I think its high. I just looked it up and it said stuff about how it is closely related to some gases.
1. Given the following equation: N2 (g) + 3 H2 (g) ↔ 2NH3 (g) ΔH = -92 kJ/mol
a. this reaction is exothermic as ΔH is -ve
b. the equilibrium will shift 2 the left if nitrogen gas is removed
c. the equilibrium shift 2 the right if the temperature is lowered
d. the equilibrium shift 2 the left if ammonia (NH3) is added
e. principle of thermodynamic potential or Gibbs energy is used to answer B-D
<span>1.02x10^2 ml
Since molarity is defined as moles per liter, the product of the molarity and volume will remain constant as mole solvent is added. So let's set up an equality to express this
m0*v0 = m1*v1
where
m0, v0 = molarity and volume of original solution
m1, m1 = molarity and volume of final solution.
Solve for v0, then substitute the known values and calculate:
m0*v0 = m1*v1
v0 = (1.75 M * 500 ml)/8.61 M
v0 = (1.75 M * 500 ml)/8.61 M
V0 = 101.6260163
Rounding to 3 significant figures gives 102 ml.
So the original volume of the 8.61 M H2SO4 solution was 102 ml or 1.02x10^2 ml.</span>
Answer:
(1) Cl₂ is the limiting reactant.
(2) 8.18 g
Explanation:
- 2Na(s) + Cl₂(g) → 2NaCl(s)
First we <u>convert the given masses of reactants into moles</u>, using their <em>respective molar masses</em>:
- Na ⇒ 12.0 g ÷ 23 g/mol = 0.522 mol Na
- Cl₂ ⇒ 5.00 g ÷ 70.9 g/mol = 0.070 mol Cl₂
0.070 moles of Cl₂ would react completely with (2 * 0.070) 0.14 moles of Na. There are more Na moles than that, so Na is the reactant in excess while Cl₂ is the limiting reactant.
Then we <u>calculate how many moles of NaCl are formed</u>, <em>using the limiting reactant</em>:
- 0.070 mol Cl₂ *
= 0.14 mol NaCl
Finally we <u>convert NaCl moles into grams</u>:
- 0.14 mol NaCl * 58.44 g/mol = 8.18 g
Answer:
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