Answer: it is soluble
Explanation: nitrates are soluble.
Answer: 3.6 L = 3600
Explanation: 3.6 L = 360 mL has a mistake. He multiplied by 100 instead of 1000.
Answer:
C. Double Replacement
Explanation:
A. Wrong because SR uses a compound and element. This equation is a compound and compound.
B. Wrong because it does not have O2 in the formula. All combustion reactions must have O2.
C. Correct because it is a compound reacting with a compound.
D. Wrong because the reactants did not form a single product. ex. (x + y > xy)
E. Wrong because the equation did not start with a single compound and break down. ex. (xy > x + y)
Answer:
0.3229 M HBr(aq)
0.08436M H₂SO₄(aq)
Explanation:
<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>
<em />
Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).
NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)
When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>
<em />
Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).
2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)
When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

Answer:
1.3 × 10²³ Atoms of Mercury
Solution:
Step 1: Calculate Mass of Mercury using following formula,
Density = Mass ÷ Volume
Solving for Mass,
Mass = Density × Volume
Putting values,
Mass = 13.55 g.cm⁻³ × 3.2 cm³ ∴ 1 cm³ = 1 cc
Mass = 43.36 g
Step 2: Calculating number of Moles using following formula;
Moles = Mass ÷ M.mass
Putting values,
Moles = 43.36 g ÷ 200.59 g.mol⁻¹
Moles = 0.216 mol
Step 3: Calculating Number of Atoms using following formula;
Number of atoms = Moles × 6.022 ×10²³
Putting value of moles,
Number of Atoms = 0.216 mol × 6.022 × 10²³
Number of Atoms = 1.3 × 10²³ Atoms of Hg