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3 years ago

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.

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.

$N_o=180 mg$

$t_{1/2}= 30 yeras$

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