A gas has a volume of 27.5 L at 302 K and 1.40 atm. How many moles are in the sample of gas?

Nutka1998 [239]

Answer: The ionic compound formed is magnesium chloride having formula $MgCl_2$

Explanation:

Ionic compound is defined as the compound which is formed by complete transfer of electrons from one atom to another atom.

The atom which looses the electron is known as electropositive atom and the atom which gains the electron is known as electronegative atom. This bond is usually formed between a metal and a non-metal.

Magnesium is the 12th element of the periodic table having electronic configuration of $1s^22s^22p^63s^2$

This element will loose 2 electrons to form $Mg^{2+}$ ion

Chlorine is the 17th element of the periodic table having electronic configuration of $1s^22s^22p^63s^23p^5$

This element will gain 1 electron to form $Cl^{-}$ ion

So, for every 1 atom of magnesium, 2 atoms of chlorine are required. Thus, the chemical formula becomes $MgCl_2$

Hence, the ionic compound formed is magnesium chloride having formula $MgCl_2$

6 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:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

3 0

3 years ago

A substance with a high pOH would likely have which of the following?

Anna [14]

Answer: a low $OH^-$ and low pH.

Explanation:

pH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pOH=-log[OH^-]$

$pOH=log\frac {1}{OH^-}$

Thus as pOH and $OH^-$ are inversely related, a solution having higher pOH will have less amount of $OH^-$ concentration. And a solution having more pOH will have less pH.

Thus a substance with a high pOH would likely have low $OH^-$ concentration and low pH.

8 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!!

kogti [31]

1 CH4 (g) + 2 O2 (g) -----> CO2 (g) + 2H2O(l) ΔH= - 890 kJ
1 mol 2 mol
1) If ΔH has minus, it means "release". We need only "release" choices.
2) From reaction
1 mol CH4 (g) "releases" ΔH= - 890 kJ - We do not have this choice.

2 mol O2 (g) "release" ΔH= - 890 kJ, so
1 mol O2 (g) "release" ΔH= - 445 kJ
Correct answer is B.

3 0

3 years ago

Read 2 more answers

[07.02]what is the concentration of a ca(oh)2 solution if 10.0 ml of 0.600 m h3po4 solution is required to completely neutralize

Mademuasel [1]

In neutralization process or titration process, we use the formula;
(M1V1)H3PO4 = M2V2(Ca(OH)2)
M1 = 0.600 m
V1 = 10.0ml
M2 = unknown
V2 = 12.5ml
M1V1 = M2V2
0.6*10 = 12.5*M2
M2 = (0.6*10)/12.5 = 0.48m
Therefore, the answer is 0.480m

3 0

3 years ago