Answer:
Following are the responses to the given question:
Explanation:
calculating the HOI:
So, in the above compund there are three bounds or we can say that one is double and two bound is single.
Please find the attachment file of the Function group:
In the above given conditon compound doesn't include the alkene functional group that is absense of
, and compound include
So, please find the attached file of the structure of the compound:
Answer:
The molar concentration of this solution is 0.0463 mol/L
Explanation:
Step 1 : Data given
Mass of a nonelectrolyte solute = 2.69 grams
Volume of water = 345 mL = 0.345 L
Temperature = 26.0°CC = 273 + 26 = 299 K
The osmotic pressure = 863 torr
⇒ 863torr /760 = 1.13553 atm
Step 2: Calculate the molar concentration of this solution
Π = i*M*R*T
⇒with Π = the osmotic pressure = 1.13553 atm
⇒with i = the van't Hoff factor of the nonelectrolyte solute = 1
⇒with M = the molar concentration = TO BE DETERMINED
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 299 K
1.13553 atm = 1 * M * 0.08206 L*atm/mol*K * 299 K
M = 1.13553 / (0.08206*299)
M = 0.0463 mol/L
The molar concentration of this solution is 0.0463 mol/L
So in order for us to know the percentage of sugar present in a 12.00 g of milk chocolate, what we are going to do is that, we just have to divide 8 by 12 and multiply in by 100 and we get 66.67. Therefore, the percentage of sugar present in 12.00 g of milk chocolate bar is 66.67%. Hope this answers your question. Have a great day!
Answer: 27.09 ppm and 0.003 %.
First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.
Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.
So, according to the <em>law of ideal gases,</em>
PV = nRT
where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)
The moles of CO will be,
n = 35 mg x
x
→ n = 0.00125 mol
We clear V from the equation and substitute P = 0.92 atm and
T = -30 ° C + 273.15 K = 243.15 K
V = 
→ V = 0.0271 L
As 1000 cm³ = 1 L then,
V = 0.0271 L x
= 27.09 cm³
<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>
c = 27 cm³ / m³ = 27 ppm
<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:
c = 27.09
x
x 100%
c = 0.003 %
So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.