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IgorC [24]
2 years ago
14

Nitrous oxide (N2O) is used as an anesthetic (laughing gas) and in aerosol cans to produce whipped cream. It is also a potent gr

eenhouse gas. The decomposition of N2O is first order. It decomposes slowly to N2 and O2:
2 N2O(g) --> 2 N2(g) + O2(g)

How many half-lives will it take for the concentration of the N20 to reach 3.13% of its original concentration?
Chemistry
1 answer:
scZoUnD [109]2 years ago
3 0

Answer:

five half lives

Explanation:

Half-life is the time required for a quantity to reduce to half of its initial value.

How many half lives it would take to reach 3.13% form 100% of it's initial concentration:

100% - 50% : First Half life

50% - 25%: Second Half life

25% - 12.5%: Third Half life

12.5% - 6.25%: Fourth Half life

6.25% - 3.125%: Fifth Half life

This means it would take five half lives to get to 3.125% (≈ 3.13%) of it's original concentration.

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Write a hypothesis to answer the lesson question
stellarik [79]

Answer:

Electric current

Hydrogen and oxygen gas

Explanation:

Electrolysis is defined as a method of separation whereby an electric current is introduced into compounds to separate them . The electric current used here is a direct electric current (DC) and it makes the compound to be spit into its' component elements.

Now, in the case of water like it says in the question. Water is a mixture of hydrogen and oxygen gas. Thus, when electric current is passed through water which has an Electrolyte, it makes the water to decompose into its compound elements which are hydrogen and oxygen.

8 0
2 years ago
Which location below would be considered a microhabitat
Anettt [7]

Answer:

Where is the pic or sum ? How can i tell ?

Explanation:

8 0
3 years ago
Find the mass of oxygen in grams produced by the decomposition of 100.0 g of CO2
SSSSS [86.1K]

The balanced chemical equation is :

2CO_2->2CO+O_2\\\\

Moles of CO_2 ,

n = \dfrac{100\ g}{44.01\ g/mol}\\\\n=2.27\ mol

Now, by given chemical equation , we can see 2 mole of CO_2 react with 1 mole of O_2.

So , 2.27 mole react with :

N=\dfrac{2.27}{2}\ mol\\\\N=1.135\ mol

Mass of oxygen is :

M = N \times 16\\\\M=1.135\times 16\ g\\\\M =18.16\ g

Therefore, mass of oxygen in grams produced is 18.16 g.

Hence, this is the required solution.

7 0
3 years ago
A nitric acid solution flows at a constant rate of 5L/min into a large tank that initially held 200L of a 0.5% nitric acid solut
leonid [27]

Answer:

x(t) = −39e

−0.03t + 40.

Explanation:

Let V (t) be the volume of solution (water and

nitric acid) measured in liters after t minutes. Let x(t) be the volume of nitric acid

measured in liters after t minutes, and let c(t) be the concentration (by volume) of

nitric acid in solution after t minutes.

The volume of solution V (t) doesn’t change over time since the inflow and outflow

of solution is equal. Thus V = 200 L. The concentration of nitric acid c(t) is

c(t) = x(t)

V (t)

=

x(t)

200

.

We model this problem as

dx

dt = I(t) − O(t),

where I(t) is the input rate of nitric acid and O(t) is the output rate of nitric acid,

both measured in liters of nitric acid per minute. The input rate is

I(t) = 6 Lsol.

1 min

·

20 Lnit.

100 Lsol.

=

120 Lnit.

100 min

= 1.2 Lnit./min.

The output rate is

O(t) = (6 Lsol./min)c(t) = 6 Lsol.

1 min

·

x(t) Lnit.

200 Lsol.

=

3x(t) Lnit.

100 min

= 0.03 x(t) Lnit./min.

The equation is then

dx

dt = 1.2 − 0.03x,

or

dx

dt + 0.03x = 1.2, (1)

which is a linear equation. The initial condition condition is found in the following

way:

c(0) = 0.5% = 5 Lnit.

1000 Lsol.

=

x(0) Lnit.

200 Lsol.

.

Thus x(0) = 1.

In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is

µ(t) = exp Z

P(t) dt

= exp

0.03 Z

dt

= e

0.03t

.

The solution is

x(t) = 1

µ(t)

Z

µ(t)Q(t) dt + C

= Ce−0.03t + 1.2e

−0.03t

Z

e

0.03t

dt

= Ce−0.03t +

1.2

0.03

e

−0.03t

e

0.03t

= Ce−0.03t +

1.2

0.03

= Ce−0.03t + 40.

The constant is found using x(t) = 1:

x(0) = Ce−0.03(0) + 40 = C + 40 = 1.

Thus C = −39, and the solution is

x(t) = −39e

−0.03t + 40.

3 0
2 years ago
The CO2 produced in one round of the citric acid cycle does not originate in the acetyl carbons that entered that round. If the
aleksley [76]

Answer:

It will require<u> second round</u> of the cycle to release 14C0_2

Explanation:

<u>Reason behind the requirement of second round of the cycle to release </u>CO_2 -:

The C4 carbon of succinyl CoA is acetyl from acetyl CoA. Succinyl CoA is converted to succinate, which is then converted to fumarate, fumarate, malate, and eventually oxaloacetate. 14C will be found in oxaloacetate at either C1 or C4. During the second round of the loop, each of these carbons will be converted to carbon dioxide.

7 0
2 years ago
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