There’s going to be 8 neutrons presented
The density of the rock is 3.314g/mL
CALCULATE DENSITY:
- According to this question, a rock weighs 23.2g. After dropping the rock into a graduated cylinder containing 55mL of water, the level changes to 62mL.
- This means that the volume of the rock can be calculated as follows:
Volume of rock = 62mL - 55mL
Volume of rock = 7mL
Density can be calculated using the formula as follows:
Density = mass ÷ volume
Density = 23.2 ÷ 7
Density = 3.314g/mL
Therefore, the density of the rock is 3.314g/mL
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Answer:
(FeSCN⁺²) = 0.11 mM
Explanation:
Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2
M (Fe(NO₃)₃ = 0.200 M
V (Fe(NO₃)₃ = 10.63 mL
n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol
M (KSCN) = 0.00200 M
V (KSCN) = 1.42 mL
n (KSCN) = 0.00200 * 1.42 = 0.00284 mmol
Total volume = V (Fe(NO₃)₃ + V (KSCN)
= 10.63 + 1.42
= 12.05 mL
Limiting reactant = KSCN
So,
FeSCN⁺² = 0.00284 mmol
M (FeSCN⁺²) = 0.00284/12.05
= 0.000236 M
Excess reactant = (Fe(NO₃)₃
n(Fe(NO₃)₃ = 2.126 mmol - 0.00284 mmol
=2.123 mmol
For standard 2:
n (FeSCN⁺²) = 0.000236 * 4.63
=0.00109
V(standard 2) = 4.63 + 5.17
= 9.8 mL
M (FeSCN⁺²) = 0.00109/9.8
= 0.000111 M = 0.11 mM
Therefore, (FeSCN⁺²) = 0.11 mM
Answer:
263.1 is exactly three half-life, so the remaining portion is (1/2 x 1/2 x 1/2) of the original sample. That's 1/8 which is 12.5%
The answer is intermediate, so E.
