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snow_lady [41]
3 years ago
10

How can you form new neural pathways that will lead to transformed thinking and a transformed life?

Chemistry
1 answer:
aleksandr82 [10.1K]3 years ago
6 0
When you understand how neural pathways<span> are created in the brain, you ... But because I had the </span>will<span> to do it, I built a </span>new<span> pathway, and I rewired or reprogrammed my brain. ... can learn </span>new<span> behaviors and attitudes and can </span>transform<span> their </span>lives<span>. ... to the habit, and see what results you're </span>creating<span> in your </span><span>life</span>
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Use the reaction below for the decomposition of sodium azide
bezimeni [28]

<u>Answer:</u> The volume of nitrogen gas at STP is 44.8 L.

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation for it follows:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}                    ...(1)

We are given:

Given mass of NaN_3 = 130.0 g

Molar mass of NaN_3 = 65.01 g/mol

Using equation 1:

Moles of NaN_3=\frac{130.0 g}{65.01g/mol}=2mol

For the given chemical equation:

2NaN_3(s) \rightarrow 2Na(s) + 3N_2(g)

By the stoichiometry of the reaction:

2 moles of NaN_3 produces 3 moles of nitrogen gas

At STP:

1 mole of a gas occupies 22.4 L of volume

So, 2 moles of nitrogen gas will occupy =\frac{22.4L}{1mol}\times 2mol=44.8L of volume

Hence, the volume of nitrogen gas at STP is 44.8 L.

8 0
3 years ago
Which three groups of the periodic table contain the most elements classified as transition elements
pantera1 [17]
The question contain these choices:
A) groups 1, 16 and 18
B) groups 5, 7 and 9
C) groups 2, 3 and 4
D) groups 15, 16 and 17
The correct answer is C) groups 2, 3 and 4
In the periodic table groups of transition elements are from group 3 to group 12
so here we have in C) group 3 which contain lanthanides and actinides so this choice contain the most elements classified as transition elements 
7 0
3 years ago
Read 2 more answers
In which type of reaction is ATP the driving force?
valentina_108 [34]
C in both anabolism and catabolism
7 0
3 years ago
Read 2 more answers
A. The reactant concentration in a zero-order reaction was 8.00×10−2 M after 155s and 3.00×10−2 M after 355s . What is the rate
irga5000 [103]

Answer:

A) The rate constant is 2.50 × 10⁻⁴ M/s.

B) The initial concentration of the reactant is 11.9 × 10⁻² M.

C) The rate constant is 0.0525 s⁻¹

D) The rate constant is 0.0294 M⁻¹ s⁻¹

Explanation:

Hi there!

A) The equation for a zero-order reaction is the following:

[A] = -kt + [A₀]

Where:

[A] = concentrationo f reactant A at time t.

[A₀] = initial concentration of reactant A.

t = time.

k = rate constant.

We know that at t = 155 s, [A] = 8.00 × 10⁻² M and at t = 355 s [A] = 3.00 × 10⁻² M. Then:

8.00 × 10⁻² M = -k (155 s) +  [A₀]

3.00 × 10⁻² M = -k (355 s) + [A₀]

We have a system of 2 equations with 2 unknowns, let´s solve it!

Let´s solve the first equation for [A₀]:

8.00 × 10⁻² M = -k (155 s) +  [A₀]

8.00 × 10⁻² M + 155 s · k = [A₀]

Replacing [A₀] in the second equation:

3.00 × 10⁻² M = -k (355 s) + [A₀]

3.00 × 10⁻² M = -k (355 s) + 8.00 × 10⁻² M + 155 s · k

3.00 × 10⁻² M - 8.00 × 10⁻² M = -355 s · k + 155 s · k

-5.00 × 10⁻² M = -200 s · k

-5.00 × 10⁻² M/ -200 s = k

k = 2.50 × 10⁻⁴ M/s

The rate constant is 2.50 × 10⁻⁴ M/s

B) The initial reactant conentration will be:

8.00 × 10⁻² M + 155 s · k = [A₀]

8.00 × 10⁻² M + 155 s · 2.50 × 10⁻⁴ M/s = [A₀]

[A₀] = 11.9 × 10⁻² M

The initial concentration of the reactant is 11.9 × 10⁻² M

C) In this case, the equation is the following:

ln[A] = -kt + ln([A₀])

Then:

ln(7.60 × 10⁻² M) = -35.0 s · k + ln([A₀])

ln(5.50 × 10⁻³ M) = -85.0 s · k + ln([A₀])

Let´s solve the first equation for ln([A₀]) and replace it in the second equation:

ln(7.60 × 10⁻² M) = -35.0 s · k + ln([A₀])

ln(7.60 × 10⁻² M) + 35.0 s · k = ln([A₀]

Replacing ln([A₀]) in the second equation:

ln(5.50 × 10⁻³ M) = -85.0 s · k + ln([A₀])

ln(5.50 × 10⁻³ M) = -85.0 s · k + ln(7.60 × 10⁻² M) + 35.0 s · k

ln(5.50 × 10⁻³ M) -  ln(7.60 × 10⁻² M) = -85.0 s · k + 35.0 s · k

ln(5.50 × 10⁻³ M) -  ln(7.60 × 10⁻² M) = -50.0 s · k

(ln(5.50 × 10⁻³ M) -  ln(7.60 × 10⁻² M)) / -50.0 s = k

k = 0.0525 s⁻¹

The rate constant is 0.0525 s⁻¹

D) In a second order reaction, the equation is as follows:

1/[A] = 1/[A₀] + kt

Then, we have the following system of equations:

1/ 0.510 M = 1/[A₀] + 205 s · k

1/5.10 × 10⁻² M = 1/[A₀] + 805 s · k

Let´s solve the first equation for 1/[A₀]:

1/ 0.510 M = 1/[A₀] + 205 s · k

1/ 0.510 M - 205 s · k = 1/[A₀]

Now let´s replace 1/[A₀] in the second equation:

1/5.10 × 10⁻² M = 1/[A₀] + 805 s · k

1/5.10 × 10⁻² M = 1/ 0.510 M - 205 s · k + 805 s · k

1/5.10 × 10⁻² M - 1/ 0.510 M = - 205 s · k + 805 s · k

1/5.10 × 10⁻² M - 1/ 0.510 M = 600 s · k

(1/5.10 × 10⁻² M - 1/ 0.510 M)/ 600 s = k

k = 0.0294 M⁻¹ s⁻¹

The rate constant is 0.0294 M⁻¹ s⁻¹

8 0
3 years ago
Write qc for the reaction of hydrogen chloride gas with oxygen gas to produce chlorine gas and water vapor.
ziro4ka [17]
Answer : When the Hydrogen chloride gas reacts with oxygen to form chlorine gas and water vapor the reaction can be as follows,

4 HCl _{(g)} + O_{2} _{(g)}  ----\ \textgreater \  2Cl_{2} _{(g)} + 2H_{2}O _{(g)}

Q_{c} = [[Cl_{2}] ^{2}      [H_{2}O]^{2}  /  [HCL]^{4}  [O_{2}].
7 0
3 years ago
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