B.sunligh
is reflected off the upper atmosphere
Answer:
molarity of acid =0.0132 M
Explanation:
We are considering that the unknown acid is monoprotic. Let the acid is HA.
The reaction between NaOH and acid will be:
Thus one mole of acid will react with one mole of base.
The moles of base reacted = molarity of NaOH X volume of NaOH
The volume of NaOH used = Final burette reading - Initial reading
Volume of NaOH used = 22.50-0.55= 21.95 mL
Moles of NaOH = 0.1517X21.95=3.33 mmole
The moles of acid reacted = 3.33 mmole
The molarity of acid will be = 
Answer:
grams H₂O produced = 8.7 grams
Explanation:
Given 2C₂H₆(g) + 7O₂(g) => 4CO₂(g) + 6H₂O(l)
7g 18g ?g
Plan => Convert gms to moles => determine Limiting reactant => solve for moles water => convert moles water to grams water
Moles Reactants
moles C₂H₆ = 7g/30g/mol = 0.233mol
moles O₂ = 18g/32g/mol = 0.563mol
Limiting Reactant => (Test for Limiting Reactant) Divide mole value by respective coefficient of balanced equation; the smaller number is the limiting reactant.
moles C₂H₆/2 = 0.233/2 = 0.12
moles O₂/7 = 0.08
<u><em>Limiting Reactant is O₂</em></u>
Moles and Grams of H₂O:
Use Limiting Reactant moles (not division value) to calculate moles of H₂O.
moles H₂O = 6/7(moles O₂) = 6/7(0.562) moles H₂O = 0.482 mole H₂O yield
grams H₂O = (0.482mol)(18g·mol⁻¹) = 8.7 grams H₂O
Climate is considered an abiotic limiting factor because it is non living . hope this helped :))
Answer is: formula of hydrate is CoCl₂· 6H₂O -c<span>obalt(II) chloride hexahydrate
</span>m(CoCl₂· xH₂O) = 1,62 g.
m(CoCl₂) = 0,88 g.
n(CoCl₂) = m(CoCl₂) ÷ M(CoCl₂)
n(CoCl₂) = 0,88 g ÷ 130 g/mol
n(CoCl₂) = 0,0068 mol.
m(H₂O) = 1,62 g - 0,88 g.
m(H₂O) = 0,74 g.
n(H₂O) = m(H₂O) ÷ m(H₂O)
n(H₂O) = 0,74 g ÷ 18 g/mol
n(H₂O) = 0,041 mol.
n(CoCl₂) : n(H₂O) = 0,0068 mol : 0,041 mol.
n(CoCl₂) : n(H₂O) = 1 : 6.
