Answer:
A. Vegetable oil
Explanation:
Solubility is determined by the principle , "like dissolves like" .
i.e. , if a compound is polar then it will dissolve in a polar compound only , and
if a compound is non - polar then it will dissolve in a non - polar compound only .
Hence , from the question ,
A. Vegetable oil is a non polar compound ,
B. Polar compound
C. ionic compound , is a polar compound .
Since , wate is polar in nature , hence it will get dissolved in a only polar compound ,
Therefore ,
Water will not get dissolved in vegetable oil , being it a non - polar compound .
Answer:
First part: The new volume of the gas is 1786 Liters.
Second part: The temperature required to change the volume of the gas sample is 347 °C
Explanation:
We assume the Charles - Gay Lussac law where, in constant pressure, volume of a gas changes directly proportional to Temperature (in Kelvin)
V1 / T1 = V2/T2
37°C + 273 = 310 K
82°C + 273 = 355 K
1560L / 310°K = V2 / 355K
(1560 / 310) . 355 = V2
1786 L = V2
1560 L / 310 K = 3120 L / T2
T2 = 3120 L . (310 K / 1560 L)
T2 = 620 K
620K - 273 = 347°C
Answer:
C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O
Explanation:
Glucose (C₆H₁₂O₆) react with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O).
The equation can be written as follow:
C₆H₁₂O₆ + O₂ —> CO₂ + H₂O
The above equation can be balance as illustrated below:
C₆H₁₂O₆ + O₂ —> CO₂ + H₂O
There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 6 in front of CO₂ as shown below:
C₆H₁₂O₆ + O₂ —> 6CO₂ + H₂O
There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H₂O as shown below:
C₆H₁₂O₆ + O₂ —> 6CO₂ + 6H₂O
There are a total of 8 atoms of O on the left side and a total of 18 atoms on the right side. It can be balance by 6 in front of O₂ as shown below:
C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O
Now, the equation is balanced.
I'm going to go with Physical.
Answer:
The graph represents an endothermic reaction.
The products have more energy than the reactants.
80kJ
160kJ
80kJ
160kJ
