3.
protium (A = 1), deuterium (A = 2), and tritium (A = 3).
Answer:
9.51 × 10⁴ kL
Explanation:
Step 1: Given data
Volume of the sample (V): 9.51 × 10⁹ cL
Step 2: Convert "V" to liters
We will use the conversion factor 1 L = 100 cL.
9.51 × 10⁹ cL × (1 L / 100 cL) = 9.51 × 10⁷ L
Step 3: Convert "V" to kL
We will use the conversion factor 1 kL = 1000 L.
9.51 × 10⁷ L × (1 kL / 1000 L) = 9.51 × 10⁴ kL
9.51 × 10⁹ cL is equal to 9.51 × 10⁴ kL.
element is made up of only one type of atoms.
it is pure form of matter
example hydrogen, oxygen, carbon etc
6. a) Team 3 (b/c their numbers were closer apart then the rest)
b)No they were not (can happen you might be precises but not accurate)
c)Team four, (on trial 4)
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.
