Answer:
4NH₃(g) +3O₂(g) ⇒2N₂(g) +6H₂O(g)
Explanation:
Answer:
The specific heat of aluminium is 0.8792 J/g °C or 0.21 Cal/g °C
Explanation:
Step 1 : Write formule of specific heat
Q=mcΔT
with Q = heat transfer (J)
with m = mass of the substance
with c = specific heat ⇒ depends on material and phase ( J/g °C)
with ΔT = Change in temperature
For this case :
Q = 1680 Calories = 7033.824 J ( 1 calorie = 4.1868 J)
m = 100.0g
c= has to be determined
ΔT = 100 - 20 = 80°C
<u>Step 2: Calculating specific heat</u>
⇒ via the formule Q=mcΔT
7033.824 J = 100g * c * 80
7033.824 = 8000 *c
c = 7033.824 /8000
c = 0,879228 J/g °C
or 0.21 Cal / g°C
The specific heat of aluminium is 0.8792 J/g °C or 0.21 Cal/g °C
thats kinda hot not even gon'
Answer: Anion. The resulting of losing or gaining electrons would be Anion.
Answer:
42.6 % CHCl₃, 13.83 % C₃H₆O and 43.5 % CH₃COBr
Explanation:
This is a solution with 3 compounds:
96.2 g of CHCl₃
31.2 g of C₃H₆O
98.1 g of CH₃COBr
Total mass of solution is : 96.2g + 31.2 g + 98.1g = 225.5 g
Percent by mass is, the mass of the compounds in 100 g of solution
Let's prepare the rule of three
225.5 g of solution have ____ 96.2 g ___ 31.2 g _____98.1 g
100 g of solution would have ____
(100 . 96.2) / 225.5 = 42.6
(100 . 31.2) / 225.5 = 13.83
(100 . 98.1) / 225.5 = 43.5
