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nasty-shy [4]
3 years ago
9

What effect does the transfer of electrons have on the nuclei of the atoms involved?

Chemistry
1 answer:
Alex777 [14]3 years ago
6 0

Answer:

Chemical processes have no effect on the nucleus otherwise we would be in deep truble. GOOD LESSONS ♡

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Calculate (a) the number of moles of hydrogen required to react with 0.0969 moles of nitrogen, and (b) the number of moles of am
Bumek [7]

Answer:

The answer to your question is:

Explanation:

Data

moles H=?

moles of N = 0.0969

moles of NH₃=?

                            N₂ (g)  + 3 H₂ (g)  ⇒   2NH₃ (g)

Process

1.- Set a rule of three to calculate the moles of hydrogen

                         1 mol of nitrogen  -------------   3 moles of hydrogen

                         0.0969 moles of N ----------     x

                         x = (0.0969 x 3) / 1

                         x = 0.2907 moles of hydrogen          

2.- Set a rule of three to calculate the moles of ammonia

                        1 mol of nitrogen --------------  2 moles of ammonia

                        0.0969 mol of N --------------   x

                        x = (0.0969 x 2) / 1

                        x = 0.1938 moles of ammonia                

5 0
3 years ago
The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be
Stels [109]

Explanation:

1) Initial mass of the Cesium-137=N_o= 180 mg

Mass of Cesium after time t = N

Formula used :

N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}

Half life of the cesium-137 = t_{1/2}=30 years[p/tex]where,
[tex]N_o = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

t_{\frac{1}{2}} = half life of the isotope

\lambda = rate constant

N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}

Now put all the given values in this formula, we get

N=180mg\times e^{-\frac{0.693}{30 years}\times t}

Mass that remains after t years.

N=180 mg\times e^{0.0231 year^{-1}\times t}

Therefore, the parent isotope remain after one half life will be, 100 grams.

2)

t = 70 years

N_o=180 mg

t_{1/2}= 30 yeras

N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

3)

N_o=180 mg

t_{1/2}= 30 yeras

N = 1 mg

t = ?

1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}

\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

7 0
3 years ago
What is the electron configuration of an element with atomic number 20?
VARVARA [1.3K]
1s2,2s2.2p6,3s2,3p6,3d4,4s2
8 0
3 years ago
Read 2 more answers
Consider the reaction: N2(g) 2 O2(g)N2O4(g) Write the equilibrium constant for this reaction in terms of the equilibrium constan
Valentin [98]

Answer : The equilibrium constant for this reaction is, K=\frac{(K_b)^2}{K_a}

Explanation :

The given main chemical reaction is:

N_2(g)+2O_2(g)\rightarrow N_2O_4(g);  K

The intermediate reactions are:

(1) N_2O_4(g)\rightarrow 2NO_2(g);  K_a

(2) \frac{1}{2}N_2(g)+O_2(g)\rightarrow NO_2(g);  K_b

We are reversing reaction 1 and multiplying reaction 2 by 2 and then adding both reaction, we get:

(1) 2NO_2(g)\rightarrow N_2O_4(g);  \frac{1}{K_a}

(2) N_2(g)+2O_2(g)\rightarrow 2NO_2(g);  (K_b)^2

Thus, the equilibrium constant for this reaction will be:

K=\frac{1}{K_a}\times (K_b)^2

K=\frac{(K_b)^2}{K_a}

Thus, the equilibrium constant for this reaction is, K=\frac{(K_b)^2}{K_a}

5 0
3 years ago
1. If the pressure of a fixed volume of gas decreased in a sealed container, what variable would you think changed? did this var
grandymaker [24]

Answer:

Temp decreased

Explanation:

If the container is sealed and the pressure DEcreases, then the temperature DEcreased

PV = n RT     n  R   V   are constant   if  P goes down then so does T    

8 0
2 years ago
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