Mass in kilograms of liquid air required = 0.78 kg
<u>Given that </u>
1 Litre of liquid air contains 1.3 grams of oxygen ( air )
<u />
<u> Determine the a</u><u>mount of liquid air</u><u> in Kg</u>
volume of air given = 600 L
mass of liquid air required = x
1 litre = 1.3 grams
600 L = x
∴ x ( mass of liquid air ) = 1.3 * 600
= 780 g = 0.78 kg
Hence we can conclude that Mass in kilograms of liquid air required = 0.78 kg
Learn more about liquid air : brainly.com/question/636295
Answer:
Explanation:
The formula for molality is:
There are 0.210 moles of KBr and 0.075 kilograms of pure water.
Substitute the values into the formula.
Divide.
The molality is <u>2.8 moles per kilogram</u>
The boiling points need to differ by 50 degrees to enable their complete separation ie of two different liquids. The separation occurs by first evaporation of one of the liquids and then its condensation and collection. It is a physical process not a chemical one.
Answer:
b) 3.10
Explanation:
HF ⇄ H
+ + F
Using Henderson-Hasselbalch Equation:
pH = pKa + log [A-]/[HA].
Where;
pKa = Dissociation constant = -log Ka
Hence, pKa of HF = -log 7.2 x 10^-4 = 3.14266
[A-] = concentration of conjugate base after dissociation = moles of base/total volume
= 0.15 x 0.3/0.8
= 0.05625 M
[HA] = concentration of the acid = moles of acid/total volume
= 0.10 x 0.5/0.8
= 0.0625 M
Note: <em>Total volume = 500 + 300 = 800 mL = 0.8 dm3</em>
pH = 3.14266 + log [0.05625/0.0625]
= 3.14267 + (-0.04575749056)
= 3.09691250944
<em>From all the available options below:</em>
<em>a) 2.97
</em>
<em>b) 3.10
</em>
<em>c) 3.19
</em>
<em>d) 3.22
</em>
<em>e) 3.32</em>
The correct option is b.
