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596277 thumbnail
Posted: 13 years ago

I really need help with this chemistry stuff, my class and I had a sub for the whole week, and we started the chapter Stoichiometry, and he never gave us any proper notes on it and never explained how to do anything at all, it was so bad, so if anyone can help me with this sheet, that would be so much appreciated. P.S chapter before this one was the Mole concept, and had a sub for that too, and I really sucked at balancing stuff.

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akhl thumbnail
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Posted: 13 years ago
@Ayeshaa
1)(a) The balanced equation is
2Al2O3 + 3C ------> 4Al + 3CO2

2 moles Al2O3 react with 3 moles of carbon
Therefore 43 moles Al2O3 react with
(3/2) * 43 = 64.5 moles carbon
Ans: 64.5 moles

(b) 3 moles C react with 2 moles Al2O3
Therefore 17.8 moles C react with
(2/3) * 17.8 = 11.87 moles Al2O3
Molar mass of Al2O3 = 27*2 + 16*3 = 102 g/mol

11.87 moles Al2O3 = 11.87 * 102 = 1210 g
Ans: 1210 g

c) 2.19 x 10^3 g Al2O3 = 2.19 x 10^3/102 = 21.47 mol Al2O3
2 moles Al2O3 react with 3 moles C
Therefore 21.47 moles Al2O3 react with
(3/2) x 21.47 = 32.21 mol C

32.21 mol C = 32.21 x 12 = 386.5 g C
Ans: 386.5 g

d) 760 kg Al = 760 x 10^3 g Al = 760 x 10^3/27 mol = 28148 mol Al
4 moles Al are produced by 2 moles Al2O3
Therefore 28148 mol Al are produced by
(2/4) * 28148 = 14074 mol Al2O3
= 14074 * 102 g Al2O3
= 1,435,548 g Al2O3
= 1436 kg Al2O3 (Answer)

akhl thumbnail
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Posted: 13 years ago
Can you solve the second problem? The balanced equation is
TiCl4 + 4Na ---> 4NaCl + Ti
596277 thumbnail
Posted: 13 years ago
^ thanks so much for the help, I think I sort of understand it now, I tried doing that Ti problem :

2. a)1 mole of Ti reacts with 4 moles of Na. So 79 moles Ti react with (4/1)*79 = 316 moles Na.

I don't get how to do b :\ And I'm not sure if I did A right,

akhl thumbnail
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Posted: 13 years ago
@Miggi

p + rho*g*h1 + 1/2*rho*v1^2 = p + rho*g*h2 + 1/2 * rho*v2^2
rho*g*h1 + 1/2*rho*v1^2 = rho*g*h2 + 1/2 * rho*v2^2
Put v1 = 0
rho*g*h1 = rho*g*h2 + 1/2 * rho*v2^2
Divide by rho
g*h1 = g*h2 + 1/2 * v2^2
v2 = sqrt[2g(h1-h2)]
v2 = sqrt(2 * 9.8 * 0.025)
v2 = 0.7 m/s
Miggi thumbnail
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Posted: 13 years ago
Hey akhl thanks for your help
Could u please look at the other questions as well? they are due tomorrow.
akhl thumbnail
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Posted: 13 years ago

Q: A solution with a concentration of 4.23mg/ml(MW= 321.4g/mol) gives a transmittence value of 28.9% in a 1.5cm cuvette at 280nm. Calculate the molar absorptivity of the sample. Show your work.

A = log(1/T) = log(1/0.289) = 0.539
c = 4.23 mg/ml = 4.23 g/l
= 4.23/321.4 mol/l
= 1.316 * 10^-2 mol/l

L = 1.5 cm

A = e * c * L
e = A/(c*L)
e = 0.539/(1.316*10^-2*1.5)
e = 27.3 liter/(mol.cm)

Edited by akhl - 13 years ago
P.Kamaljit.Sean thumbnail
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Posted: 13 years ago
Can Anyone Please Please Please Help Me Out .- I am Stuck On My Homework .-

I got this Homework for my Compositon Class .
and the teacher said that : give me 5 colorful words , 5 colorless word and 5 colored word.

i googled it but no luck

and can anyone please help me
Miggi thumbnail
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Posted: 13 years ago
I have one more question
Engine oil that is classified as SAE 5W-30 has a density of 859kg/m^3 and a viscosity of 6.6Pa.s at a temperture of -30 degree celcius. The oil is poured through a funnel with a neck that is 10.0cm long and 1.00cm diameter. If the level of the oil is kept at a height of 10.0cm above the top of the top of the neck, how long will it take to pour 1.00Liter of oil through the funnel. Hint: YOu may assume that viscous effects need only be considered along the length of the neck of the funnel. Note. Your anaswer will be surprisingly large!
Miggi thumbnail
Explorer Thumbnail
Posted: 13 years ago
A 0.50kg object , suspended from an ideal spring of spring constant of 25N/m, is oscillating vertically with amplitude of 7.5cm. How much change of kinetic energy occurs while the object moves from the equilibrium point to a point 5.9cmlower?

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