Which is the balanced equation for magnesium added to copper(i) chloride yielding copper and magnesium chloride?

Karolina [17]

Answer:

for anapproximate result , divide the pressure value by 7.501

8 0

2 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

2 years ago

As a boy was it a balloon its volume increases

goldenfox [79]

What is Avogadro’s Law

5 0

3 years ago

Calculate the concentration of chloride ions when 100.0 mL of 0.233 M sodium chloride is mixed with 250.0 mL of 0.150 M calcium

serg [7]

The dissociation of both salts NaCl and CaCl₂ are as follows;
NaCl --> Na⁺ + Cl⁻

CaCl₂ --> Ca²⁺ + 2Cl⁻

the molar ratio of NaCl to Cl⁻ is 1:1
therefore number of NaCl moles is equal to number of Cl⁻ ions dissociated from NaCl
then number of Cl⁻ ion moles - 0.233 mol/L x 0.1000 L = 0.0233 mol

molar ratio of CaCl₂ to Cl⁻ ions is 1:2
1 mol of CaCl₂ gives out 2 mol of Cl⁻ ions.
number of CaCl₂ moles - 0.150 mol/L x 0.2500 L = 0.0375 mol
then the number of Cl⁻ ion moles - 0.0375 x 2 = 0.0750

total number of Cl⁻ ion moles = 0.0233 mol + 0.0750 mol = 0.0983 mol
volume of solution - 100.0 + 250.0 = 350.0 mL

concentration of Cl⁻ = 0.0983 mol / 0.3500 L = 0.281 M
concentration of Cl⁻⁻ is 0.281 M

4 0

2 years ago

The table shows the properties of some organic compounds.

Shalnov [3]

Answer:

option C is correct

..............,

3 0

2 years ago