<span>0.0165 m
The balanced equation for the reaction is
AgNO3 + MgCl2 ==> AgCl + Mg(NO3)2
So it's obvious that for each Mg ion, you'll get 1 AgCl molecule as a product. Now calculate the molar mass of AgCl, starting with looking up the atomic weights.
Atomic weight silver = 107.8682
Atomic weight chlorine = 35.453
Molar mass AgCl = 107.8682 + 35.453 = 143.3212 g/mol
Now how many moles were produced?
0.1183 g / 143.3212 g/mol = 0.000825419 mol
So we had 0.000825419 moles of MgCl2 in the sample of 50.0 ml. Since concentration is defined as moles per liter, do the division.
0.000825419 / 0.0500 = 0.016508374 mol/L = 0.016508374 m
Rounding to 3 significant figures gives 0.0165 m</span>
Answer:
The law is observed in the given equation.
Explanation:
CaCO₃ + 2HCI → CaCI₂ +H₂O + CO₂
In order to find out if the law of conservative mass is followed, we need to <u>count how many atoms of each element are there in both sides of the equation</u>:
- Ca ⇒ 1 on the left, 1 on the right.
- C ⇒ 1 on the left, 1 on the right.
- O ⇒ 3 on the left, 3 on the right.
- H ⇒ 2 on the left, 2 on the right.
- Cl ⇒ 2 on the left, 2 on the right.
As the numbers for all elements involved are the same, the law is observed in the given equation.
Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
______
NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
Answer:
Covalent, or, elements on the right side of the ladder (on the periodic table). Covalent compounds satisfy the octet rule by sharing electrons.
