Answer:
Cp = 0.093 J.g⁻¹.°C⁻¹
Solution:
The equation used for this problem is as follow,
Q = m Cp ΔT ----- (1)
Where;
Q = Heat = 300 J
m = mass = 267 g
Cp = Specific Heat Capacity = ??
ΔT = Change in Temperature = 12 °C
Solving eq. 1 for Cp,
Cp = Q / m ΔT
Putting values,
Cp = 300 J / (267 g × 12 °C)
Cp = 0.093 J.g⁻¹.°C⁻¹
According to Kepler's second law of orbital motion, a plane's orbital speed changes , depending on how far it is from the sun. The closer a planet is to the sun, the stronger the sun's gravitational pull on it, and the faster the planet moves. The farther away from the sun, the weaker the sun's gravitational pull and the slower it moves in its orbit.
The orbit of a planet around the sun is not a perfect circle, but an ellipse - a flattened circle.
Don’t eat or drink in labs
Dress for the lab; don’t wear open toed shoes
Dispose of lab waste properly
No horse play
Don’t taste or sniff things in the lab
Tire your hair back
The given chemical reaction given above is already balanced such that the number of atoms in the left hand side of the equation is equal to that of the right hand side. Using the dimensional analysis, proper conversion factors and the molar masses,
mass of nitrogen = (0.129 g H₂)(1 mol H₂/2 g H₂)(1 mol N₂/3 mol H₂)(28 g N₂/1 mol N₂)
mass of nitrogen = 0.602 g N₂
Therefore, 0.602 g of nitrogen will be required for he reaction.
Answer:
114mL.
Explanation: hope this helped