How many oxygen atoms are there for three molecules of glucose (3C6H1206)?
1 answer:
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Answer : The pH of the solution is, 2.67
Explanation :
The equilibrium chemical reaction is:
Initial conc. 0.450 0 0
At eqm. (0.450-x) x x
As we are given:
The expression for equilibrium constant is:
Now put all the given values in this expression, we get:
The concentration of = x = 0.00212 M
Now we have to calculate the pH of solution.
Therefore, the pH of the solution is, 2.67
<h3>Answer:</h3>
1.43 × 10⁻²⁰ mol Li
<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>Atomic Structure</u>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Step 1: Define</u>
8.63 × 10³ atoms Li
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.43355 × 10⁻²⁰ mol Li ≈ 1.43 × 10⁻²⁰ mol Li
Answer:
1223.38 mmHg
Explanation:
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 =
Also,
Moles = mass (m) / Molar mass (M)
Density (d) = Mass (m) / Volume (V)
So, the ideal gas equation can be written as:
Given that:-
d = 1.80 g/L
Temperature = 32 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (32 + 273.15) K = 305.15 K
Molar mass of nitrogen gas = 28 g/mol
Applying the equation as:
P × 28 g/mol = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K
⇒P = 1223.38 mmHg
<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>