<span>a) movement of heat through space </span>
Answer:
The Gulf Stream is influential on the climate of the Florida peninsula.East winds moving over this warm water move warm air from over the Gulf Stream inland, helping to keep temperatures milder across the state than elsewhere across the Southeastern United States during the winter.
Explanation:
hope this helps!
The precipitate that is most likely formed from a solution containing Ba+2, Li+, OH-1, and CO3^-2 is BaCO3.
This is because carbonates of all metals except sodium, Lithium potassium (group 1) and ammonium are insoluble in water. Hydroxides of sodium, Lithium, potassium and ammonium are very soluble in water, calcium and barium are moderately soluble. Ba(CO3) is insoluble in water and therefore forms a precipitate.
Answer:
1) 1.235 g.
2) 0.61 g.
Explanation:
- From the balanced equation:
<em>Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O.</em>
1.0 mol of Al(OH)₃ reacts with 3.0 moles of HCl to produce 1.0 mol of AlCl₃ and 3.0 moles of H₂O.
<em>1) How many grams of HCl can a tablet with 0.880 g of Al(OH)₃ consume? </em>
- To calculate the amount of HCl needed to consume 0.880 g of Al(OH)₃, we need to calculate the no. of moles of Al(OH)₃:
no. of moles of Al(OH)₃ = mass/molar mass = (0.880 g)/(78.0 g/mol) = 1.13 x 10⁻² mol.
∵ Every 1.0 mol of Al(OH)₃ needs 3.0 moles of HCl to be consumed.
∴ 1.13 x 10⁻² mol of Al(OH)₃ needs (3 x 1.13 x 10⁻² = 3.385 x 10⁻² mol) of HCl.
The no. of grams of HCl = no. of moles of HCl x molar mass of HCl = (3.385 x 10⁻² mol)(36.5 g/mol) = 1.235 g.
<em>2) How much H₂O?</em>
∵ Every 1.0 mol of Al(OH)₃ produces 3.0 moles of H₂O.
∴ 1.13 x 10⁻² mol of Al(OH)₃ produces (3 x 1.13 x 10⁻² = 3.385 x 10⁻² mol) of H₂O.
<em>The no. of grams of H₂O = no. of moles of H₂O x molar mass of H₂O </em>= (3.385 x 10⁻² mol)(18.0 g/mol) = <em>0.6092 g ≅ 0.61 g.</em>
Answer:
It takes 5.83s to decrease the concentration of the reactant from 0.537M to 0.100M
Explanation:
A zero-order reaction follows the equation:
[A] = [A]₀ - kt
<em>Where [A] is actual reaction of the reactant = 0.100M</em>
<em>[A]₀ the initial concentration = 0.537M</em>
<em>k is rate constant = 0.075Ms⁻¹</em>
<em>And t is time it takes:</em>
<em />
0.100M = 0.537M -0.075Ms⁻¹t
-0.437M = -0.075Ms⁻¹t
5.83s = t
It takes 5.83s to decrease the concentration of the reactant from 0.537M to 0.100M
