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:
84.24 g
Explanation:
Given data:
Mass of oxygen = 75 g
Mass of Al required to react = ?
Solution:
Chemical equation:
4Al + 3O₂ → 2Al₂O₃
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 75 g/ 32 g/mol
Number of moles = 2.34 mol
Now we will compare the moles of oxygen with Al.
O₂ : Al
3 : 4
2.34 : 4/3×2.34 = 3.12 mol
Mass of Al required:
Mass = number of moles × molar mass
Mass = 3.12 mol × 27 g/mol
Mass = 84.24 g
FALSE
There are other limiting factors like lack of space, diseases, and com petition.
Answer:
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Explanation:
Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.
H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
We will balance it using the trial and error method.
First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
Finally, we will get the balanced equation by multiplying H₂O by 6.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Answer:
True
Explanation:
True definition of a mixture itself
