Volume of the tank is 5.5 litres.
Explanation:
mass of the CO2 is given 8.6 grams
Pressure of the gas is 89 Kilopascal which is 0.8762 atm
Temperature of the gas is 29 degrees ( 0 degrees +273.5= K) so (29+273)
R = gas constant 0.0821 liter atmosphere per kelvin)
FROM THE IDEAL GAS LAW
PV=nRT ( P Pressure, V Volume, n is number of moles of gas, R gas constant, Temperature in Kelvin)
no of moles = mass/atomic mass
= 8.6/44
= 0.195 moles
now putting the values in equation
V=nRT/P
= 0.195*0.0821*302/ 0.8762
= 5.5 litres.
As the carbon dioxide gas occupies the volume os the tank hence volume of tank is 5.5 litres.
Answer:
100%
Explanation:
My teacher just asked this question.
During selection of indicator. We choose an indicator which have pH range equivalent to the pH change of reaction to give better result and better observation.
So there are some different indicator are used in table 2 as compared to the table 1.
- Alizarin and phenolphthalein are basic indicator and their pH range is more than 8 so they are used in table 2
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Answer:
4.214 × 10^23 molecules.
Explanation:
Number of molecules in a substance can be calculated by multiplying the number of moles in that substance by Avagadro's number, which is 6.02 × 10^23.
That is, no. of molecule = n × Avagadro constant
In this case, there are 0.7 moles of fructose. Hence;
number of molecules = 0.7 × 6.02 × 10^23
no. of molecule = 4.214 × 10^23 molecules.
Limitations of Van der waal's equation. (i) The value of 'b' is not constant but varies with pressure and temperature. (ii) The value of is not equal to 3b, but actually it is equal to, in some case; and in other cases 2b. (iii) The value of is not equal to but it is usually more than 3 for most of the gases.
