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3 years ago

According to the concept of the tragedy of the commons, what happens to shared resources over time?

Chemistry

2 answers:

mario62 [17]3 years ago

6 0

Another APEX Answer:

They are overused.

Maslowich3 years ago

5 0

I would believe the answer to this question is D. According to the concept of the tragedy of the commons, shared resources are used by more than one organism. Due to the large consumption of shared resources they start to be fewer and fewer in number and over time if we are not careful they will be depleted.

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John dalton proposed a theory that all matter is made up of individuals particles which cannot be divided. what were those parti

juin [17]

The particles are atoms, his theory is the atomic theory

3 0

3 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 $M^{-2}s^{-1}$

(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 $M^{-2}s^{-1}$

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

Vitamin C contains the elements C, H, and O. It is known to contain 40.9% C and 4.58% H by mass. The molar mass of vitamin C has

soldier1979 [14.2K]

Answer:

C₆H₈O₆

Explanation:

First off, the percent of oxygen by mass of vitamin C is:

100 - (40.9+4.58) = 54.52 %

Assume we have one mol of vitamin C. Then we would have 180 grams, of which:

180 * 40.9/100 = 73.62 grams are of Carbon

180 * 4.58/100 = 8.224 grams are of Hydrogen

180 * 54.52/100 = 98.136 grams are of Oxygen

Now we convert each of those masses to moles, using the elements' respective atomic mass:

C ⇒ 73.62 g ÷ 12 g/mol = 6.135 mol C ≅ 6 mol C

H ⇒ 8.224 g ÷ 1 g/mol = 8.224 mol H ≅ 8 mol H

O ⇒ 98.136 g ÷ 16 g/mol = 6.134 mol O ≅ 6 mol O

So the molecular formula for vitamin C is C₆H₈O₆

8 0

3 years ago

What is the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen

natta225 [31]

Carbon(C):
number of moles= mass/molar mass(Mr)
=65.5/12
=5.5 moles

Hydrogen(H):
number of moles=mass/molar mass (Mr)
=5.5/1
=5.5 moles

Oxygen (O):
number of moles = mass/molar mass (Mr)
=29.0/16
=1.8 moles

EF= lowest number of moles over each of the elements

So,
C= 5.5/1.8 = 3
H= 5.5/1.8 = 3
O= 1.8/1.8 = 1

Therefore Emperical formula= C3H3O

6 0

2 years ago

A sample of lemon juice is found to have a pH of 2.3. What is the concentration of hydrogen ions in the lemon juice?

Elenna [48]

Answer: It's equal to 10^(-2.3), or 0.00501 M, or 5.01 * 10^-3 moles/Liter

Explanation:

Well, pH = - log[H+]

Or, in words, pH is equal to -1 multiplied by the logarithm (base 10) of the hydrogen ion concentration. So you have 2.3 = -log[H+]. We want to isolate the H+, so let's start simplifying the right hand side of the equation. First, we multiply both sides by -1. -2.3=log[H+] Now, the definition of a logarithm says that if the log (base 10) of [H+] is -2.3, then 10 raised to the -2.3 power is [H+] So on each side of the equation, we raise 10 to the power of that side of the equation. 10^(-2.3) = 10^(log[H+]) and because 10^log cancels out... 10^(-2.3) = [H+] Now we've solved for [H+], the hydrogen ion concentration!

7 0

3 years ago