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3241004551 [841]
2 years ago
8

Which type of molecule is octanal? A. Alcohol B. Ketone C. Amine D. Aldehyde

Chemistry
1 answer:
tia_tia [17]2 years ago
4 0
It is an aldehyde.

the ending (al) tells you that. alcohols end in “ol”, ketones end in “one”, amines have 2 different ways of naming so i suggest finding which one you want to use, and aldehyde end in “al” :)
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A substance that increases the rate of a chemical reaction is called a(n) _____. answer
rodikova [14]
Catalyst increases the rate of a chemical reaction
3 0
3 years ago
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The difference in an area with high concentration and an area with low concentration is called.
Varvara68 [4.7K]

The difference in an area with high concentration and an area with low concentration is called the concentration gradient.

<h3>What is Concentration Gradient ?</h3>

A concentration gradient occurs when the concentration of particles is higher in one area than another.

In passive transport, particles will diffuse down a concentration gradient, from areas of higher concentration to areas of lower concentration, until they are evenly spaced.

This difference in an area with high concentration and an area with low concentration is called the concentration gradient.

Learn more about diffusion here ;

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5 0
2 years ago
What is the parameter for finding the ∆H of a body...?​
11111nata11111 [884]

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Explanation: hope this helps im sorry if i didnt

7 0
3 years ago
4 g of ag2so4 will dissolve in 1l of water. calculate the solubility product (ksp) for silver (i) sulfate.
Bas_tet [7]
Reaction of dissociation: Ag₂SO₄ → 2Ag⁺ + SO₄²⁻.
m(Ag₂SO₄) = 4 g.
V(Ag₂SO₄) = 1 l.
n(Ag₂SO₄) = m(Ag₂SO₄) ÷ M(Ag₂SO₄).
n(Ag₂SO₄) = 4 g ÷ 311,8 g/mol.
n(Ag₂SO₄) = 0,0128 mol.
n(Ag⁺) = 2 · 0,0128 mol = 0,0256 mol.
n(Ag₂SO₄) = n(SO₄²⁻) = 0,0128 mol.
c(Ag⁺) = n ÷ V = 0,0256 mol ÷ 1 l = 0,0256 mol/l.
Ksp = c(Ag⁺)² · c(SO₄²⁻).
Ksp = (0,0256 mol/l)² · 0,0128 mol/l.
Ksp = 8,3·10⁻⁶.

7 0
3 years ago
How many grams of oxygen are there in 45.7 grams of Ba(NO2),?
Flauer [41]
I’m assuming you mean barium nitrite, Ba(NO2)2.

First convert grams of Ba(NO2)2 to moles using the molar mass of Ba(NO2)2. Then use the mole ratio of 4 moles of oxygen per 1 mole of Ba(NO2)2 to convert to moles of oxygen. Then use the molar mass of oxygen to convert to grams of oxygen.

45.7 g Ba(NO2)2 • 1 mol Ba(NO2)2 / 229.35 g Ba(NO2)2 • 4 mol O / 1 mol Ba(NO2)2 • 16.0 g O / 1 mol O = 12.8 g oxygen
4 0
3 years ago
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