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Reil [10]
3 years ago
13

Does Silicon-32 has a half-life of 170 years and can be used to find the age of rocks it is contained in.

Chemistry
1 answer:
ioda3 years ago
8 0
Silicon-32's half-life is actually 710 years.
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Which explanation best compares the movement of particles in the three states of matter?
ohaa [14]

Answer:solid: still boi liquid: bouncy boi has: crazy boi

Explanation:

3 0
3 years ago
Identify reagents necessary to convert cyclohexane into 1,3-cylohexadiene: notice that the starting material has no leaving grou
Vinvika [58]
<span>1,3-cylohexadiene i synthesized starting from cyclohexane in following 4 steps.

1) Free Radical Substitution Rxn:
 Halogenation of cyclohexane in the presence of UV yield chlorocyclohexane.

2) Elimination Rxn: Dehydrohalogenation of chlorocyclohexane yields cyclohexene.

3) Halogenation of Cyclohexene (Electrophillic Addition Rxn) gives 1,2-dihalocyclohexane.

4) Elemination Rxn: When dibromocyclohexane is treated with KOH and heated it gives 1,3-cyclohexadiene as shown below,</span>

4 0
3 years ago
Why do reactivity increases on moving from top to bottom in group 2 of modern periodic table?
Nikolay [14]

Answer:

that's because....

group 1 (e.g Na, K) those tend to lose one electron to gain noble gas electron configuration.

they can achieve that by just losing one electron from their outer shell.

as you go down the group 1, element gets bigger in size, which means there is more space between nucleus (which is in center of atom) and electron of outer shell. the more far away they are the less attraction force between them.

so its easier for potassuim to lose one electron than for lithuim.

so that means potassium will easily give up 1 electron to react with non metal or other element therefore it is more reactive than lithuim

but in case of non metal, the opposite happens but simple to understand.

as you go down the group 7 (halogen- Cl, Br, I) element will get bigger therefore force between nucleus and outer electron is getting smaller. they have to gain 1 electron in order to fill the outer shell (to gain noble gas electron configuration.)

as florine is more smaller in size than clorine it is more reactive because florine has more tendency to pull extra electron from metal or other element towards its side. so it easily gain 1 electron to react.

3 0
4 years ago
2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
azamat

Answer:

K=1.12x10^9

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

K=\frac{[NO_2]^2}{[NO]^2[O_2]}

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0
3 years ago
The heat of reaction for the combustion of propane is –2,045 kJ. This reaction is: C3H8(g) +502 (g) 3CO2 (g) + 4H2O (g). Determi
Ede4ka [16]

Answer:

\Delta _fH_{C_3H_8}=-102.7kJ/mol

Explanation:

Hello,

In this case, we can consider that the given heat of combustion is indeed the heat of reaction since it corresponds to the combustion of propane, which is computed by using the heat formation of all the involved species as shown below:

\Delta _cH=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-5*\Delta _fH_{O_2}

Thus, since the heat of formation of gaseous carbon dioxide is -393.5 kJ/mol, water -241.8 kJ/mol and oxygen 0 kJ/mol, the heat of formation of propane is:

\Delta _fH_{C_3H_8}=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-5*\Delta _fH_{O_2}-\Delta _cH\\\\\Delta _fH_{C_3H_8}=3*(-393.5)+4*(-241.8)-5*0-(-2045)\\\\\Delta _fH_{C_3H_8}=-102.7kJ/mol

Best regards.

8 0
3 years ago
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