Answer:
800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .
Explanation:
3000 lb of 13% solution is required .
Total adhesive in weight = 3000 x .13 = 390 lb of adhesive
Available = 500 lb of 10% solution = 50 lb of adhesive
Rest = 390 - 50 = 340 lb required .
rest mass of solution = 3000 - 500 = 2500 lb
mass of adhesive required = 340 lb
Let the mass of 20% required be V
mass of adhesive = .20 V
.20 V = 340
V = 1700
rest of the volume = 2500 - 1700 = 800 lb which will be of pure solvent
So 800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .
The columns of the periodic table, also referred to as "groups" contain elements with similar reactive properties, due to these elements having a similar configuration of electrons in their outer shell.
<h3>Answer:</h3>
64 g O₂
<h3>General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 36 g H₂O
[Solve] x g O₂
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol O₂ → 2 mol H₂O
[PT] Molar Mass of O - 16.00 g/mol
[PT] Molar Mas of H - 1.01 g/mol
Molar Mass of O₂ - 2(16.00) = 32.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Stoichiometry</u>
- Set up conversion:

- Divide/Multiply [Cancel Units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
63.929 g O₂ ≈ 64 g O₂
Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen
if it occupies 31.6 mL at
.
Explanation:
Given : Mass of oxygen = 0.023 g
Volume = 31.6 mL
Convert mL into L as follows.

Temperature = 
As molar mass of
is 32 g/mol. Hence, the number of moles of
are calculated as follows.

Using the ideal gas equation calculate the pressure exerted by given gas as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the value into above formula as follows.

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen
if it occupies 31.6 mL at
.
<span>The superscripts in an electron configuration represents the number of electrons and protons in an element. </span>