Answer:
0.6941 mg
Explanation:
First we <u>calculate how many LiNO₃ moles there are</u>, using the <em>given concentration and volume</em>:
- 1.0 mL * 0.10 M = 0.10 mmol LiNO₃
As 1 mol of LiNO₃ contains 1 mol of Li,<em> in the problem solution there are 0.10 mmol of Li</em> (the only metallic ion present).
Now we<u> convert Li milimoles into miligrams</u>, using its <em>atomic mass</em>:
- 0.10 mmol Li * 6.941 mg/mmol = 0.6941 mg
Answer:

Explanation:
We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.
We will use the following formula to calculate heat energy.

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.
- ΔT = final temperature - inital temperature
The aluminum block was heated from 23.0 °C to 73.5 °C.
- ΔT= 73.5 °C - 23.0 °C = 50.5 °C
Now we know all three variables and can substitute them into the formula.
- m= 225 g
- c= 0.897 J/g° C
- ΔT= 50.5 °C

Multiply the first two numbers. The units of grams cancel.



Multiply again. This time, the units of degrees Celsius cancel.


The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

Multiply by the answer we found in Joules.




The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

Approximately <u>10.2 kilojoules</u> of energy would be required.
The rate constant : k = 9.2 x 10⁻³ s⁻¹
The half life : t1/2 = 75.3 s
<h3>Further explanation</h3>
Given
Reaction 45% complete in 65 s
Required
The rate constant and the half life
Solution
For first order ln[A]=−kt+ln[A]o
45% complete, 55% remains
A = 0.55
Ao = 1
Input the value :
ln A = -kt + ln Ao
ln 0.55 = -k.65 + ln 1
-0.598=-k.65
k = 9.2 x 10⁻³ s⁻¹
The half life :
t1/2 = (ln 2) / k
t1/2 = 0.693 : 9.2 x 10⁻³
t1/2 = 75.3 s
Answer:
Look at the picture.
Explanation:
(2S,3S)-2-Bromo-3-phenylbutane will undergo E2 reaction and form trans product of elimination due to its thermodynamic stability.
A phylogenetic tree may be built using morphological (body shape), biochemical, behavioral, or molecular features of species or other groups. In building a tree, we organize species into nested groups based on shared derived traits (traits different from those of the group's ancestor).