Answer:
I think its D
Explanation:
.........................
Methanol is prepared by reacting Carbon monoxide and Hydrogen gas,
CO + 2 H₂ → CH₃OH
Calculating Moles of CO:
According to equation,
32 g (1 mole) of CH₃OH is produced by = 1 Mole of CO
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of CO
Solving for X,
X = (3.60 × 10² g × 1 Mole) ÷ 32 g
X = 11.25 Moles of CO
Calculating Moles of H₂:
According to equation,
32 g (1 mole) of CH₃OH is produced by = 2 Mole of H₂
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of H₂
Solving for X,
X = (3.60 × 10² g × 2 Mole) ÷ 32 g
X = 22.5 Moles of H₂
Result:
3.60 × 10² g of CH₃OH is produced by reacting 11.25 Moles of CO and 22.5 Moles of H₂.
<u>Answer:</u> The final temperature of water is 32.3°C
<u>Explanation:</u>
When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid water)
The equation used to calculate heat released or absorbed follows:
![m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
......(1)
where,
q = heat absorbed or released
= mass of solution 1 (liquid water) = 50.0 g
= mass of solution 2 (liquid water) = 29.0 g
= final temperature = ?
= initial temperature of solution 1 = 25°C = [273 + 25] = 298 K
= initial temperature of solution 2 = 45°C = [273 + 45] = 318 K
c = specific heat of water= 4.18 J/g.K
Putting values in equation 1, we get:
![50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K](https://tex.z-dn.net/?f=50.0%5Ctimes%204.18%5Ctimes%20%28T_%7Bfinal%7D-298%29%3D-%5B29.0%5Ctimes%204.18%5Ctimes%20%28T_%7Bfinal%7D-318%29%5D%5C%5C%5C%5CT_%7Bfinal%7D%3D305.3K)
Converting this into degree Celsius, we use the conversion factor:
Hence, the final temperature of water is 32.3°C
Answer:
The Ideal Gas Law cannot be applied to liquids. The Ideal Gas Law is #PV = nRT#. That implies that #V# is a variable. But we know that a liquid has a constant volume, so the Ideal <u><em>Gas Law cannot apply to a liquid.</em></u>
Explanation:
this is my awnser soory if it was a multiple choice question plz mark brainliest
