Answer:
Explanation:
Types of Attractive Intermolecular Forces. Dipole-dipole forces: electrostatic interactions of permanent dipoles in molecules; includes hydrogen bonding.
This is all I can think of, I hope this has helped you.
-QueenBeauty666-
From the stoichiometry of the combustion reaction, we can see that 7.4 L of oxygen is consumed.
<h3>What is combustion?</h3>
Combustion is a reaction in which a substance is burnt in oxygen. The equation of the reaction is; C4H10O(l) + 6O2 (g) → 4CO2 (g) + 5H2O(l)
We can obtain the number of moles of CO2 from;
PV = nRT
n = 1.02 atm * 7.15 L/0.082 atm LK-1mol-1 * (125 + 273) K
n = 7.29 /32.6
n = 0.22 moles
If 6 moles of oxygen produces 4 moles of CO2
x moles of oxygen produces 0.22 moles of CO2
x = 0.33 moles
1 mole of oxygen occupies 22.4 L
0.33 moles of oxygen occupies 0.33 moles * 22.4 L/ 1 mole
= 7.4 L of oxygen
Learn more about stoichiometry: brainly.com/question/13110055
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Answer:
A budget is an estimation of revenue and expenses over a specified future period of time and is usually compiled and re-evaluated on a periodic basis. Budgets can be made for a person, a group of people, a business, a government, or just about anything else that makes and spends money.
Data Given:
Pressure = P = ?
Volume = V = 3.0 L
Temperature = T = 115 °C + 273 = 388 K
Mass = m = 75.0 g
M.mass = M = 44 g/mol
Solution:
Let suppose the Gas is acting Ideally. Then According to Ideal Gas Equation,
P V = n R T
Solving for P,
P = n R T / V ------ (1)
Calculating Moles,
n = m / M
n = 75.0 g / 44 g.mol⁻¹
n = 1.704 mol
Putting Values in Eq. 1,
P = (1.704 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 388 K) ÷ 3.0 L
P = 18.08 atm
Answer:
Mass of carbon dioxide produced = 52.8 g
Explanation:
Given data:
Mass of carbon react = 14.4 g
Mass of oxygen = 56.5 g
Mass of oxygen left = 18.1 g
Mass of carbon dioxide produced = ?
Solution:
C + O₂ → CO₂
Number of moles of C:
Number of moles = mass/ molar mass
Number of moles = 14.4 g/ 12 g/mol
Number of moles = 1.2 mol
18.1 g of oxygen left it means carbon is limiting reactant.
Now we will compare the moles of C with CO₂.
C : CO₂
1 : 1
1.2 : 1.2
Mass of CO₂:
Mass = number of moles × molar mass
Mass = 1.2 mol × 44 g/mol
Mass = 52.8 g
