I believe cc would result in a recessive pair.
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
Answer:
531.6g
Explanation:
Total moles of glucose in this case is: 886/180= 4.922 (mole)
For every 1 mole glucose we get 6 mole water
-> Mole of water is: 4.922 * 6= 29.533 (mole)
weight of water is 18. Therefore, total weight of water that we will have from 886g of glucose are: 25.933*18= 531.6g
Lithium atomic number = 3
Li atom has 3 protons,3 electrons.Then it has no charge.
So the charge of Li atom here is zero.
<u>Option b: 0 is the right answer.</u>
Of course, at STP, dioxygen is a gas, but 10.0 g is still 10.0 g. We could calculate its volume at STP, which is 22.4 L × its molar quantity, approx. 8⋅L . There are 1.51×1023molecules O2 in 10.0 g O2 .
