Before............................
I think it is d because it talks about other things the other ones finest talk about so go with d or c I hope you get it right good luck
The answer is true . hope this helps!
Answer:
Carbon dioxide is a very soluble gas. It dissolves readily in water. As the oceans formed, carbon dioxide dissolved to form soluble carbonate compounds so its amount in the atmosphere decreased. Carbonate compounds were then precipitated as sedimentary rocks, eg limestone.
Uptake by living organisms
Carbon dioxide was also absorbed from the oceans into photosynthetic algae and plants. Many of these organisms, and the simple organisms in the food chains that they supported were turned into fossil fuels, eg crude oil, coal and natural gas, which all contain carbon.
Coal is a fossil fuel which was formed from trees which were in dense forests in low-lying wetland areas. Flooding caused the wood from these forests to be buried in a way that prevented oxidation taking place. Compression and heating over millions of years turned the wood into coal.
Crude oil and natural gas were formed from simple plants and tiny animals which were living in oceans and lakes. These small organisms died and their remains sank to the bottom where they were buried under sediments. The lack of oxygen prevented oxidation from occurring.
Over millions of years, heat and pressure turned the remains of the organisms into crude oil and natural gas. Natural gas contains the smallest molecules and is often found on top of crude oil, trapped under sedimentary rock.
Hope this helps..Pls mark as Brainliest!!
Answer:
1. 35 mg of H₃PO₄
2. 27 mol AlF₃; 82 mol F⁻
3. 300 mL of stock solution.
Explanation:
1. Preparing a solution of known molar concentration
Data:
V = 80 mL
c = 4.5 × 10⁻³ mol·L⁻¹
Calculations:
(a) Moles of H₃PO₄
Molar concentration = moles of solute/litres of solution
c = n/V
n = Vc = 0.080L × (4.5 × 10⁻³ mol/1 L) = 3.60 × 10⁻⁴ mol
(b) Mass of H₃PO₄
moles = mass/molar mass
n = m/MM
m = n × MM = 3.60 × 10⁻⁴ mol × (98 g/1 mol) = 0.035 g = 35 mg
(c) Procedure
Dissolve 35 mg of solid H₃PO₄ in enough water to make 80 mL of solution,
2. Moles of solute.
Data:
V = 4900 mL
c = 5.6 mol·L⁻¹
Calculations:
Moles of AlF₃ = cV = 4.9 L AlF₃ × (5.6 mol AlF₃/1L AlF₃) = 27 mol AlF₃
Moles of F⁻ = 27 mol AlF₃ × (3 mol F⁻/1 mol AlF₃) = 82 mol F⁻.
3. Dilution calculation
Data:
V₁= 750 mL; c₁ = 0.80 mol·L⁻¹
V₂ = ? ; c₂ = 2.0 mol·L⁻¹
Calculation:
V₁c₁ = V₂c₂
V₂ = V₁ × c₁/c₂ = 750 mL × (0.80/2.0) = 300 mL
Procedure:
Measure out 300 mL of stock solution. Then add 500 mL of water.
