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Nadusha1986 [10]
2 years ago
11

In an isotope, which subatomic particle varies in number ?

Chemistry
1 answer:
wolverine [178]2 years ago
8 0
The number of neutrons vary in isotopes
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How many grams are there in 3.30x10 23 grams of n2i6
qwelly [4]
Technically, the answer should be 3.30 * 10^23 grams. But I think you mean either molecules, atoms, moles or grams of Ni2I6 with that number of molecules .

1 mole of Ni2 I6  = 6.02 * 10^23 molecules
x  [mole] = 3.30 * 10^23 molecules

1/x = 6.02 * 10^23 / 3.30 * 10^23 Cancel the 10^23 on the right side
1/x = 6.02 / 3.30 Cross multiply
3.30 = 6.02 x Divide by 6.02
3.30 / 6.02 = x 
x = 0.548 moles 

what to do from here?

1 mole of Ni2I6 is
2 * Ni = 2 * 59 = 118 grams
6 *  I = 6 * 131 = <u>786 grams</u>
Total  =               904 grams

Set up a proportion.
<u>1 mole Ni2I6 </u>  = <u>904 grams</u>  
0.548 moles    = x

1/0.548 = 904/x Cross multiply
x = 0.548 * 904
x = 495.4 grams of Ni2I6 <<<<<< Answer.

8 0
3 years ago
Which of the following is unchanged at the end of the CNO cycle?
Musya8 [376]

Answer:

The correct option is: A. carbon-12

Explanation:

The CNO cycle, the abbreviation for the carbon-nitrogen-oxygen cycle, is a catalytic cycle by which the stars produce helium from elemental hydrogen, via a series of nuclear fusion reactions.

This cycle involves the fusion of four protons with carbon (_{6}^{12}\textrm{C}), nitrogen isotope (_{7}^{13}\textrm{N}), and oxygen isotope (_{8}^{15}\textrm{O}), to give an alpha particle and two electron neutrinos and positrons.

The reaction involves the regeneration of carbon (_{6}^{12}\textrm{C}) nucleus in the last step.

_{6}^{12}\textrm{C} → _{7}^{13}\textrm{N} → _{6}^{13}\textrm{C} → _{7}^{14}\textrm{N} → _{8}^{15}\textrm{O} → _{7}^{15}\textrm{N} → _{6}^{12}\textrm{C}

3 0
3 years ago
What type of reaction is P4 + 3O2 = 2P2O3
valentina_108 [34]

Answer:

synthesis

Explanation:

hope this helps

4 0
3 years ago
Of rational function<br>1<br>3. + 3 &lt; 0<br>3x-1<br>nctions.​
pshichka [43]

Answer:

I didn't understand questions properly. Sorry

4 0
2 years ago
The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed
Oksanka [162]

Answer:

(a) The rate of disappearance of O_{2} is: 4.65*10^{-5} M/s

(b) The value of rate constant is: 0.83036 M^{-2}s^{-1}

(c) The units of rate constant is:  M^{-2}s^{-1}

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

2NO(g)+O_{2}->2NO_{2}

rate = -\frac{1}{2} \frac{d}{dt}[NO] = -\frac{d}{dt}[O_{2}] = \frac{1}{2}\frac{d}{dt}[NO_{2}] -----(1)

According to the question, the reaction is second order in NO and first order in  O_{2}.

Then we can say that, rate = k[NO]^{2}[O_{2}] -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

-\frac{d}{dt}[NO] = 9.3*10^{-5} M/s.

(a) From (1), we can get the rate of disappearance of O_{2}.

    Rate of disappearance of  O_{2} = -\frac{d}{dt}[O_{2}] = (0.5)*(9.3*10^{-5}) M/s = 4.65*10^{-5} M/s.

(b) The rate of the reaction can be obtained from (1).

    rate = -\frac{1}{2} \frac{d}{dt}[NO] = (0.5)*(9.3*10^{-5})

    rate = 4.65*10^{-5} M/s

   The value of rate constant can be obtained by using (2).

    rate constant = k = \frac{rate}{[NO]^{2}[O_{2}]}

    k = \frac{4.65*10^{-5}}{(0.040)^{2}(0.035)} = 0.83036 M^{-2}s^{-1}

(c) The units of the rate constant can be obtained from (2).

    k = \frac{rate}{[NO]^{2}[O_{2}]}

    Substituting the units of rate as M/s and concentrations as M, we get:

\frac{Ms^{-1} }{M^{3}} = M^{-2}s^{-1}

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

     rate\alpha [NO]^{2}

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of (1.8)^{2} = 3.24

     

5 0
3 years ago
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