Answer:
The answer to your question is: letter D. 1.33 L
Explanation:
Data
V1 = 50 ml
C1 = 19.3
To solve this problem use the formula C₁V₁ = C₂V₂
C2 = C1V1 / V2
C = concentration
V = volume
a) 1.15 L
C2 = (19.3)(50) / 1150
C2 = 0.84 M
b) No right answer
c) V2= 0.80 L
C2 = (19.3)(50) / 800
C2 = 1.2 M
d) V2 = 1.33 L
C2 = (19.3)(50) / 1330
C2 = 0.72 M
e) V2 = 350 ml
C2 = (19.3)(50) / 350
C2 = 2.75 M
Answer:
The compound contains lauryl sulfate and ammoium ions. Lauryl sulfate contains lauric acid (in black and white) , the fatty acid formed by the covalent bonds between C-C attached to hydrogens, and sulfate ions attached to lauric acid (in red) with C-S covalent bond. Sulfer is attached to oxygen by covalent bonds. In Ammonium ions, N is surrounded by four hydrogen atoms.
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Sandy would reach an incorrect outcome.that's because she will repeat this mistake in the future too.
Answer:
The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol
Explanation:
The chemical reaction of the combustion of methane is given as follows;
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)
Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor
Where:
CH₄ (g): Hf = -74.6 kJ/mol
CO₂ (g): Hf = -393.5 kJ/mol
H₂O (g): Hf = -241.82 kJ/mol
Therefore, the combustion of 1 mole of methane releases;
-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol
Hence the combustion of 2 moles of methae will rellease;
2 × -802.54 kJ/mol or 1,605.08 kJ/mol.
