Answer:
V = 85.2
Explanation:
STP = 273K and 1 atm
Considering what we know about STP, we get the moles, temperature, and pressure. Using the ideal gas law we can find the volume (PV = nRT). Plug in our variables: (1 * V = 3.80 * R * 273). Since we are dealing with atm and not kPA or mmHg, we use the constant for atm (0.0821) which we use for R. (So.. now our equation is 1 * V = 3.80 * 0.0821 * 273). We now multiply the right side to get 85.17054. So... V = 85.2 considering sigificant figures (this is the part where I am the least sure of, since I havent done sig figs in a while)
<h3>
Answer: b) 0.250 mol</h3>
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Work Shown:
Using the periodic table, we see that
- 1 mole of carbon = 12 grams
- 1 mole of oxygen = 16 grams
These are approximations and these values are often found underneath the atomic symbol. For example, the atomic weight listed under carbon is roughly 12.011 grams. I'm rounding to 2 sig figs in those numbers listed above.
So 1 mole of CO2 is approximately 12+2*16 = 44 grams. The 2 is there since we have 2 oxygens attached to the carbon atom.
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Since 1 mole of CO2 is 44 grams, we can use that to convert from grams to moles.
11.0 grams of CO2 = (11.0 grams)*(1 mol/44 g) = (11.0/44) mol = 0.250 mol of CO2
In short,
11.0 grams of CO2 = 0.250 mol of CO2
This is approximate.
We don't need to use any of the information in the table.
Answer:
The answer to your question is: ΔH = -283 kJ/mol, first option
Explanation:
Reaction
CO + O₂ ⇒ CO₂
ΔH = ∑H products - ∑H products
ΔH = -393.5 - (-110.5 + 0)
ΔH = -393.5 + 110.5
ΔH = -283 kJ/mol
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
Row or periods have in common is the valence electron count. The valence electron count goes up as you move across the periodic table. Also atomic size gets smaller as you move from left to right
