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# A mixture of nitrogen, oxygen and helium contains 0.25, 0.15 and 0.4 mole of these gases respectively. If the pressure contribution due to oxygen was 2.5 atm. The partial pressure of helium is.

### 2. The isomer of a compound C5H10 which does NOT decolorize bromine water is

a) 2-methylbutane

b) 2,2-dimethylpropane

c) 2-methylbut-1-ene

d) Methylcyclobutane

C5H10 is an hydrocarbon by inspection expected to be an alkene. Bromine water is used to test for unsaturation in organic compounds but for C5H10 to test negative to bromine water includes that it is a saturated organic compounds with one double bond equivalent and must be a cyclic compound. Option A and B have molecular formula C5H10 a saturated alkane, option C is an alkene and will test positive to bromine water but the isomer of a compound that does NOT decolorize bromine water is option (D) Methylcyclobutane

### 3. A mixture of nitrogen, oxygen and helium contains 0.25, 0.15 and 0.4 mole of these gases respectively. If the pressure contribution due to oxygen was 2.5 atm. The partial pressure of helium is.

a) 4.0 atm

b) 0.8 atm

c) 3.33 atm

d) 6.67,atm

Solution

We have 3 element Nitrogen, Oxygen and Helium

Nitrogen has 0.25 moles

Oxygen has 0.15 moles

Helium has 0.4 moles

Total no. Of moles = no of moles of ( nitrogen+ oxygen + helium)

Total no. Of moles = 0.25+0.15+0.4

= 0.8 moles

##### If the pressure contribution due to oxygen was 2.5 atm, find The partial pressure of helium is.

For us to find the partial pressure of helium we must first find the total pressure of mixture

Partial pressure of oxygen = No. Of moles of oxygen/ Total No. Of moles × total pressure of mixture

Make ( Total pressure of mixture) the subject of the formula

Total pressure of mixture = ( Total No. Of moles/ no. Of moles of oxygen ) × pressure contribution of oxygen)

= ( 0.8 × 2.5)/0.15
= 13.3 atm
The partial pressure of helium is.
Partial pressure of helium = No. Of moles of helium/ Total No. Of moles × total pressure of mixture
= 0.4/0.8 × 13.3
= 6.66 atm [D]

### 4. Element P has atomic number of 12 while element Q has an atomic number 15. Combination of P and Q gave a compound PmQn. The respective values of m and n are

• 2 and 2
• 2 and 3
• 3 and 2
• 2 and 1

Let’s find the electronic configuration of each element
P = 12 = 1S2 2S2 2P6 3S2
Q = 15 = 1S2 2S2 2P6 3S2 3P3
Element P have 2-electrons in it’s outermost shell i.e P2+ Element Q have 5-electrons in its outermost shell. I.e Q3+. Element Q needs 3-electrons to achieve octet configuration. It will need 6-electrons from 3-atoms of P to complete its configuration while 2 atoms of Q is needed, the compound is P3Q2 ( Exchange of valency)

p3Q2. Where m is 3 and n is 2

Therefore the final answer is P3Q2 (3 and 2)

### 5. The hydrocarbon, CxHY in the reaction above is most likely

• an alkane
• a benzene
• an alkene
• an alkyne

Let’s balance the equation. If you look at the equation very well, you’ll observe that at the second side carbon is 6, oxygen is 18 and hydrogen is 12 i.e( 2×6 = 12)
So therefore CxHY = C6H12 ( it’s an alkene)
It would have been an alkane is it’s C6H14
The final answer is (C) alkene

YOU CAN DO IT

### Oluwamuyide Peter

On the 4th of November I officially became a member of the exclusive 1st student with distinction after five years of no such record, in the history of The Polytechnic Ibadan, Faculty of Engineering to graduate with distinction as a DPP students since its establishment in 2011. My unrelenting power to solve problems, have made me to create a platform where student can get valid information anywhere, anyplace at anytime Evolving education world wide 🌎

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