20.7 Degree celsius is the temperature of Zippy's room.
Explanation:
The initial conditions of the kookie cola will be depicted by P1,V1 and T1.
P1= 36.5 psi or 2.4816 atm
V1= 355ml or 0.35 L
T1= 275.5K
The final conditions are given P2,V2 and T2
P2= 38.9 psi or 2.6469 atm
V2 = 355 ml or 0.35 L
also the solubility of the carbon dioxide does not change with temperature.
Applying the formula
P1VI/T1=P2V2/T2
since volume does not change and remains constant.
It can be written as:
P1/T1=P2/T2
= 2.4816/275.5=2.6469/T2
T2= 293.85 K
Answer:
A) Crystal lattice
Answer is: specific gravity of glucose is 1,02.
d(glucose) = 1,02 g/ml.
d(water) = 1,00 g/ml.
Specific gravity of glucose = density of glucose ÷ density of water.
Specific gravity of glucose = 1,02 g/ml ÷ 1,00 g/ml.
Specific gravity of glucose = 1,02.
Specific gravity<span> is the ratio of the </span>density<span> of a substance (in this case glucose) to the density of a reference substance (water).</span>
The answer is :
2H₂O (l) ------>2H₂ (g) + O₂ (g)
That is ,
Oxidation number of oxygen increases in reaction :
2H₂O (l) ------>2H₂ (g) + O₂ (g)
In H₂O, oxidation number of oxygen is -2 but in O₂ molecule it is 0.
So, oxidation number of oxygen increases in this reaction that is from -2 to 0.
So in the reaction that is 2H₂O (l) ------>2H₂ (g) + O₂ (g) , oxidation number of oxygen increases .
Answer:
32
Step-by-step explanation:
There are two ways you can count the valence electrons.
A. From the Periodic Table
1 × P (Group 15) = 5
4 × O (Group 16) = 4 × 6 = 24
+3 e⁻ (for the charges) = <u> 3</u>
Total = 32
B. From the Lewis structure
In the <em>Lewis structure</em> (below), each line (bond) represents a pair of bonding electrons, and each dot represents an unbound electron (half a lone pair).
5 lines (bonds) = 5 × 2 = 10
3 single-bonded O atoms = 3 × 6 = 18
1 double-bonded O atom = <u> 4</u>
Total = 32
