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igor_vitrenko [27]
3 years ago
8

(blank) from digested carbohydrates is an important substance that contains stored chemical energy.

Chemistry
1 answer:
Debora [2.8K]3 years ago
5 0
E because glucose is a very important energy to help an organism
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Formulación del dióxido de cobalto
Svetlanka [38]

Answer:

Co2O3 esa es la formula creo espero que te sirva

4 0
2 years ago
How many grams of NaOH are produced from 20.0 grams of Na2CO3?
natita [175]

Answer:

Hope this helps!

Explanation:

Ans: 15.1 grams

Given reaction:

Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3

Mass of Na2CO3 = 20.0 g

Molar mass of Na2CO3 = 105.985 g/mol

# moles of Na2CO3 = 20/105.985 = 0.1887 moles

Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH

# moles of NaOH produced = 0.1887*2 = 0.3774 moles

Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol

Mass of NaOH produced = 0.3774*39.996 = 15.09 grams

4 0
3 years ago
At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average power of approximately 8.00
Archy [21]
Energy transferred per second in given area:
8000 x 3.9 / 10000
Energy transferred = 3.12 J

Using Planck's equation:
E = nhc/λ
n = (3.12 x 510 x 10⁻⁹) / (3 x 10⁸ x 6.63 x 10⁻³⁴)
n = 8 x 10¹⁸ photons
3 0
3 years ago
Reactants → products
uranmaximum [27]

Answer:

they are equal

Explanation:

the Law of Conservation of Mass states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time, as the system's mass cannot change

3 0
3 years ago
When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed?
Ad libitum [116K]

You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.

For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.

6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2

For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.

13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2

As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.

In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!

6 0
3 years ago
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