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Nastasia [14]
3 years ago
10

The process of cellular respiration, which converts simple sugars such as glucose into CO2 and water, is an example of _____. Se

e Concept 8.1 (Page 144) View Available Hint(s) The process of cellular respiration, which converts simple sugars such as glucose into CO2 and water, is an example of _____. See Concept 8.1 (Page 144) a pathway in which the entropy of the system decreases a catabolic pathway an endergonic pathway a pathway that occurs in animal cells but not plant cells a pathway that converts organic matter into energy
Chemistry
1 answer:
Paha777 [63]3 years ago
7 0

a catabolic pathway. Cellular respiration is a catabolic pathway.  

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The reaction for the combustion of methane is shown below. If 28.70g of methane
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Omg hard question but it’s also amazing imma think about that question
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A 25 liter balloon holding 1.5 moles of carbon dioxide leaks. If we are able to determine that 0.9 moles of carbon dioxide escap
ad-work [718]

Answer:

10 Litre

Explanation:

Given that ::

v1 = 25L ; n1 = 1.5 mole ; v2 =? ; n2 = (1.5-0.9) = 0.6 mole

Using the relation :

(n2 * v1) / n1 = (n2 * v2) / n2

v2 = (n2 * v1) / n1

v2 = (0.6 mole * 25 Litre) / 1.5 mole

v2 = 15 / 1.5 litre

v2 = 10 Litre

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2 years ago
Which of the following is considered a renewable resource?
Mashutka [201]

Explanation:

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1) water

2) biomass

3)Soil

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5 0
3 years ago
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Which pair of elements would most likely form a covalent bond
denpristay [2]

Answer:

Nonmetals and nonmetals tend to form covalent bonds.

or

P and S

Explanation:

5 0
2 years ago
Consider the reaction. 2 HBr(g) ¡ H2(g) + Br2(g) a. Express the rate of the reaction in terms of the change in concentration of
Studentka2010 [4]

Answer :

(A) The rate expression will be:

Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}

(B) The average rate of the reaction during this time interval is, 0.00176 M/s

(C) The amount of Br₂ (in moles) formed is, 0.0396 mol

Explanation :

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The given rate of reaction is,

2HBr(g)\rightarrow H_2(g)+Br_2(g)

The expression for rate of reaction :

\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}

\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}

\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}

<u>Part A:</u>

The rate expression will be:

Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}

<u>Part B:</u>

\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}

\text{Average rate}=-\frac{1}{2}\frac{(0.512-0.600)M}{(25.0-0.0)s}

\text{Average rate}=0.00176M/s

The average rate of the reaction during this time interval is, 0.00176 M/s

<u>Part C:</u>

As we are given that the volume of the reaction vessel is 1.50 L.

\frac{d[Br_2]}{dt}=0.00176M/s

\frac{d[Br_2]}{15.0s}=0.00176M/s

[Br_2]=0.00176M/s\times 15.0s

[Br_2]=0.0264M

Now we have to determine the amount of Br₂ (in moles).

\text{Moles of }Br_2=\text{Concentration of }Br_2\times \text{Volume of solution}

\text{Moles of }Br_2=0.0264M\times 1.50L

\text{Moles of }Br_2=0.0396mol

The amount of Br₂ (in moles) formed is, 0.0396 mol

8 0
3 years ago
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