Answer:
V ≈ 646.50 L
General Formulas and Concepts:
<u>Chemistry - Gas Laws</u>
- Reading a Periodic Table
- Stoichiometry
- Combined Gas Law: PV = nRT
- R constant - 62.4 (L · torr)/(mol · K)
- Kelvin Conversion: K = °C + 273.15
Explanation:
<u>Step 1: Define</u>
RxN: N₂H₄ (g) + O₂ (g) → N₂ (g) + 2H₂O (l)
Given: 34.9 °C, 755.08 torr, 914.894 g H₂O
<u>Step 2: Identify Conversions</u>
Kelvin Conversion
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Convert</u>
Stoichiometry: 
= 25.3955 mol N₂
Temperature: 34.9 + 273.15 = 308.05 K
<u>Step 4: Find Volume</u>
- Substitute variables: (755.08 torr)V = (25.3955 mol)(62.4 (L · torr)/(mol · K))(308.05 K)
- Multiply: (755.08 torr)V = 488160 L · torr
- Isolate <em>V</em>: V = 646.502 L
<u>Step 5: Check</u>
<em>We are given 5 sig figs as our lowest. Follow sig fig rules and round.</em>
646.502 L ≈ 646.50 L
Answer:
15.95
Explanation:
This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.
Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.
In 100 grams of the binary compound we have
30.46 g N
69.54 g E
The number of moles is the mass divided by atomic weight:
mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N
mol E = 65.54 g / A.W E
Thus,
4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18
2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g
So the A.W is 15.94 g/mol which is close the atomic weight of O.
Answer:
A
Explanation:
The answer A is the best answer because it contains the most general characteristic of a chemical change.
N = given mass/ molar mass.
n = number of moles
given mass = 2.47 g
molar mass = 197 g/mol
n = 2.47 / 197
n = 0.01253 moles.
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Answer:
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Explanation:
