Aluminum is a substance because it is a particular type of matter that has particular and unique properties. aluminum is a unique element with a place on the periodic table
Mass percent= grams solute/ grams of solution x 100
Mass Percent= (50/ 150)x100= 33.3%
The final temperature in Celsius of the metal block is 49°C.
<h3>How to find the number of moles ?</h3>
Moles water = 
= 
= 0.0266 moles
Heat lost by water = 0.0266 mol x 44.0 kJ/mol
= 1.17 kJ
= 1170 J [1 kJ = 1000 J]
Heat lost = Heat gained
Heat gained by aluminum = 1170 J
1170 = 55 x 0.903 (T - 25) = 49.7 T - 1242
1170 + 1242 = 49.7 T
T = 48.5°C (49°C at two significant figures)
Thus from the above conclusion we can say that The final temperature in Celsius of the metal block is 49°C.
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Answer:
E
Explanation:
A catalyst is a substance which alters the rate of a chemical reaction. We check for the correctness of each of the options as follows:
A. Is wrong.
A catalyst can increase the rate of the forward and backward reaction
B. Is wrong
A catalyst does not slow the reverse reaction only. This particular case is the case of a negative catalyst
C is wrong
A catalyst has no effect on the equilibrium nor the equilibrium constant
D is wrong
Catalyst has no effect on equilibrium value
E is correct
Although Catalysts has no effect on equilibrium or its constant value, it can increase the the rate at which equilibrium is achieved by speeding up the reaction through bringing down the activation energy
<h3>
Answer:</h3>
3.67 mol Al
<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>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
2.21 × 10²⁴ atoms Al
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:

- Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
3.66988 mol Al ≈ 3.67 mol Al