Hey There:
First order half life equation :
T 1/2 = ln ( 2 ) /K
T 1/2 = 0.693 / 3.40
T 1/2 = 0.204 s
Answer B
Answer:

Explanation:
We are asked to find the specific heat capacity of a liquid. We are given the heat added, the mass, and the change in temperature, so we will use the following formula.

The heat added (q) is 47.1 Joules. The mass (m) of the liquid is 14.0 grams. The specific heat (c) is unknown. The change in temperature (ΔT) is 1.80 °C.
- q= 47.1 J
- m= 14.0 g
- ΔT= 1.80 °C
Substitute these values into the formula.

Multiply the 2 numbers in parentheses on the right side of the equation.


We are solving for the heat capacity of the liquid, so we must isolate the variable c. It is being multiplied by 25.2 grams * degrees Celsius. The inverse operation of multiplication is division, so we divide both sides of the equation by (25.2 g * °C).



The original measurements of heat, mass, and temperature all have 3 significant figures, so our answer must have the same. For the number we found that is the hundredth place. The 9 in the thousandth place to the right tells us to round the 6 up to a 7.

The heat capacity of the liquid is approximately 1.87 J/g°C.
Mass of Gold = 267.165 × 0.01552494829
⇒ 4.1477228099
The amount of heat(q) required to raise m grams of a substance-specific C from T1 to T2 is given by
q=m C (T2-T1) ........1
Given : q= 2.1200 J
the initial temperature of gold, T1 = 22.0Celcius
the final temperature of gold, T2 = 1064.4Celcius
specific heat of gold = 0.131
putting values in eq 1:
⇒ 2.1200 = m × 0.131 × (1064.4-22)
⇒ 2.1200 = m × 0.131 × 1042.4
⇒ 2.1200 / 136.5544
⇒ 0.01552494829
Since 1g= 0.01552494829 Pounds
Mass of Gold = 267.165 × 0.01552494829
⇒ 4.1477228099
Learn more about temperature here: brainly.com/question/11464844
#SPJ9
Answer: 3) 39.96 amu
Explanation:
Mass of isotope Ar- 36 = 35.97 amu
% abundance of isotope Ar- 36= 0.337% = 
Mass of isotope Ar- 38 = 37.96 amu
% abundance of isotope 2 = 0.063 % = 
Mass of isotope Ar- 40 = 39.96 amu
% abundance of isotope 2 = 99.600 % = 
Formula used for average atomic mass of an element :

![A=\sum[(35.97\times 3.37\times 10^{-3})+(37.96\times 6.3\times 10^{-4})+(39.96\times 0.996)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%2835.97%5Ctimes%203.37%5Ctimes%2010%5E%7B-3%7D%29%2B%2837.96%5Ctimes%206.3%5Ctimes%2010%5E%7B-4%7D%29%2B%2839.96%5Ctimes%200.996%29%5D)

Therefore, the average atomic mass of argon is 39.96 amu