One of the many awe-inspiring things about algae, Professor Greene explains, is that they can grow between ten and 100 times faster than land plants. In view of this speedy growth rate – combined with the fact they can thrive virtually anywhere in the right conditions – growing marine microalgae could provide a variety of solutions to some of the world’s most pressing problems.
Take, global warming. Algae sequesters CO2, as we have learned, but owing to the fact they grow faster than land plants, can cover wider areas and can be utilised in bioreactors, they can actually absorb CO2 more effectively than land plants. AI company Hypergiant Industries, for instance, say their algae bioreactor was 400 times more efficient at taking in CO2 than trees.
And it’s not just their nutritional credentials which could solve humanity’s looming food crisis, but how they are produced. Marine microalgae grow in seawater, which means they do not rely on arable land or freshwater, both of which are in limited supply. Professor Greene believes the use of these organisms could therefore release almost three million km2 of cropland for reforestation, and also conserve one fifth of global freshwater
1. The empirical formula of the hydrocarbon is CH₃
2. The molecular formula of the hydrocarbon is C₂H₆
<h3>How to determine the mass of Carbon </h3>
- Mass of CO₂ = 1.47 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 1.47
Mass of C = 0.4 g
<h3>How to determine the mass of H</h3>
- Mass of compound = 0.5 g
- Mass of C = 0.4 g
- Mass of H = ?
Mass of H = (mass of compound) – (mass of C)
Mass of H = 0.5 – 0.4
Mass of H =0.1 g
<h3>1. How to determine the empirical formula </h3>
- C = 0.4 g
- H = 0.1 g
- Empirical formula =?
Divide by their molar mass
C = 0.4 / 12 = 0.03
H = 0.1 / 1 = 0.1
Divide by the smallest
C = 0.03 / 0.03 = 1
H = 0.1 / 0.03 = 3
Thus, the empirical formula of the compound is CH₃
<h3>2. How to determine the molecular formula</h3>
- Empirical formula = CH₃
- Molar mass = 30 g/mol
- Molecular formula =?
Molecular formula = empirical × n = mass number
[CH₃]n = 30
[12 + (3×1)]n = 30
15n = 30
Divide both side by 15
n = 30 / 15
n = 2
Molecular formula = [CH₃]n
Molecular formula = [CH₃]₂
Molecular formula = C₂H₆
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Flamability boiling point color
Answer:
I Believe it is 4 orbitals s,p,p,p or aka sp^3
Explanation:
Answer: Ag^+ (aq) + NO3^-(aq) + Na^+ (aq) +Cl^- (aq) ⇒ AgCl(s) + NaNO3(aq)
