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3 years ago

Lonic bonds form between oppositely charged

Chemistry

1 answer:

fomenos3 years ago

4 0

answer:

ionic bonds form between oppositely charged ions

explanation:

an ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions.
ionic bonds are formed between a cation, which is usually a metal, and an anion, which is usually a nonmetal.
a covalent bond involves a pair of electrons being shared between atoms.

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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

Which of the above solutions could be used to remove lead from lead (II) nitrate?

alisha [4.7K]

Ni solution could be used to remove lead from lead (II) nitrate.

Explanation:

The removal of lead from lead nitrate will take place by displacement reaction.

In displacement reaction less reactive element is displaced by more reactive element form it compound.

The reactivity is decided by the placement of metal in the activity series.

A metal which is at higher position in the activity series will be able to displace the metal or element having lower position.

From the options given we will check their position in the activity series in comparison to Pb

Cu is lower in series than Pb hence cannot displace.

Hg is placed lower in the series than Pb hence cannot displace Pb.

Ag is placed lower than Pb in the series hence cannot displace Pb.

Ni is placed above the Pb in activity series hence can displace lead.

8 0

3 years ago

Which of the following is NOT a basic feature of an electric circuit? a. resistor c. single throw b. cell d. wire

Naya [18.7K]

C. Single Throw - I am not for sure

5 0

4 years ago

Heat is added to a 1.0-kg block of ice at OC. Determine if the process is

Tatiana [17]

Answer:

endothermic

Explanation:

Heat is added to make the process possible.

5 0

3 years ago

The balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI is Pb(ClO3)2(aq)+2NaI(aq)⟶PbI2(s)+2NaClO3(aq) What

ryzh [129]

Answer : The mass of $PbI_2$ precipitate produced will be, 9.681 grams.

Explanation : Given,

Molarity of NaI = 0.210 M

Volume of solution = 0.2 L

Molar mass of $PbI_2$ = 461.01 g/mole

First we have to calculate the moles of $NaI$ .

$\text{Moles of }NaI=\text{Molarity of }NaI\times \text{Volume of solution}=0.210M\times 0.2L=0.042moles$

Now we have to calculate the moles of $PbI_2$ .

The balanced chemical reaction is,

$Pb(ClO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaClO_3(aq)$

From the balanced reaction we conclude that

As, 2 moles of $NaI$ react to give 1 mole of $PbI_2$

So, 0.042 moles of $NaI$ react to give $\frac{0.042}{2}=0.021$ moles of $PbI_2$

Now we have to calculate the mass of $PbI_2$ .

$\text{Mass of }PbI_2=\text{Moles of }PbI_2\times \text{Molar mass of }PbI_2$

$\text{Mass of }PbI_2=(0.021mole)\times (461.01g/mole)=9.681g$

Therefore, the mass of $PbI_2$ precipitate produced will be, 9.681 grams.

6 0

4 years ago