Im going for d i think i took this last year
Answer:
ºC
Explanation:
We have to start with the variables of the problem:
Mass of water = 60 g
Mass of gold = 13.5 g
Initial temperature of water= 19 ºC
Final temperature of water= 20 ºC
<u>Initial temperature of gold= Unknow</u>
Final temperature of gold= 20 ºC
Specific heat of gold = 0.13J/gºC
Specific heat of water = 4.186 J/g°C
Now if we remember the <u>heat equation</u>:


We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

Now we can <u>put the values into the equation</u>:

Now we can <u>solve for the initial temperature of gold</u>, so:

ºC
I hope it helps!
Answer:
NaOBr (or) Na⁺ ⁻OBr
Explanation:
The Oxo-Acids of Bromine are as follow,
Hypobromous Acid = HOBr
Bromous Acid = HOBrO
Bromic Acid = HBrO₃
Perbromic Acid = HBrO₄
When these acids are converted to their conjugate bases their names are as follow,
Hypobromite = ⁻OBr
Bromite = ⁻OBrO
Bromate = ⁻OBrO₂
Perbromate = ⁻OBrO₃
According to rules, the positive part of ionic compound is named first and the negative part is named second. So, Sodium Hypobromite has a chemical formula of Na⁺ ⁻OBr or NaOBr.
I’m not sure what the answer is but I hope someone can help you