Which part of homeostasis is this adorable puppy balancing out with his little stuffed animal buddy?

How many Mol are in 4.000 grams of MgCl2 ?

grigory [225]

Answer:

The number of mol is: 0, 042 mol in 4 grams of MgCl2

Explanation:

We calculate the weight of 1 mol of MgCl2:

Weight 1mol of MgCl2= weight Mg + (weight Cl)x 2=

24, 3 grams + 2 x 35, 5 grams = 95, 3 grams/mol MgCl2

95, 3 grams------1 mol MgCl2

4 grams -------x = (4 grams x1 mol MgCl2)/ 95, 3 grams= 0, 04197 mol MgCl2

4 0

3 years ago

Which of the following is true about the elements on the Periodic Table? There may be 1 or more correct answers.

nlexa [21]

Answer: 1. Block

2. True ( I'm unsure. He arranged it according to mass, but he is credited for the periodic table)

3. Noble gasses

4. technetium

5. Alkaline earth metals

6. Number of protons

7. False

8. Not true: generally decreases as atomic number increases within a period

9. Argon

10. Four

11. False

12. False

Explanation:

4 0

2 years ago

Pls help me out with this question!

jeka94

At STP, copper (Cu) would be the only substance here that will exist in the solid state.

4 0

2 years ago

What is the chemical formula for light energy

Goshia [24]

There is no chemical formula for light energy because light is not a chemical.

5 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