write the chemical equation for the reaction that occurs when 6 m hcl solution is added to the reaction mixture.

Chemistry
1 answer:
Semenov [28]3 years ago
6 0
When u add the solution to the chemical's
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PH=4
-log(0.0001)=4
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How can the existence of spectra help to prove that energy levels in atoms exist?
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The existence of spectra can help the existence of atoms by expanding and multiplying in the law of London Dispersion
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An electrochemical cell has the following standard cell notation.
Alona [7]

Explanation:

The given following standard cell notation.  

Mg(s) | Mg^2+ (aq) || Aq^+(aq) | Aq(s)

Oxidation:

Mg(s)\rightarrow Mg^{2+}+2e^-....(1)

Magnesium metal by loosing 2 electrons is getting converted into magnesium cation. Hence, getting oxidized

Reduction:

Ag^+(aq)+1e^-\rightarrow Ag(s)...(2)

Silver ion by gaining 1 electrons is getting converted into silver metal. Hence, getting reduced.

Overall redox reaction: (1)+2 × (2)

Mg(s)+2Ag^+(aq)\rightarrow Mg^{2+}+2Ag(s)

4 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\%.

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3 years ago
Due to the change in the number of electrons, the metal and nonmetal atoms both become _____
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The answer is ions , i need to have 20 words so yeah dhajzbajs dhakabxbsjsjs
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