B: SO2
S=Sulfur
O=Oxygen
Dioxide if i am correct means 2 oxygen.
so thats one Sulfur and 2 oxygen.
<h3>
Answer:</h3>
1.827 × 10²⁴ molecules H₂S
<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>Compounds</u>
- Writing Compounds
- Acids/Bases
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
103.4 g H₂S (Sulfuric Acid)
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of H - 1.01 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 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 4 sig figs.</em>
1.82656 × 10²⁴ molecules H₂S ≈ 1.827 × 10²⁴ molecules H₂S
Answer:
11.72 grams
Explanation:
Let the equilibrium concentration of BrCl be y
Initial concentration of Br2 = number of moles ÷ volume = (0.979×1000/160) ÷ 199 = 0.031 M
Initial concentration of Cl2 = (1.075×1000/71) ÷ 199 = 0.076 M
From the equation of reaction
1 mole of Br2 reacted with 1 mole of Cl2 to form 2 moles of BrCl
Therefore, equilibrium concentration of Br2 = (0.031 - 0.5y) M while that of Cl2 = (0.076 - 0.5y) M
Kp = [BrCl]^2/[Br2][Cl2]
1.1×10^-4 = y^2/(0.031 - 0.5y)(0.076 - 0.5y)
y^2/0.002356-0.0535y+0.25y^2 = 0.00011
y^2/0.00011 = 0.002356-0.0536y+0.25y^2
9090.9y^2-0.25y^2+0.0536y-0.002356 = 0
9090.65y^2+0.0535y-0.002356 = 0
The value of y must be positive and is obtained using the quadratic formula
y = [-0.0535 + sqrt(0.0535^2 - 4×9090.65×-0.002356)] ÷ 2(9090.65) = 9.2025/18181.3 = 0.00051 M
Mass of BrCl = concentration×volume×MW = 0.00051×199×115.5 = 11.72 grams
Answer: 1.311 × 10^18 photons are produced in each pulse
Explanation: Please see the attachments below
Answer:
d. Adding 5% HCl solution to the crude reaction mixture will protonate aniline increasing its solubility in the aqueous solution.
Explanation:
On this case, we have to check the <u>structures of each compound</u> (figure 1). For naphthalene we dont have <u>any functional groups</u> therefore, the addition of HCl or NaOH it will not affect naphthalene so <u>we can discard "B" and"C".</u>
When we add HCl solution we will have the production
the presence of this <u>hydronium ion will protonate the acid</u>, so we can <u>discard a.</u>
<u />
Finally, for d when we add
the <u>hydronium ion will react with aniline</u> (a base) and will produce an <u>ammonium ion</u>. This ammonium ion have a <u>positive charge</u>, therefore the <u>polarity will increase</u> and the molecule would be more soluble on water (figure 2).
I hope it helps!