Answer is a
bacteria doesn't have any cellular function in the cell.
Since there is one carbon with 4 Fluorines attached to it, and both compounds are no metals, we use the covalent method for naming,
Here we ignore the prefix for the first element if it is 1. Mono. Then pay attention to the second one, it would be tetra, because tetra means 4. Here there are 4 fluorines.
Drop ine and place ide
CF4 = carbon tetrafluoride.
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: 18.65L
Explanation:
Given that,
Original volume of oxygen (V1) = 30.0L
Original temperature of oxygen (T1) = 200°C
[Convert temperature in Celsius to Kelvin by adding 273.
So, (200°C + 273 = 473K)]
New volume of oxygen V2 = ?
New temperature of oxygen T2 = 1°C
(1°C + 273 = 274K)
Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law
V1/T1 = V2/T2
30.0L/473K = V2/294K
To get the value of V2, cross multiply
30.0L x 294K = 473K x V2
8820L•K = 473K•V2
Divide both sides by 473K
8820L•K / 473K = 473K•V2/473K
18.65L = V2
Thus, the new volume of oxygen is 18.65 liters.
B. combination of all wavelengths of visible light
