Answer:
In the same way, a solution with a pH of 5 contains 10-5mol/l of hydrogen ions, a solution with a pH of 6 contains 10-6mol/l of hydrogen ions, while the solution with a pH of 7 contains 10-7mol/l of hydrogen ions.
Explanation:
Reactant C is the limiting reactant in this scenario.
Explanation:
The reactant in the balanced chemical reaction which gives the smaller amount or moles of product is the limiting reagent.
Balanced chemical reaction is:
A + 2B + 3C → 2D + E
number of moles
A = 0.50 mole
B = 0.60 moles
C = 0.90 moles
Taking A as the reactant
1 mole of A reacted to form 2 moles of D
0.50 moles of A will produce
= 
thus 0.50 moles of A will produce 1 mole of D
Taking B as the reactant
2 moles of B reacted to form 2 moles of D
0.60 moles of B reacted to form x moles of D
= 
x = 2 moles of D is produced.
Taking C as the reactant:
3 moles of C reacted to form 2 moles of D
O.9 moles of C reacted to form x moles of D
= 
= 0.60 moles of D is formed.
Thus C is the limiting reagent in the given reaction as it produces smallest mass of product.
<h3>
Answer:</h3>
0.125 mol Ca
<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>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
5.00 g Ca
<u>Step 2: Identify Conversions</u>
Molar mass of Ca - 40.08 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.12475 mol Ca ≈ 0.125 mol Ca
Answer:
85.34g of NH3
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Step 2:
Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.
Therefore, 5.02 moles of NH3 is produced from the reaction.
Step 3:
Conversion of 5.02 moles of NH3 to grams. This is illustrated below:
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Number of mole of NH3 = 5.02 moles
Mass of NH3 =..?
Mass = mole x molar Mass
Mass of NH3 = 5.02 x 17
Mass of NH3 = 85.34g
Therefore, 85.34g of NH3 is produced.