Ask question

Ask question

All categories

2 years ago

14

When 0.0801 mol of an unknown hydrocarbon is burned in a bomb calorimeter, the calorimeter increases in temperature by 2.19°C. I

f the heat capacity of the bomb calorimeter is 1.229 kJ/°C, what is the heat of combustion for the unknown hydrocarbon, in kJ/mol?

1 answer:

Andreyy892 years ago

8 0

Answer:

S A T. NV

Explanation:

You might be interested in

The work function for metallic potassium is 2.3 eV. Calculate the velocity in km/s for electrons ejected from a metallic potassi

mote1985 [20]

Answer:

932.44 km/s

Explanation:

Given that:

The work function of the magnesium = 2.3 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, work function = $2.3\times 1.6022\times 10^{-19}\ J=3.68506\times 10^{-19}\ J$

Using the equation for photoelectric effect as:

$E=\psi _0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\psi _0+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

$\lambda$ is the wavelength of the light being bombarded

$\psi _0=Work\ function$

v is the velocity of electron

Given, $\lambda=260\ nm=260\times 10^{-9}\ m$

Thus, applying values as:

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{260\times 10^{-9}}=3.68506\times 10^{-19}+\frac{1}{2}\times 9.11\times 10^{-31}\times v^2$

$3.68506\times \:10^{-19}+\frac{1}{2}\times \:9.11\times \:10^{-31}v^2=\frac{6.626\times \:10^{-34}\times \:3\times \:10^8}{260\times \:10^{-9}}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=7.64538\times 10^{-19}-3.68506\times 10^{-19}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=3.96032462\times 10^{-19}$

$v^2=0.869446\times 10^{12}$

v = 9.3244 × 10⁵ m/s

Also, 1 m = 0.001 km

<u>So, v = 932.44 km/s</u>

5 0

2 years ago

In cellular respiration, living things use sugar to produce energy.<br> True or false

wel

I pretty sure it true I’m not completely sure

3 0

2 years ago

What is an orbital shell

Lyrx [107]

<span>orbital shell is the</span><span> the circular paths around the nucleus of an atom along which the electrons traverse.</span>

4 0

2 years ago

What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11

lutik1710 [3]

The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m

4 0

3 years ago

Which combination of reagents used in the indicated order with benzene will give m-nitropropylbenzene? a. 1) CH3CH2CH2Cl/AlCl3,

VladimirAG [237]

Answer:

1) HNO3/H2SO4, 2) CH3CH2CH2Cl/AlCl3

Explanation:

Benzene is a stable aromatic compound hence it undergoes substitution rather than addition reaction.

When benzene undergoes substitution reaction, the substituent introduced into the ring determines the position of the incoming electrophile.

If I want to synthesize m-nitropropylbenzene, I will first carry out the nitration of benzene using HNO3/H2SO4 since the -nitro group is a meta director. This is now followed by Friedel Craft's alkykation using CH3CH2CH2Cl/AlCl3.

6 0

2 years ago