5.0g sample of dry ice turned into 2400cm3 of carbon dioxide gas at RTP.What is the percentage purity of the dry ice?
1 answer:
Answer:
Explanation:
Hello,
In this case, since the dry ice is the solid phase of carbon dioxide gas, via the yielded volume we can compute the grams that were sublimated via the ideal gas equation at RTP (1 atm and 25 °C) considering the volume in liters (2.1 L):
Those previously computed mass are the pure grams, therefore, the percentage purity is:
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The structure of Alanine is shown below,
Except the carbon atom of carbonyl group which is Sp² Hybridized, all remaining atoms are Sp³ Hybridized. The hybridization of each element is depicted in a box below,
Except the carbon atom of carbonyl group which is Sp² Hybridized, all remaining atoms are Sp³ Hybridized. The hybridization of each element is depicted in a box below,
Answer:
a)4.51
b) 9.96
Explanation:
Given:
NaOH = 0.112M
H2S03 = 0.112 M
V = 60 ml
H2S03 pKa1= 1.857
pKa2 = 7.172
a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.
Therefore, the half points will also be the middle point.
Solving, we have:
pH = (½)* pKa1 + pKa2
pH = (½) * (1.857 + 7.172)
= 4.51
Thus, pH at first equivalence point is 4.51
b) pH at second equivalence point:
We already know there is a presence of SO3-2, and it ionizes to form
SO3-2 + H2O <>HSO3- + OH-
[HSO3-] = x = [OH-]
mmol of SO3-2 = MV
= 0.112 * 60 = 6.72
We need to find the V of NaOh,
V of NaOh = (2 * mmol)/M
= (2 * 6.72)/0.122
= 120ml
For total V in equivalence point, we have:
60ml + 120ml = 180ml
[S03-2] = 6.72/120
= 0.056 M
Substituting for values gotten in the equation
We noe have:
=4.04
pH = 14- pOH
= 14 - 4.04
= 9.96
The pH at second equivalence point is 9.96