Answer:
(Z)-4-methyl-2-pentene
Explanation:
Hydrogenation is a syn addition. This implies that the hydrogen atoms add to the same face of the multiple bond.
Hence, if we have the alkyne, 4-methyl-2-pentyne which reacts with excess hydrogen in the presence of a platinum catalyst, we obtain (Z)-4-methyl-2-pentene since the hydrogen atoms add on the same side of the multiple bond.
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Answer:
P2 = 0.935 atm
Explanation:
initial conditions:
∴ V1 = 3.41 L
∴ T1 = - 91.5 °C + 273 = 181.5 K
∴ P1 = 3990 torr * ( atm / 760 torr ) = 5.25 atm
∴ R = 0.082 atm.L / K.mol
⇒ n = P1V1 / RT1 = ((5.25 atm)*(3.41 L)) / ((0.082 atm.L/K.mol)*(181.5 K))
⇒ n = 1.107 mol
final conditions:
∴ V2 = 3.3 L
∴ T2 = - 239 °C = 34 K
∴ n = 1.107 mol
⇒ P2 = nRT2 / V2
⇒ P2 = ((1.107 mol)*(0.082 atm.L/K.mol)*(34 K)) / 3.3 L
⇒ P2 = 0.935 atm
Answer:
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Explanation:
This is ur answer....
<em>1) All matter is made of atoms. Atoms are indivisible and indestructible. </em><em>3) Compounds are formed by a combination of two or more different kinds of atoms. 4) A chemical reaction is a rearrangement of atoms.</em>
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Answer:
6.78 × 10⁻³ L
Explanation:
Step 1: Write the balanced equation
Mg₃N₂(s) + 3 H₂O(g) ⇒ 3 MgO(s) + 2 NH₃(g)
Step 2: Calculate the moles corresponding to 10.2 mL (0.0102 L) of H₂O(g)
At STP, 1 mole of H₂O(g) has a volume of 22.4 L.
0.0102 L × 1 mol/22.4 L = 4.55 × 10⁻⁴ mol
Step 3: Calculate the moles of NH₃(g) formed from 4.55 × 10⁻⁴ moles of H₂O(g)
The molar ratio of H₂O to NH₃ is 3:2. The moles of NH₃ produced are 2/3 × 4.55 × 10⁻⁴ mol = 3.03 × 10⁻⁴ mol.
Step 4: Calculate the volume corresponding to 3.03 × 10⁻⁴ moles of NH₃
At STP, 1 mole of NH₃(g) has a volume of 22.4 L.
3.03 × 10⁻⁴ mol × 22.4 L/mol = 6.78 × 10⁻³ L
