Mass of metal piece is 611 g and volume of graduated cylinder is 25.1 mL. When metal piece is placed in the graduated cylinder water level increases to 56.7 mL. The increase in volume is due to volume of metal piece that gets added to the volume of water.
Thus, volume of metal piece can be calculated by subtracting initial volume from the final one.
Thus, volume of metal piece will be 31.6 mL. The mass of metal piece is given 611 g, density of metal can be calculated as follows:
Therefore, density of metal is 19.33 g/mL.
Answer:
The new pressure is 53.3 kPa
Explanation:
This problem can be solved by this law. when the volume remains constant, pressure changes directly proportional as the Aboslute T° is modified.
T° increase → Pressure increase
T° decrease → Pressure decrease
In this case, temperature was really decreased. So the pressure must be lower.
P₁ / T₁ = P₂ / T₂
80 kPa / 300K = P₂/200K
(80 kPa / 300K) . 200 K = P₂ → 53.3 kPa
Answer:
The correct option is b
Explanation:
Firstly, the compound is ClF₃ and not ClF₃ClF₃. The name of the compound ClF₃ is chlorine trifluoride. It's electron geometry is trigonal bipyramidal (with the chlorine at the center and the atoms of the fluorine forming a triangular bipyramid around it) with a bond angle of 175° with an hybridization of sp³d.
