Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, 
molecule.
Solution :
At STP,
22.4 L volume of ethane present in 1 mole of ethane gas
64.28 L volume of ethane present in 
of ethane gas
And, as we know that
1 mole of ethane molecule contains 
molecules of ethane
2.869 moles of ethane molecule contains 
molecules of ethane
Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, molecule.
Base + acid = salt + water
Example :
<span>HCl + NaOH = NaCl + H</span>₂<span>O
</span>
Answer A
hope this helps!
I would suggest convection !!!
Answer : Option a) True.
Explanation : A placebo is a substance that is unlikely to affect the dependent variable. True as in a clinical trial for any change in the placebo arm is known as the placebo response, and the difference between this placebo response and the result of no treatment is called as the placebo effect. Therefore, a placebo may be given to any person in a clinical context in order to deceive that recipient into thinking that it is an part of active treatment. It actually resembles a drug but is not an actual active drug, so, it is clear that it will not affect dependent variable.
Answer:
pH = 7.233
Explanation:
Initially, the buffer contains 0.208 moles of NaHSO₃ and 0.134 moles of Na₂SO₃.
NaHSO₃ reacts with NaOH thus:
NaHSO₃ + NaOH → Na₂SO₃ + H₂O
50.0 mL of 1.00 M NaOH are:
0.0500L × (1mol / 1L) = 0.0500moles of NaOH added. That means after the addition are produced 0.0500moles of Na₂SO₃ and consumed 0.0500moles of NaHSO₃. That means final moles of the buffer are:
NaHSO₃: 0.208 mol - 0.050 mol = <em>0.158 mol</em>
Na₂SO₃: 0.134 mol + 0.050 mol = <em>0.184 mol</em>
<em> </em>
As pKa of this buffer is 7.167, it is possible to use H-H equation to find pH, thus:
pH = pKa + log₁₀ [Na₂SO₃] / [NaHSO₃]
pH = 7.167 + log₁₀ [0.184] / [0.158]
<em>pH = 7.233</em>
