Answer:
Q = 28.9 kJ
Explanation:
Given that,
Mass of Aluminium, m = 460 g
Initial temperature, 
Final temperature, 
We know that the specific heat of Aluminium is 0.9 J/g°C. The heat required to raise the temperature is given by :

So, 28.9 kJ of heat is required to raise the temperature.
Answer:
When writing equation the mass on left side of equation must be equal to the mass on right side. True
Explanation:
The chemical reactions always follow the law of conservation of mass.
Law of conservation of mass:
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
Explanation:
This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
For example:
In photosynthesis reaction:
6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂
there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass because total mass is equal on both side of equation.
Answer:
The International Date Line passes through the mid-Pacific Ocean and roughly follows a 180 degrees longitude north-south
line on the Earth. It is located halfway round the world from the prime meridian—the zero degrees longitude established in Greenwich
50 grams or 50,000 mili grams is the mass of solute in 1000 grams of a solution having a concentration of 5 parts per million.
Explanation:
Total mass of solution = 1000 grams or 1000 ml since 1 gram = 1 ml
concentration is 5 parts per million ( 5 mg in 1000 ml solution or 0.005 gram in 1000 ml)
the formula used for parts per million:
parts per million = 
putting the values in the equation:
parts per million = 
0.005 x 1000 = mass of solute
50 grams= mass of solute
converting this into mg
50,000 mg. is the total mass of solute in 5ppm of 1000 ml solution.
Answer:
Empirical formula is Cr₂O₃.
Explanation:
Given data:
Percentage of Cr = 68.4%
Percentage of O = 31.6%
Empirical formula = ?
Solution:
Number of gram atoms of Cr = 68.4 / 52 = 1.3
2
Number of gram atoms of O = 31.6 / 16 = 1.98
Atomic ratio:
Cr : O
1.32/1.32 : 1.98/1.32
1 : 1.5
Cr : O = 1 : 1.5
Cr : O = 2(1 : 1.5)
Empirical formula is Cr₂O₃.