175.8 g NaCl to moles:
(175.8 g)/(58.44 g/mol) = 3.008 mol NaCl
Molarity = moles of solute/volume of solution in liters
(3.008 mol NaCl)/(1.5 L) = 2.0 M.
The molarity of this solution would be 2.0 M.
The answer is C to achieve an octet of valence electrons so they become stable
2) carbon and hydrogen only.
Answer:
46.0g of Iron are produced
Explanation:
Based on the chemical reaction:
FeO(l) + Mg(l) → Fe(l) + MgO(s)
<em>1 mole of Iron (II) oxide reacts per mole of Mg to produce 1 mole of iron</em>
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To solve this question we need to convert each mass of reactant to moles using its respectives molar masses in order to find limitng reactant. Moles of limiting reactant = Moles of iron produced:
<em>Moles FeO (Molar mass: 71.85g/mol):</em>
80.0g * (1mol / 71.85g) = 1.11moles FeO
<em>Moles Mg (Molar mass: 24.305g/mol)</em>
20.0g * (1mol / 24.305g) = 0.823 moles Mg
As moles of Mg < Moles FeO, Mg is limiting reactant and the moles of Fe are 0.823 moles.
The mass of Iron produced is:
0.823 moles Fe * (55.845g/mol) =
46.0g of Iron are produced
Answer:
B. 13.7 mm
Explanation:
by similarity of rectangular triangles, we have:
(1) Δ ABC and Δ AB'C'
∴ AB/AB' = AC/AC'
∴ hypotenuse (AB)² = AC² + BC².....Pythagorean theorem
⇒ AB² = (51 mm)² + (13 mm)²
⇒ AB² = 2770 mm²
⇒ AB = 52.63 mm
(2) Δ AB'C' :
∴ hypotenuse (AB')² = AC'² + B'C'²
∴ B'C' = 3.5 mm
from (1):
⇒ AB' = (AB)(AC') / AC = (52.63)(AC') / (51 mm)
⇒ AB' = 1.032AC'.......(3)
(3) in (2):
⇒ (1.032AC')² = AC'² + (3.5)²
⇒1.065AC'² = AC'² + 12.25
⇒ 0.065AC'² = 12.25
⇒ AC'² = 188.462
⇒ AC' = √188.462
⇒ AC' = 13.72 mm
