4 Living organisms contain relatively large amounts of oxygen, carbon, hydrogen,nitrogen, and sulfur (these five elements are known as the bulk elements), along with sodium, magnesium, potassium, calcium,chlorine, and phosphorus (these six elements are known as macro minerals).
Answer:
0.940mol &
0.000301mol respectively.
Explanation:
number of moles = given mass / molar mass
given mass of Nacl = 55g Molar mass = 23 + 35.5
n=m/M = 55g/58.5g/mol = 0.940mol
note- (add the atomic weights of sodium and chlorine to get the molar mass of Nacl.) = 58.5g/mol
similarly, NaCO3 = 23 + 12 + 16*3 = 83g/mol
n=m/M = 0.025g/83g/mol = 3.01 * 10^-4 = 0.000301mol
extra: If you ever get asked to put it in number of particles just use the relation of 1mole = 6.02 * 10^23 particles.
The question is incomplete as it does not have the options which are:
A) photosystem II
B) photosystem I
C) cyclic electron flow
D) linear electron flow
E) chlorophyll
Answer:Cyclic electron flow
Explanation:
The plants produce ATP molecule by the process of a light-dependent phase of photosynthesis which produces ATP and NADPH molecules.
The ATP and NADPH are produced by the non-cyclic flow of electrons called Z-scheme but when the plant needs extra ATP molecules, they produce more ATP by the cyclic electron flow.
The cyclic electron flow begins when the P₇₀₀ activates electron which then transferred to ferredoxin and then to cyt b₆f and then to plastocyanin. This is repeated and produce ATP molecules escaping the production of NADPH.
Thus, Cyclic electron flow is correct.
The mass in grams of NH₃ produced from the reaction is 3.4 g
<h3>Balanced equation</h3>
We'll begin by writing the balanced equation for the reaction. This illustrated below:
N₂ + 3H₂ -> 2NH₃
From the balanced equation above,
1 dm³ of N₂ reacted to produced 2 dm³ NH₃
<h3>How to determine the volume of NH₃ produced</h3>
From the balanced equation above,
1 dm³ of N₂ reacted to produced 2 dm³ NH₃
Therefore,
2.24 dm³ of N₂ will react to produce = 2.24 × 2 = 4.48 dm³ of NH₃
<h3>How to determine the mass of NH₃ produced</h3>
We'll begin by obtained the mole of 4.48 dm³ of NH₃. Details below:
22.4 dm³ = 1 mole NH₃
Therefore,
4.48 dm³ = 4.48 / 22.4
4.48 dm³ = 0.2 mole of NH₃
Finally, we shall determine the mass of NH₃ as follow:
- Molar mass of NH₃ = 17 g/mol
- Mole of NH₃ = 0.2 mole
- Mass of NH₃ =?
Mass = mole × molar mass
Mass of NH₃ = 0.2 × 17
Mass of NH₃ = 3.4 g
Learn more about stoichiometry:
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