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Jlenok [28]
2 years ago
12

Glucose is a simple sugar made up of carbon, hydrogen, and oxygen. What kind of molecule is

Chemistry
2 answers:
Flauer [41]2 years ago
4 0
The answer is <span>C. Glucose is an organic molecule.

Glucose is carbohydrate so it cannot be a protein or a nucleic acid. It is an organic molecule. The organic molecule is compound consisting of carbon to which are attached hydrogen, oxygen, and nitrogen. Since, g</span><span>lucose is a simple sugar made up of carbon, hydrogen, and oxygen, it, as well as any other carbohydrate, is the organic molecule.</span>
charle [14.2K]2 years ago
3 0

Answer:

C

Explanation:

Edg

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Balance the following redox reaction in acidic solution: H+(aq)+Zn(s)→H2(g)+Zn2+(aq)
erma4kov [3.2K]
2H⁺(aq) + Zn(s) ----> H2(g) + Zn²⁺(aq)

2H⁺(aq) +2e⁻ ----->H2(g)            reduction
<span><u>Zn(s)            ------>Zn²⁺ +2e⁻      oxidation</u>
</span>2H⁺(aq) + Zn(s) -----> H2(g) + Zn²⁺
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3 years ago
PLEASE HELP!!!!!!!!!!!!!!!!!!!!!! DUE TOMORROW 40 POINTS!!!!!!!!!!!!!!!!!!
ollegr [7]

Answer: I believe this is the answer: The vertical columns on the periodic table are called groups or families because of their similar chemical behavior. All the members of a family of elements have the same number of valence electrons and similar chemical properties. The horizontal rows on the periodic table are called periods.

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3 years ago
Read 2 more answers
Cacti and succulents are better adapted
masha68 [24]

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answer is C

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8 0
3 years ago
For each of the following ionic bonding examples, show the transfer of electrons, the ions resulting with the charges and the fi
fgiga [73]

Answer:

BaO

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Al₂O₃

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6 0
2 years ago
An enzyme with molecular weight of 310 kDa undergoes a change in shape when the substrate binds. This change can be characterize
oee [108]

Answer:

(a) r = 6.26 * 10⁻⁷cm

(b) r₂ = 6.05 * 10⁻⁷cm

Explanation:

Using the sedimentation coefficient formula;

s =  M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle

s = M ( 1 - Vρ) / N*6πnr

making r sbjct of formula, r =  M (1 - Vρ) / N*6πnrs

Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s

r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)

r = 6.26 * 10⁻⁷cm

b. Using the formula r₂/r₁ = s₁/s₂

s₂ = 0.035 + 1s₁ = 1.035s₁

making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁

r₂ = 6.3 * 10⁻⁷cm / 1.035

r₂ = 6.05 * 10⁻⁷cm

8 0
3 years ago
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