Answer:
5 g
Explanation:
The heat required to vaporize ice is the sum of
i) Heat required to melt ice at 0°C
ii) Heat required to raise the temperature from 0°C to 100°C
iii) Heat required to vaporize water at 100°C
Thus;
H = nLfus + ncθ + nLvap
H= n(Lfus + cθ + Lvap)
Lfus = 6.01 kJ/mol
Lvap = 41 kJ/mol
c = 75.38
n =?
2100 = n(6.01 + 75.38(100) + 41)
n = 2100 KJ/7585.01 kJ/mol
n = 0.277 moles
Mass of water = number of moles * molar mass
Mass of water = 0.277 moles * 18 g/mol
Mass of water = 5 g
Answer:
q1 = mCpΔT
= 18.016g × 1.84J/g.K × (418.15-373.15)
= 1491.72 J
q2 = n×ΔH vap
= 1mol ×44.0kJ/mol
= 44KJ
∴ qtotal = q1+ q2
= 1.498kJ + 44.0kJ
= 45.498KJ
Explanation: The heat flow can be separated into steps.all that is being observed at a constant pressure,the heat flow is equal to the enthalpy.
The molar concentration is 1.11M.
<h3>What is molar concentration?</h3>
The phrase "molar concentration" (also known as "molarity," "amount concentration," or "substance concentration") refers to the amount of a substance per unit volume of solution and is used to describe the concentration of a chemical species, specifically a solute, in a solution. The most frequent measure of molarity in chemistry is the number of moles per liter, denoted by the unit symbol mol/L or mol/dm3 in SI units. A solution with a concentration of 1 mol/L is referred to as 1 molar, or 1 M.
<h3>Given : </h3>
Volume of the solution = 2L
Mass of glucose given = 200g
Concentration of glucose= ?
<h3>Formula use: </h3>
Molarity = no. of moles of solute / volume of the solution (L)
Moles of solute = given mass of solute / molar mass of the solute
<h3>Solution: </h3>
No. of moles of solute( glucose ) = 200 / 180 = 1.11 moles'
Molarity = 1.11 / 2 = 0.5555 mol L ^(-1)
Therefore, the molar concentration of glucose in the solution = 0.555 mol L ^(-1)
To learn more about molar concentration :
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Answer:
2-ethoxy-2-methylpropan-1-ol
Explanation:
On this reaction, we have an "<u>epoxide"</u> (2-methyl-1,2-epoxypropane). Additionally, we have <u>acid medium</u> (due to the sulfuric acid
). The acid medium will produce the <u>hydronium ion</u> (
). This ion would be attacked by the oxygen of the epoxide. Then a <u>carbocation</u> would be produced, in this case, the most stable carbocation is the <u>tertiary one</u>. Then an <u>ethanol</u> molecule acts as a nucleophile and will attack the carbocation. Finally, a <u>deprotonation </u>step takes place to produce <u>2-ethoxy-2-methylpropan-1-ol</u>.
See figure 1
I hope it helps!
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Pure Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Br and Br,
E.N of Bromine = 2.96
E.N of Bromine = 2.96
________
E.N Difference
0.00 (Non Polar/Pure Covalent)
For N and O,
E.N of Oxygen = 3.44
E.N of Nitrogen = 3.04
________
E.N Difference
0.40 (Non Polar/Pure Covalent)
For P and H,
E.N of Hydrogen = 2.20
E.N of Phosphorous = 2.19
________
E.N Difference 0.01 (Non Polar/Pure Covalent)
For K and O,
E.N of Oxygen = 3.44
E.N of Potassium = 0.82
________
E.N Difference 2.62 (Ionic)