The peaks at 3300 cm-1 and 2200 cm-1 are due to terminal alkyne group (triple bond between two carbons) and triple bonded C-H stretching
So
The answer is : terminal alkyne
Another peak near 3430 cm-1 must be due to an -OH group
so the answer is : alcohol
Answer:
C4H8O4
Explanation:
To determine the molecular formula, first, let us obtain the empirical formula. This is illustrated below:
From the question given, we obtained the following information:
C = 45.45%
H = 6.12%
O = 48.44%
Divide the above by their molar mass
C = 45.45/12 = 3.7875
H = 6.12/1 = 6.12
O = 48.44/16 = 3.0275
Divide by the smallest
C = 3.7875/3.0275 = 1
H = 6.12/3.0275 = 2
O = 3.0275/3.0275 = 1
The empirical formula is CH2O
The molecular formula is given by [CH2O]n
[CH2O]n = 132.12
[12 + (2x1) + 16]n = 132.12
30n = 132.12
Divide both side by the coefficient of n i.e 30
n = 132.12/30 = 4
The molecular formula is [CH2O]n = [CH2O]4 = C4H8O4
<h3>
Answer:</h3>
23.95 °C
<h3>
Explanation:</h3>
We are given;
- Mass of the sample is 0.5 gram
- Quantity of heat released as 50.1 Joules
- Specific heat capacity is 4.184 J/g°C
We are required to calculate the change in temperature;
- Quantity of heat absorbed is given by the formula;
- Q = mass × specific heat capacity × Change in temperature
That is, Q = mcΔT
Rearranging the formula;
ΔT = Q ÷ mc
Therefore;
ΔT = 50.1 J ÷ (0.5 g × 4.184 J/g°C)
= 23.95 °C
Therefore, the expected change in temperature is 23.95 °C
Q1)
molarity is defined as the number of moles of solute in 1 L of solution.
the NaCl solution volume is 1.00 L
number of moles NaCl = NaCl mass present / molar mass of NaCl
NaCl moles = 112 g / 58.5 g/mol = 1.91 mol
the number of moles of NaCl in 1.00 L of solution is - 1.91 mol
therefore molarity of NaCl is 1.91 M
Q2)
molality is defined as the number of moles of solute in 1 kg of solvent.
density is mass per volume.
density of the solution is 1.08 g/mL.
therefore mass of the solution is = density x volume
mass = 1.08 g/mL x 1000 mL = 1080 g
since we have to find the moles in 1 kg of solvent
mass of solvent = 1080 g - 112 g = 968 g
number of moles of NaCl in 968 g of solvent - 1.91 mol
therefore number of NaCl moles in 1000 g - (1.91 mol / 968 g) x 1000 g/kg = 1.97 mol/kg
molality of NaCl solution is 1.97 mol/kg
Q3)
mass percentage is the percentage of mass of solute by total mass of the solution
mass percentage of solution = mass of solute / total mass of the solution
mass of solute = 112 g
total mass of solution = 1080 g
mass % of NaCl = 112 g / 1080 g x 100%
therefore mass % of NaCl = 10.4 %
answer is 10.4 %
Answer:
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Explanation:
the balanced equation is
2NaBr +Cl2 =2NaCl + Br2
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