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Romashka [77]
3 years ago
6

Find the volume of 2.40 moles of gas whose temperature is 50.0°C and whose pressure is 2.00 atm.

Chemistry
1 answer:
DanielleElmas [232]3 years ago
5 0
PV=nRT
n=2.4 moles
T=273.15+50=323.15K
P=2*101325=202650 Pa
R=8.31


Solve for V:
V=nRT/P=2.4*8.31*323.15/202650=.032m^3
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A piece of glass is immersed in a liquid. It proceeds to float on the liquid's surface. This shows that the density of the glass
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Answer: b) Less dense

Explanation:

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An object more dense than the fluid in which it is immersed will sink, while objects less dense than the fluid in which it is immersed will float to the surface.

But objects floats at constant level if the density is equal to the density of the fluid in which it is immersed; it neither rises nor sinks in the fluid in this case.

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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.

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  • 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
I need help with Hess law
ICE Princess25 [194]

Answer:

alright

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Hess law is  the enthalpy change for a reaction that is carried out in a series of steps is equal to the sum of the enthalpy changes for the individual steps. Consistent with the law of conservation of energy. Starting and final conditions must be the same.

6 0
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makkiz [27]

Answer:

The mass of 0.100 mole of neon is 2.018 grams.

Explanation:

As we know the formula to find mass:

                           Number of moles         =  Mass/ Molar mass

                                        0.100                = Mass/ 20.17

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Hope it helps!

7 0
2 years ago
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