This question requires the knowledge of density.
The density of ethyl alcohol = 789 kg m⁻³
The density of water = 1000 kg m⁻³
Density = Mass / Volume
By applying ethyl alcohol,
789 kg m⁻³ = Mass / 0.9 m³
Mass = 710.1 kg
hence the mass of 0.9 m³ ethyl alcohol is 710.1 kg.
Then by applying water,
1000 kg m⁻³ = 710.1 kg / Volume
Volume = 0.7101 m³
= 0.7 m³
hence the equal water volume is 0.7 m³
When an acid is neutralized by a base, that means moles of H+ = moles of OH-
moles of H+ = 0.5 M * 0.025 L HCl = 0.0125 moles H+
moles of OH- should be equal to 0.0125 moles, so
0.0125 moles = (x) * 0.025 L NaOH
x is the concentration of NaOH, which we want to find.
x = 0.5 M
The correct answer is C) 0.5 M.
I believe Winter is <span>your answer.</span>
Answer: B or C
Explanation: The question does not include the variable or steps Brian is using so either one could be correct. It has to be the one that he is controlling though. This is because a control group is used to rule out any alternate explanations. Therefore the answer should be the one that he is trying to test out.
Answer:
1.71x10²⁷
Explanation:
If we sum 1/2 of (3) + 1/2 of (1):
1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷ = 4.58x10²³
1/2 (1) 1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8
C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>
K' = 4.58x10²³ * 11.8 = 5.42x10²⁴
+1/2 (2):
<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2
C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)
K'' = 5.42x10²⁴* 316.2 =
<h3>1.71x10²⁷</h3>