Equation: M1V1 = M2V2
Where M = concentration & V = volume
Step 1: Write down what is given and what you are trying to find
Given: M1 = 6.00M, V1 = 2.49mL, and V2 = 50.0mL
Find: M2
Step 2: Plug in the values into the equation
M1V1 = M2V2
(6.00M)(2.49mL) = (M2)(50.0mL)
Step 3: Isolate the variable (Divide both sides by 50.0mL so M2 is by itself)
(6.00M)(2.49mL) / (50.0mL) = M2
Answer: M2 = 0.30M
*Don't forget sig figs & units!
Answer:
The answer to your question is the letter C) 5648 kJ/mol
Explanation:
Data
C₁₂H₂₂O₁₁ + 12 O₂ ⇒ 12 CO₂ + 11 H₂O
H° C₁₂H₂₂O₁₁ = -2221.8 kJ/mol
H° O₂ = 0 kJ / mol
H° CO₂ = -393.5 kJ/mol
H° H₂O = -285.8 kJ/mol
Formula
ΔH° = ∑H° products - ∑H° reactants
Substitution
ΔH° = 12(-393.5) + 11(-285.8) - (-2221.8) - (0)
ΔH° = -4722 - 3143.8 + 2221.8
Result
ΔH° = -5644 kJ/mol
Answer:
A. Alternating single and double bonds
Explanation:
Hydrocarbons i.e. hydrogen and carbon containing compounds, are grouped into two namely: aliphatic and aromatic hydrocarbons. Aliphatic hydrocarbons contains straight or branched chains of carbon and hydrogen atoms in their structure e.g. alkanes, alkenes etc.
On the other hand, aromatic hydrocarbons are cyclic hydrocarbons containing one or more cyclic rings. The benzene ring is the basis of all aromatic hydrocarbons and one characteristics of benzene is that it possesses an alternating single (-) and double bonds (=) in their structure.
Since benzene is a building constituent of aromatic hydrocarbons, an "alternating single and double bond" is a characteristics of aromatic hydrocarbons.
Answer:
Explanation:
We can use oxidation numbers to decide which substance is reduced.
The oxidation number of Na changes from 0 in Na to +1 in Na⁺.
The oxidation number of H changes from +1 in H₂O to 0 in H₂.
1 and 4 are wrong because H₂ and Na⁺ are products.
2. is wrong because there is no H⁺ to be oxidized or reduced.
