The answer is- The energy of 1 L water at temperature 347.78 °C have more energy as 1 L of water at temperature 65°C.
Heat is a type of energy that causes a person's body to feel hot or cold.
While the temperature of an object is a parameter that indicates how hot or cold the object is.
How is the temperature in degree Fahrenheit converted to degree celsius?
- To convert the temperature in Fahrenheit to Celsius, subtract 32 and multiply by 5/9.
°
- Now, heat is a form of energy that flows from hotter object to colder object and temperature indicates whether the object is hot or cold by measuring its average kinetic energy.
- Now, the given temperature of 1 L water is 658 °F. This temperature in degree celsius is calculated as-
°C 
- Now, higher the temperature, higher is the energy of water. Thus, the energy of 1 L water at 347.78 °C have more energy as 1 L of water at 65°C.
To learn more about heat and temperature, visit:
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White phosphorus melts and then vaporizes at high temperatures. The gas effuses at a rate that is 0.404 times that of neon in the same apparatus under the same conditions-There are 4 atoms of P in the molecule
Explanation:
Ar=30,97g/mol
/
=
=0,404
0,404=
=20,18/30,97*x
X=20,18/30,97*0,163
X=4
There are 4 atoms of P in the molecule
White phosphorus melts and then vaporizes at high temperatures. The gas effuses at a rate that is 0.404 times that of neon in the same apparatus under the same conditions-There are 4 atoms of P in the molecule
Maybe to not get rained on.
Hahhahahaha I ain't sure tho
The oxidation number of elements in equation below are,
4NH₃ + 3Ca(ClO)₂ → 2N₂ + 6H₂O + 3CaCl₂
O.N of N in NH₃ = -3
O.N of Ca in Ca(ClO)₂ and CaCl₂ = +2
O.N of N in N₂ = 0
O.N of Cl in Ca(ClO)₂ = +1
O.N of Cl in CaCl₂ = -1
Oxidation:
Oxidation number of Nitrogen is increasing from -3 (NH₃) to 0 (N₂).
Reduction:
Oxidation number of Cl is decreasing from +1 [Ca(ClO)₂] to -1 (CaCl₂).
Result:
<span>N is oxidized and Cl is reduced.</span>
Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
