<span>Mg + O2 > MgO. In reactant side, 2 O atoms and 1 Mg are present. In product side, 1 Mg and O atoms are present. Put 2 in product side to balance O atoms and 2 at Mg in reactant side to balance Mg atoms. Therefore the balanced equation becomes, 2Mg + O2 ----> 2MgO. Hope it helps.</span>
pH of 0.048 M HClO is 4.35.
<u>Explanation:</u>
HClO is a weak acid and it is dissociated as,
HClO ⇄ H⁺ + ClO⁻
We can write the equilibrium expression as,
Ka = ![$\frac{[H^{+}] [ClO^{-}] }{[HClO]}](https://tex.z-dn.net/?f=%24%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%20%5BClO%5E%7B-%7D%5D%20%20%7D%7B%5BHClO%5D%7D)
Ka = 4.0 × 10⁻⁸ M
4.0 × 10⁻⁸ M = 
Now we can find x by rewriting the equation as,
x² = 4.0 × 10⁻⁸ × 0.048
= 1.92 × 10⁻⁹
Taking sqrt on both sides, we will get,
x = [H⁺] = 4.38 × 10⁻⁵
pH = -log₁₀[H⁺]
= - log₁₀[ 4.38 × 10⁻⁵]
= 4.35
The second stage of photosynthesis also called Calvin stage produces glucose
Answer:
3) NaCl.
Explanation:
<em>∵ ΔTf = iKf.m</em>
where, <em>i</em> is the van 't Hoff factor.
<em>Kf </em>is the molal depression freezing constant.
<em>m</em> is the molality of the solute.
<em>The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. </em>
<em></em>
- For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.
<em>So, for sugar: i = 1.</em>
<em>∴ ΔTf for sugar = iKf.m = (1)(Kf)(2.0 m) = 2 Kf.</em>
<em></em>
- For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.
For NaCl, it is electrolyte compound which dissociates to Na⁺ and Cl⁻.
<em>So, i for NaCl = 2.</em>
<em>∴ ΔTf for NaCl = iKf.m = (2)(Kf)(1.0 m) = 2 Kf.</em>
<em></em>
<em>So, the right choice is: 3) NaCl.</em>
<em></em>
Answer: E = 2.455 x 10^5 N/C
Explanation:
q1 = 1.2x10^-7C
q2 = 6.2x10^-8C
Electric field, E = kQ/r²
where k = 9.0x10^9
since the location is (27 - 5)cm from q1
hence electric field, E1 = k*q1/r²
E1= (9x10^9 x 1.2x10^-7)/(0.22)² = 22314.05 N/C
for q2:
E1 = k*q2/r²
E2 at 5cm
E2 = (9x10^9 x 6.2x10^-8)/(0.05)² = 223200 N/C
Hence, the total electric field at 5cm position is
E = E1 + E2
E = 22314.05 + 223200 = 245514.05 N/C
E = 2.455 x 10^5 N/C
