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3 years ago

Can Someone help me pls?

Chemistry

1 answer:

Ilia_Sergeevich [38]3 years ago

8 0

Answer:

For H3O concentration you do 10^-pH so if pH is 5 then H3O+ is 10^-5= 1*10^-5 H3O+ ions

For OH is one extra step. First find H3o+ ions using equation above then you have to use that to divide 1*10^-14

So if pH is 5....the H3O+ is 1*10^-5 then OH- = (1*10^-14)/(1*10^-5) = 1*10^-9 OH ions

as far as acid/base pH 0-6 is Acid 8-14 is Base. pH of 7 is neutral. Recheck your work *hint* *hint* water is neutral. Spit is above 7 so is base.

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A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

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