According the VSEPR theory the molecular geometry for CH3+ is triagonal planar
Answer:
See the answer below
Explanation:
<u>A scientist B might want to replicate the experiment of another scientist A in order to assess the claims made by scientist A.</u> In other words, replication of the experiment of another scientist is done to see if a similar outcome would be arrived at or there would be variations.
<em>The claims made by a scientist while reporting the outcome of a particular experiment must be reproducible by another scientist under similar conditions. Otherwise, the claims are said to be false.</em>
Answer:The new volume is 5mL
Explanation:
The formular for Boyles Law is; P1 V1 = P2 V2
Where P1 = 1st Pressure V1 = First Volume
P2 = 2nd Pressure V2 = Second Volume
From the question; P1 = 5atm, V1 = 10ml
P2 = 2 x P1 (2 x 5) = 10 atm V2 =?
Using the Boyles Law Formular; P1 V1 = P2 V2, we make V2 the subject of formular; P1 V1/ P2 = V2
∴ 5 x 10/ 10 = 5
∴ V2 = 5mL
Answer:
(A) 4.616 * 10⁻⁶ M
(B) 0.576 mg CuSO₄·5H₂O
Explanation:
- The molar weight of CuSO₄·5H₂O is:
63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol
- The molarity of the first solution is:
(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M
The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.
- Now we use the dilution factor in order to calculate the molarity in the second solution:
(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M
To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:
- 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
- 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O
