Answer:
4.14 x 10²⁴ molecules CO₂
Explanation:
2 C₄H₁₀ + 13 O₂ --> 8 CO₂ + 10 H₂O
To find the number of CO₂ molecules, you need to start with 100 grams of butane (C₄H₁₀), convert to moles (using the molar mass), convert to moles of CO₂ (using coefficients from equation), then convert to molecules (using Avagadro's number). The molar mass of C₄H₁₀ is calculated using the quantity of each element (subscript) multiplied by the number on the periodic table. The ratios should be arranged in a way that allows for units to be cancelled.
4(12.011g/mol) + 10(1.008 g/mol) = 58.124 g/mol C₄H₁₀
100 grams C₄H₁₀ 1 mol C₄H₁₀ 8 mol CO₂
-------------------------- x ---------------------- x ---------------------
58.124 g 2 mol C₄H₁₀
6.022 x 10²³ molecules
x ------------------------------------ = 4.14 x 10²⁴ molecules CO₂
1 mol CO₂
Answer:
1.44 x 10²⁵ ions of Na⁺
Explanation:
Given parameters:
Mass of NaCl = 1.4kg = 1400g
Unknown:
Number of ions of sodium = ?
Solution:
The compound NaCl in ionic form can be written as;
NaCl → Na⁺ + Cl⁻
In 1 mole of NaCl we have 1 mole of sodium ions
Now, let us find the number of moles in NaCl;
Number of moles = 
Molar mass of NaCl = 23 + 35.5 = 58.5g/mol
Number of moles = 
= 23.93mol
So;
Since 1 mole of NaCl gives 1 mole of Na⁺
In 23.93 mole of NaCl will give 23.93 mole of Na⁺
1 mole of a substance = 6.02 x 10²³ ions of a substance
23.93 mole of a substance = 6.02 x 10²³ x 23.93
= 1.44 x 10²⁵ ions of Na⁺
Answer:
The answer to your question is ΔH° rxn = -1343.9 kJ/mol
Explanation:
P₄O₆ (s) + 2 O₂ (g) ⇒ P₄O₁₀
ΔH°rxn = ?
Formula
ΔH°rxn = ∑H° products - ∑H° reactants
H° P₄O₆ = -1640.1 kJ/mol
H° O₂ = 0 kJ/mol
H° P₄O₁₀ = -2984 kJ/mol
-Substitution
ΔH° rxn = (-2984) - (-1640.1) - (0)
-Simplification
ΔH° rxn = -2984 + 1640.1
ΔH° rxn = -1343.9 kJ/mol
Answer:
To calculate the theoretical yield, determine the number of moles of each reactant, in this case the sole reactant ethanol. Convert the 100 g to moles; the molecular weight of ethanol is 46 g/mole, therefore: Since there is only one reactant, it is also the limiting reagent.
Explanation:
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