We will use boiling point formula:
ΔT = i Kb m
when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35
and Kb is the boiling point constant =5.03
and m = molality
i = vant's Hoff factor
so by substitution, we can get the molality:
1.35 = 1 * 5.03 * m
∴ m = 0.27
when molality = moles / mass Kg
0.27 = moles / 0.015Kg
∴ moles = 0.00405 moles
∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol
Answer:
Part A:
"360 grams of NaCl can be dissolved in 1 L water. So, 2000 grams sugar can be dissolved in 1 L water then we can say that the solubility of salt is lesser in water as to sugar and both heightened by increasing the temperature. If we make a batch of 800 L we can add sugar, 1600 kg at 25 0c. We can add salt is 288 kg at 25 0c and the ingredient tomato is having low solubility."
Read more at Answer.Ya.Guru – https://answer.ya.guru/questions/8061-describe-the-sequence-of-adding-ingredients-to-make-the-recipe.html
Part B:
'Manufacturers can generate new value minimize cost and increase operational stability by focusing on 4 broad areas; Management, Supply Circle, Product Design, and Value Recovery.'
Read more at Answer.Ya.Guru – https://answer.ya.guru/questions/2807911-what-changes-could-be-made-to-optimize-the-manufacturing-process.html
The gas is confined in 3.0 L container ( rigid container) ⇒ the volume remains constant when the temperature is increased from from 27oC to 77oC and therefore V1=V2 .
<h2>
Hope it helps you please mark as brainlist</h2>
Answer:
4244.48 g to the nearest hundredth.
Explanation:
The molar mass of Glucose = 6*12.011 + 12*1.008+ 6*15.999
= 180.156.
So 23.56 moles = 180.156 * 23.56 = 4244.48 g
My guess is b for the question
