Physics, Chemistry and Biology: Post your doubts here!

[Phoenix Blood]

if it was conc thn hydrogen would be released at the electrode instead of sodium(metal)........question had asked to liberate the metal........

A concentrated solution ... does it mean that it contains water?

 

[exploded diper]

A concentrated solution ... does it mean that it contains water?

yes...concenterated and dilute are aqueous solutions.....

 

[exploded diper]

explain with the help of a diagram sumone plz....

 

[Phoenix Blood]

These two...? eureka!

 

[Phoenix Blood]

Help me out.
eureka!

 

[pyaparus]

What are ionic equations?

 

[exploded diper]

Help me out.
eureka!

20 volume of hydrocarbon + 150vol. o2 = 100 vol. co2 + (20 vol) h20
so 2hydc--> 10 co2
1:5...whic mean there r fice carbons in the hydrocarbon....
15*2=30..o2--->10*2(C)o2+10(h2)o
30-20=10>>(h2)o
so 2hydc-->10(h2)
so 1hydc--->5(h2)...tht equals to 10h
so hydrocarbon contains c5h10....
so the equation is 2c5h10+15O2--->10co2+10h2o

 

[Phoenix Blood]

20 volume of hydrocarbon + 150vol. o2 = 100 vol. co2 + (20 vol) h20
so 2hydc--> 10 co2
1:5...whic mean there r fice carbons in the hydrocarbon....
15*2=30..o2--->10*2(C)o2+10(h2)o
30-20=10>>(h2)o
so 2hydc-->10(h2)
so 1hydc--->5(h2)...tht equals to 10h
so hydrocarbon contains c5h10....
so the equation is 2c5h10+15O2--->10co2+10h2o

Thanks!

 

[eureka!]

These two...? eureka!

Sorry! Didn't see this until after the exam. Anyway, I guess you've to simply describe it the way you'd describe SiO2. Giant macromolecule. Tetrahedral, each Ge atom covalently bonded to 4 O (1Ge : 4O), and 1 O atom bonded to 2 Ge atoms (1O : 2Ge). Very strong covalent bonds.

 

[eureka!]

0625_s13_qp_12 Question 33?

Yeah, I mucked up in that one too. Dunno why that's the answer, sorry about it!

 

[eureka!]

What are ionic equations?

Equations in which you've gotta mention the oxidation numbers and exclude all spectator ions.
Eg., in the displacement of tin ions by magnesium atoms.
Mg + Sn2+ -> Mg2+ + Sn

 

[Phoenix Blood]

Sorry! Didn't see this until after the exam. Anyway, I guess you've to simply describe it the way you'd describe SiO2. Giant macromolecule. Tetrahedral, each Ge atom covalently bonded to 4 O (1Ge : 4O), and 1 O atom bonded to 2 Ge atoms (1O : 2Ge). Very strong covalent bonds.

You know what? - this idiocy had come in the paper I'd seen the answer in the MS, but i dunno... Do we have to KNOW this or DERIVE it?

 

[exploded diper]

0625_s13_qp_12 Question 33?

since x and y are parallel there resistance is less.........so when y is removed the resistance across x increeases..and so does the voltage...as x and z are arranged in series so voltage across z decreases so it becomes less brighter......brightness depends on voltage

 

[exploded diper]

You know what? - this idiocy had come in the paper I'd seen the answer in the MS, but i dunno... Do we have to KNOW this or DERIVE it?

derive it......

 

[princess sofia]

cn any1 plz ans thx

 

[exploded diper]

cn any1 plz ans thx View attachment 38906

1. 20cm3/1000=0.02 dm3/24dm3=0.00083 moles of butyne
2 mole -----> 11mole o2 .... so 0.00083--->11*0.00083/2=0.0045 moles o2
so 0.0045*24=0.11 dm3*1000=110 cm3 o2
tell me if i m right
i will answer rest later

 

[exploded diper]

2. 2 moles butyne ---> 8 mole co2
so 0.00083 ---> 8* 0.00083/2= 0.00332 moles co2
volume= o.oo332*24000=79.68 cm3 co2

 

[exploded diper]

cn any1 plz ans thx View attachment 38906

3.o2 left is 150 -110= 40 cm3
co2 left is 80 cm3
so total volume = 40 + 80 = 120 cm3

 

[Awesome12]

View attachment 38869
since x and y are parallel there resistance is less.........so when y is removed the resistance across x increeases..and so does the voltage...as x and z are arranged in series so voltage across z decreases so it becomes less brighter......brightness depends on voltage

In a voltage circuit, the voltage remains the same. Thus when we add or remove the lamp, the total voltage at X will still remain the same. As a result, Z will light with the same brightness

 

[3abder7man]

In a voltage circuit, the voltage remains the same. Thus when we add or remove the lamp, the total voltage at X will still remain the same. As a result, Z will light with the same brightness

Nope. Assume each bulb has a resistance of 2 ohms, so before Y is removed, X and Y have a combined resistance of (1/2)+(1/2) = 1 ohm
while after Y is removed X would have a resistance of 2 ohms, and so a higher voltage. Consequently, the voltage across Z will decrease and it gets dimmer