Kepler did not study the speed of the planets, rather, he studied how the planets move in the solar system. He proposed three laws. As a summary, he described that the planets move around the sun in the shape of an ellipse (orbit), and the Sun being one of the foci. Then, he proposed the period for the planet to complete one revolution around the Sun.
On the other hand, Newton studied the forces acting on the planet (or any object in space) that explain how the planets move around the solar system as described by Kepler. Also, Kepler's observations only apply to planets and not the moons or satellites. Thus, Kepler only made laws from observations, while Newton based it from underlying principles that led him to mathematical equations such as the law of universal gravitation.
The balanced chemical reaction is:
<span>CuCl2 + 2Na → 2NaCl + Cu
We are given the amount of sodium to be used up in the reaction. This will be the starting point for our calculations.
15 g Na ( 1 mol / 22.99 ) ( 1 mol Cul2 / 2 mol Na ) (134.45 g / 1 mol ) = 43.86 g CuCl2 needed to be able to obtain the maximum amount of copper.</span>
C is not a property of metals
Th actual yield of the reaction is 24.86 g
We'll begin by calculating the theoretical yield of the reaction.
2Na + Cl₂ → 2NaCl
Molar mass of Na = 23 g/mol
Mass of Na from the balanced equation = 2 × 23 = 46 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl from the balanced = 2 × 58.5 = 117 g
From the balanced equation above,
46 g of Na reacted to produce 117 g of NaCl.
Therefore,
11.5 g of Na will react to produce = (11.5 × 117) / 46 = 29.25 g of NaCl.
Thus, the theoretical yield of NaCl is 29.25 g.
Finally, we shall determine the actual yield of NaCl.
- Theoretical yield = 29.25 g
Actual yield = Percent yield × Theoretical yield
Actual yield = 85% × 29.25
Actual yield = 0.85 × 29.25 g
Actual yield = 24.86 g
Learn more about stoichiometry: brainly.com/question/25899385
In this compound (Phosgene) the central atom (carbon is Sp² Hybridized).
Sp, Sp² and Sp³ can be calculated very simply by doing three steps,
Step 1:
Assume triple bond and double bond as one bond and assign s or p to it. In this example carbon double bond oxygen is considered once and let suppose it is s. Now we are having our s.
Step 2:
Count lone pair of electron, each lone pair counts for s and p. In this case there is no lone pair of electron on carbon, so not included.
Step 3:
Count single bonds for s and p. As we have already assigned s to the double bond, now one p for one single bond, and other p for the other single bond.
Result:
So, we counted 1 s for double bond, 1 p for one single and other p for second single bond. As a whole we got,
Sp²
Practice:
You can practice for hybridization of Oxygen in this molecule. Oxygen has 2 lone pair of electrons. (Hint: Sp² Hybridization)
