Answer:
Diagram 1
Explanation:
The solubility of the oxygen gas in water has to do with the interaction of the oxygen with the dipoles in water.
Water is a polar molecule having oxygen as the negative dipole and hydrogen as the positive dipole.
Water can interact with the oxygen atoms in the molecule via intermolecular hydrogen bonds with molecular oxygen as shown in diagram 1.
8.03 solutions report is described below.
Explanation:
8.03 Solutions Lab Report
In this laboratory activity, you will investigate how temperature, agitation, particle size, and dilution affect the taste of a drink. Fill in each section of this lab report and submit it and your pre-lab answers to your instructor for grading.
Pre-lab Questions:
In this lab, you will make fruit drinks with powdered drink mix. Complete the pre-lab questions to get the values you need for your drink solutions.
Calculate the molar mass of powered fruit drink mix, made from sucrose (C12H22O11).
Using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 M solution of 100 mL.
Answer:
3.82 x 10²¹ molecules As₂O₃
Explanation:
To find the amount of molecules arsenic (III) oxide (As₂O₃), you need to (1) convert kg to lbs, then (2) convert g As₂O₃ to moles As₂O₃ (via molar mass), and then (3) convert moles to molecules (via Avogadro's number).
1 kilogram = 2.2 lb
Molar Mass (As₂O₃): 2(74.992 g/mol) + 3(15.998 g/mol)
Molar Mass (As₂O₃): 197.978 g/mol
Avogadro's Number:
6.022 x 10²³ molecules = 1 mole
0.0146 g As₂O₃ 1 kg 189 lb
------------------------ x --------------- x ------------------ x ................
1 kg 2.2 lb
1 mole 6.022 x 10²³ molecules
x ------------------ x --------------------------------------- = 3.82 x 10²¹ molecules As₂O₃
197.978 g 1 mole
Answer:
The answer is 105.98844.
<h3>Explanation: </h3>
We assume you are converting between grams Na2CO3 and mole. You can view more details on each measurement unit: molecular weight of Na2CO3 or mol This compound is also known as Sodium Carbonate.
I am going to go with,
0.10 g/mL
0.0700 g/mL
0.0447 g/mL
I don't know if this is the correct answer, but I am 80% sure that it may be.
:) :)
