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krok68 [10]
3 years ago
13

a gas has a pressure of 0.410 atm and a volume of 32L. At what pressure would the volume of the gas change to 28L?

Chemistry
1 answer:
densk [106]3 years ago
8 0

Answer:

The new pressure is 0,469 atm

Explanation:

We apply Boyle Mariotte's Law: for a given mass of gas at constant temperature, its pressure and volume vary inversely proportionally (the higher the pressure, the lower the volume and vice versa):

P1 x V1 = P2 xV2               We solve P2:

P2 = (P1 x V1)/V2

P2= (0,410 atm x 32 L) / 28 L

<em>P2= 0,469 atm</em>

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Explanation:

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By observing (measuring) the dependent variable, while only one independent variable changes you can understandhow such independent variable explains (determines) the dependent variable, leading to a conclusion.

Conversely, if two or more independent variables change at a time, then there is no way that you can tell how the output (dependent variable) is related with one or other of the changes of the indipendent variables. You wolud not be able to discriminate (distinguish) the effect of one or other variable, making the experiment inconclusive

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3 years ago
Why do reactivity increases on moving from top to bottom in group 2 of modern periodic table?
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Answer:

that's because....

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they can achieve that by just losing one electron from their outer shell.

as you go down the group 1, element gets bigger in size, which means there is more space between nucleus (which is in center of atom) and electron of outer shell. the more far away they are the less attraction force between them.

so its easier for potassuim to lose one electron than for lithuim.

so that means potassium will easily give up 1 electron to react with non metal or other element therefore it is more reactive than lithuim

but in case of non metal, the opposite happens but simple to understand.

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3 years ago
Why do we need to standardize EDTA in determination of water hardness?​
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3 years ago
In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe
const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = 10^{-9} m.

Hence, 0.283 nm = 0.283 \times 10^{-9} m

  • Formula for coulombic energy is as follows.

             U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}

where,   e = 1.6 \times 10^{-19} C

            \epsilon_{o} = 8.85 \times 10^{-12}

          U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}

                         = 1.423 \times 10^{-18} J

  • As 1 eV = 1.6 \times 10^{-19} J

So,       1 J = \frac{1 eV}{1.6 \times 10^{-19}}

Hence,    U = \frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}

                   = 8.9 eV

  • Also,   1 J = \frac{10^{-3} kJ}{6.022 \times 10^{23}mol}

                = 1.67 \times 10^{-27} kJ/mol

Therefore, U = 1.423 \times 10^{-18} J \times 1.67 \times 10^{-27} kJ/mol

                     = 2.37 \times 10^{-45} kJ/mol

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