Answer:
3. 116.5 V
4. 119.6 V
Explanation:
3. Determination of the voltage.
Resistance (R) = 25 Ω
Current (I) = 4.66 A
Voltage (V) =?
V = IR
V = 4.66 × 25
V = 116.5 V
Thus, the voltage is 116.5 V
4. Determination of the voltage.
Current (I) = 9.80 A
Resistance (R) = 12.2 Ω
Voltage (V) =?
V = IR
V = 9.80 × 12.2
V = 119.6 V
Thus, the voltage is 119.6 V
Answer:
0.109 g.
Explanation:
Equation of the reaction:
Na3PO4 + 3HCl --> 3NaCl + H3PO4
Number of moles of HCl = molar concentration × volume
= 0.1 × 0.04
= 0.004 mol.
By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3
= 0.0013 mol
Mass of Na3PO4 = molar mass × number of moles
= 0.0013 × 164
= 0.219 g
Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance
= 0.219 × 50 g/100 g
= 0.109 g.
Answer:
7.12 mm
Explanation:
From coulomb's law,
F = kqq'/r².................... Equation 1
Where F = force, k = proportionality constant, q and q' = The two point charges, r = distance between the two charges.
Make r the subject of the equation,
r = √(kqq'/F).......................... Equation 2
Given: q = q' = 75.0 nC = 75×10⁻⁹ C, F = 1.00 N
Constant: k = 9.0×10⁹ Nm²/C².
Substitute into equation 2
r = √[ (75×10⁻⁹ )²9.0×10⁹/1]
r = 75×10⁻⁹.√(9.0×10⁹)
r = (75×10⁻⁹)(9.49×10⁴)
r = 711.75×10⁻⁵
r = 7.12×10⁻³ m
r = 7.12 mm
Hence the distance between the point charge = 7.12 mm
<span>If you look at the chlorine box, with the symbol Cl, you see the atomic mass is equal to 35.453 atomic mass units. This is the weighted average mass of chlorine, including its isotopes, as found in nature. This also means that one mole of chlorine atoms has a mass of 35.453 grams.</span>
For it to be the same element it must contain the same number of protons
