Answer:
The specific heat of molybdenum is 0.254 joules per gram-Celsius.
Explanation:
We consider the system formed by the molybdenum metal and water as our system, a control mass inside an insulated cup, that is, a container that avoids any energy and mass interactions between system and surroundings.
From statement we notice that metal is cooled down whereas water is heated. According to the First Law of Thermodynamics, we know that:
Where:
- Heat received by water, measured in joules.
- Heat released by metal, measured in joules.
Now we expand this identity by definition of sensible heat:
The specific heat of the metal is cleared within equation above:
If we know that 
, 
, 
, 
, 
and 
, the specific heat of molybdenum is:
The specific heat of molybdenum is 0.254 joules per gram-Celsius.
ANSWER:
4 a) Specific elements have more than one oxidation state, demonstrating variable valency.
For example, the following transition metals demonstrate varied valence states: 
, 
, 
, etc.
Normal metals such as 
also show variable valencies. Certain non-metals are also found to show more than one valence state 
4 b) Isotopes are members of a family of an element that all have the same number of protons but different numbers of neutrons.
For example, Carbon-14 is a naturally occurring radioactive isotope of carbon, having six protons and eight neutrons in the nucleus. However, C-14 does not last forever and there will come a time when it loses its extra neutrons and becomes Carbon-12.
5 a) 
→ 
5 b) 
→ 
5 c) 
→ 
(already balanced so don't need to change)
5 d) 
→ 
5 e) 
→ 
EXPLANATION (IF NEEDED):
1. Write out how many atoms of each element is on the left (reactant side) and right (product side) of the arrow.
2. Start multiplying each side accordingly to try to get atoms of the elements on both sides equal.
EXAMPLE OF BALANCING:
Because things that are burnt have been changed into something else like moldy bread thats cooked.
It is chemical change because you cannot unburn something.
The decomposition time : 7.69 min ≈ 7.7 min
<h3>Further explanation</h3>
Given
rate constant : 0.029/min
a concentration of 0.050 mol L to a concentration of 0.040 mol L
Required
the decomposition time
Solution
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time
For first-order reaction :
[A]=[Ao]e^(-kt)
or
ln[A]=-kt+ln(A0)
Input the value :
ln(0.040)=-(0.029)t+ln(0.050)
-3.219 = -0.029t -2.996
-0.223 =-0.029t
t=7.69 minutes
Answer:
the same is what is this question like what did u exame
