Answer:
<u><em>A. They don't form compounds</em></u>
Explanation:
- High density: The strong bonding between the atoms in transition metals cause them to be tightly held together, creating a high density. Colored compounds: Transition metals tend to form more colored compounds than other elements, either in solid form or dissolved in a solvent.
No disadvantage, it’s important have a lid on a crucible to hand extreme temperatures. The lid may also be used to contain chemicals being heated whilst in a lab.
Zn (s) -> Zn+2 (aq) + 2e-
Zn (s) with a neutral charge is oxidized and looses two electrons in the process to form ZnCl2 (aq) where Zn has a charge of 2+.
Answer:
F' = 3 N
Explanation:
Given that,
Two objects of equal mass have a force of gravity of 6 N between them.
If the mass of one is cut in half and the other stays the same such that,
m₁' = m₁ and m₂' = (m₂/2)
We need to find the new force. The gravitational force between two objects is given by :
We have, F = 6 N
New force,
So, the new force becomes 3 N.
Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
