STP condition mean we have P=1 atm. T=273K. R=ideal gas constant, but make sure you use the one that has the same units of pressure, temperature that you are using. In this case R=0.0821 L*atm K^-1mol^1. You are provided with n=2.1 moles.
V=nRTP
Input your values and solve.
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Answer:
Option A = 2.2 L
Explanation:
Given data:
volume of one mole of gas = 22.4 L
Volume of 0.1 mole of gas at same condition = ?
Solution:
It is known that one mole of gas at STP occupy 22.4 L volume. The standard temperature is 273.15 K and standard pressure is 1 atm.
For 0.1 mole of methane.
0.1/1 × 22.4 = 2.24 L
0.1 mole of methane occupy 2.24 L volume.
Answer:
[H₃O⁺] = 0.05 M & [OH⁻] = 2.0 x 10⁻¹³.
Explanation:
- HNO₃ is completely ionized in water as:
<em>HNO₃ + H₂O → H₃O⁺ + NO₃⁻.</em>
- The concentration of hydronium ion is equal to the concentration of HNO₃:
[H₃O⁺] = 0.05 M.
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] </em>= 10⁻¹⁴/0.05 = <em>2.0 x 10⁻¹³.</em>
Answer:
All of the above.
Explanation:
In positive deviation from Raoult's Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.
When a solution is non ideal then it shows positive or negative deviation.
Let two solutions A and B to form non- ideal solutions.let the vapour pressure of component A is 
and vapour pressure of component B is 
.
= Vapour pressure of component A in pure form
= Vapour pressure of component B in pure form
=Mole fraction of component A
=Mole fraction of component B
The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.
,
Therefore, 
Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.
Hence, option a,b,c and d are true.
