Answer:
C. It decreases by a factor of 4
Explanation:
F1 = kq1*q2/r²
F2 = kq1*q2/(2r)² = kq1*q2/(4r²) = kq1*q2/(r²*4) = F1/4
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
The products are on the right side of the equation. For this one it would be 2AlPO4 + 3CaSO4
C) is higher for nonpolar substances than for polar ones
One thing to notice in the question is, we are asked about molecular oxygen that has formula O2 not atomic oxygen O.
As we are asked about molecular oxygen, we will answer the question in terms of number of molecules that are present in 16 grams of molecular oxygen.
To get the number of molecules present in 16 grams of O2, we will use the formula:
No. of molecules = no. of moles x Avogadro's number (NA)----- eq 1)
As we know:
The number of moles = mass/ molar mass of molecule
Here we have been given mass already, 16 grams and the molar mass of O2 is 32 grams.
Putting the values in above formula:
= 16/32
= 0.5 moles
Putting the number of moles and Avogadro's number (6.02 * 10^23) in eq 1
No. of molecules = 0.5 x 6.02 * 10^23
=3.01 x 10^23 molecules
or 301,000,000,000,000,000,000,000 molecules
This means that 16 grams of 3.01 x 10^23 molecules of oxygen.
Hope it helps!
