When the guard cell is filled with water and it becomes turgid, the outer wall balloons outward, drawing the inner wall with it and causing the stomate to enlarge.
Answer:
True
Explanation:
This is because, The hydroboration oxidation of an alkene which is isobutene in the presence of a catalyst will result to alcohol as the product . Therefore, the OH group will attach or link itself to the carbon which is less obstructed. Thus this reaction is in accordance to Anti-Markownikoff's rule.
So isobutene on hydroboration oxidation will produce ter isobutyl alcohol.
Answer:
a) 0.525 mol
b) 0.525 mol
c) 0.236 mol
Explanation:
The combustion reactions (partial and total) will be:
C₇H₁₆ + (15/2)O₂ → 7CO + 8H₂O
C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O
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2C₇H₁₆ + (37/2)O₂ → 7CO + 7CO₂ + 16H₂O
It means that the reaction will form 50% of each gas.
a) 0.525 mol of CO
b) 0.525 mol of CO₂
c) The molar mass of heptane is: 7*12 g/mol of C + 16*1 g/mol of H = 100 g/mol
So, the number of moles is the mass divided by the molar mass:
n = 11.5/100 = 0.115 mol
For the stoichiometry:
2 mol of C₇H₁₆ -------------- (37/2) mol of O₂
0.115 mol of C₇H₁₆ --------- x
By a simple direct three rule:
2x = 2.1275
x = 1.064 mol of O₂
Which is the moles of oxygen that reacts, so are leftover:
1.3 - 1.064 = 0.236 mol of O₂
Answer:
molar mass M(s) = 65.326 g/mol
Explanation:
- M(s) + H2SO4(aq) → MSO4(aq) + H2(g)
∴ VH2(g) = 231 mL = 0.231 L
∴ P atm = 1.0079 bar
∴ PvH2O(25°C) = 0.03167 bar
Graham´s law:
⇒ PH2(g) = P atm - PvH2O(25°C)
⇒ PH2(g) = 1.0079 bar - 0.03167 bar = 0.97623 bar = 0.9635 atm
∴ nH2(g) = PV/RT
⇒ nH2(g) = ((0.9635 atm)(0.231 L))/((0.082 atmL/Kmol)(298 K))
⇒ nH2(g) = 9.1082 E-3 mol
⇒ n M(s) = ( 9.1082 E-3 mol H2(g) )(mol M(s)/mol H2(g))
⇒ n M(s) = 9.1082 E-3 mol
∴ molar mass M(s) [=] g/mol
⇒ molar mass M(s) = (0.595 g) / (9.1082 E-3 mol)
⇒ molar mass M(s) = 65.326 g/mol
The heat is measured with degree Celsius and degree Fahrenheit. The heat capacity depends on both it's mass and it's chemical composition.
