Answer:
here
Explanation:
0.000141 to kilowatt-hours. hope this helped
<h3>
Answer:</h3>
1.83 × 10⁻⁷ mol Au
<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>
3.60 × 10⁻⁵ g Au (Gold)
<u>Step 2: Identify Conversions</u>
Molar Mass of Au - 196.97 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>
1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au
From the information given:
- The volume of the graduated cylinder = 50.0 mL
- when a sterling silver pendant is added, the volume increases to = 61.3 mL
∴
The volume of the sterling silver pendant is:
= 61.3 mL - 50.0 mL
= 11.3 mL
Since, 1 mL = 1cm³
Then;
11.3 mL = 11.3 cm³
- the density of the sterling silver = 10.25 g/cm³
Using the relation for Density; i.e.
mass = 10.25 g/cm³× 11.3 cm³
mass of the sterling silver = 115.825 grams
Recall that sterling silver has:
- 92.5% silver and;
- 7.5% copper
∴
The mass of the copper contained in the sterling silver pendant can be calculated as:
= 8.687 grams
Therefore, we can conclude that the mass of the copper contained in the sterling silver pendant is 8.687 grams
Learn more about the relation between Density, Mass, and Volume here:
brainly.com/question/24386693?referrer=searchResults
Answer: 1) Maximum mass of ammonia 198.57g
2) The element that would be completely consumed is the N2
3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g
Explanation:
- In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:
N2(g) + 3H2(g) ⟶2NH3(g)
Both equal amount of atoms side to side.
- Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)
Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3
Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3
- As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
- We proceed calculating the maximum mass of NH3 from the 163.3g of N2.
11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3
- In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:
163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2
That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left
Explanation:
Let the mass of isoamyl acetate be 100g.
Moles of Carbon = 60.58/12 = 5.048mol
Moles of Hydrogen = 7.07/1 = 7.07mol
Moles of Oxygen = 32.28/16 = 2.018mol
Mole Ratio of C : H : O
= 5.048 : 7.07 : 2.018
= 5 : 7 : 2.
Hence the empirical formula of isoamyl acetate is C5H7O2.
