• Take a tablespoonful of turmeric powder. Add a little water and make a paste. Apply turmeric paste on a sheet of plane white p
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<h3>Answer:</h3>
2.04 mol CBr₄
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Organic</u>
- Writing Organic Compounds
- Writing Covalent Compounds
- Organic Prefixes
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Step 1: Define</u>
675 g CBr₄
<u>Step 2: Identify Conversions</u>
Molar Mass of C - 12.01 g/mol
Molar Mass of Br - 79.90 g/mol
Molar Mass of CBr₄ - 12.01 + 4(79.90) = 331.61 g/mol
<u>Step 3: Convert</u>
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<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
2.03552 mol CBr₄ ≈ 2.04 mol CBr₄
Answer:
2.24 L of hydrogen gas, measured at STP, are produced.
Explanation:
Given, Moles of magnesium metal, = 0.100 mol
Moles of hydrochloric acid, = 0.500 mol
According to the reaction shown below:-
1 mole of Mg reacts with 2 moles of
0.100 mol of Mg reacts with 2*0.100 mol of
Moles of must react = 0.200 mol
Available moles of = 0.500 moles
Limiting reagent is the one which is present in small amount. Thus, is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
1 mole of on reaction forms 1 mole of
0.100 mole of on reaction forms 0.100 mole of
Mole of = 0.100 mol
At STP,
Pressure = 1 atm
Temperature = 273.15 K
Volume = ?
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
1 atm × V L = 0.100 × 0.0821 L.atm/K.mol × 273.15 K
<u>⇒V = 2.24 L</u>
2.24 L of hydrogen gas, measured at STP, are produced.