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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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You are planning an investigation that explores properties of matter.
Elena-2011 [213]

Answer:

Placing a powder into a beaker that contains liquid, resulting the beaker to get hotter.

Explanation:

Physical property is something that you can observe that does not affect the mixture/solution/substance, and that includes temperature

7 0
2 years ago
Do you think the reflected waves would ever return to the airplane that transmitted them? explain
Ket [755]
No, it is very unlikely for that to happen.
8 0
2 years ago
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If the half life of iridium-182 is 15 years, how much of a 3 gram sample is left after 2 half-lives?
padilas [110]

Answer:

D. 0.75 grams

Explanation:

The data given on the iridium 182 are;

The half life of the iridium 182, t_{(1/2)} = 15 years

The mass of the sample of iridium, N₀ = 3 grams

The amount left, N(t) after two half lives is given as follows;

N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}} }

For two half lives, t = 2 × t_{(1/2)}

∴ t = 2 × 15 = 30

\dfrac{t}{t_{(1/2)}} = \dfrac{30}{15} = 2

\therefore N(t) = 3 \times\left (\dfrac{1}{2} \right )^2 = 0.75

∴ The amount left, N(t) = 0.75 grams

4 0
2 years ago
What else is produced during the combustion of butane, C4H10?
denpristay [2]

Another product: CO₂

<h3>Further explanation</h3>

Given

Reaction

2C₄H₁₀ + 13O₂⇒ 8__+ 10H₂O

Required

product compound

Solution

In the combustion of hydrocarbons there can be 2 kinds of products

If there is excess Oxygen, you will get Carbon dioxide(CO₂) and water in the product

If Oxygen is low, you'll get Carbon monoxide(CO) and water

Or in other ways, we can use the principle of the law of conservation of mass which is also related to the number of atoms in the reactants and in the products

if we look at the reaction above, there are C atoms on the left (reactants), so that in the product there will also be C atoms with the same number of C atoms on the left

2C₄H₁₀ + 13O₂⇒ 8CO₂+ 10H₂O

5 0
2 years ago
Read 2 more answers
2. Calculate the mass of 3.47x1023 gold atoms.
lapo4ka [179]

3.47 x 10^{23} atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x 10^{23}

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X 10^{23}

from the relation,

1 mole of element contains 6.022 x 10^{23} atoms.

so no of moles of gold given = \frac{3.47 X 10^{23}  }{6.022 X 10^{23} }

0.57 moles of gold.

from the relation:

number of moles = \frac{mass}{atomic mass of 1 mole}

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x 10^{23} atoms of gold have mass of 113.44 grams

3 0
3 years ago
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