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3 years ago

is it possible to gain or lose mass or energy after a chemical reaction has occurred in a closed system?

Chemistry

1 answer:

Ivenika [448]3 years ago

5 0

Answer:

The mass will not change during a chemical reaction.

Explanation:

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I need this answer quick please show work

Ainat [17]

Answer:

The answer to your question is 25.2 g of acetic acid.

Explanation:

Data

[Acetic acid] = 0.839 M

Volume = 0.5 L

Molecular weight = 60.05 g/mol

Process

1.- Calculate the number of moles of acetic acid

Molarity = moles / volume

-Solve for moles

moles = Molarity x volume

-Substitution

moles = (0.839)(0.5)

-Result

moles = 0.4195

2.- Calculate the mass of acetic acid using proportions and cross multiplications

60.05 g ----------------------- 1 mol

x ----------------------- 0.4195 moles

x = (0.4195 x 60.05) / 1

x = 25.19 g

3.- Conclusion

25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M

4 0

3 years ago

If the following elements were to combine with each other, choose which demonstrate the Law of Multiple Proportion. Select all t

Ilia_Sergeevich [38]

Missing question: Z:X = 7:1 A:Z = 2.5:1 A:Z = 2.2:1 Y:X = 11:1.

Answer is:<span> Z:X = 7:1 and Y:X = 11:1.
Law of multiple proportions or Dalton's Law said that the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers.
In this example ratios are whole numbers, A:Z = 2.5 : 1 and A:Z = 2.2 : 1 are not whole number ratios.</span>

3 0

3 years ago

Identify three reasons why

BARSIC [14]

The radioactive decay of unstable isotopes continually generates new energy within Earth's crust and mantle, providing the primary source of the heat that drives mantle convection. Plate tectonics can be viewed as the surface expression of mantle convection.

6 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

3 years ago

How do this I’m in school really don’t know what to do

avanturin [10]

Answer:

you gotta rotate it 90 degrees and get that screen fixed

Explanation:

4 0

2 years ago