<span>The correct answer is( A) blood.
when the buffer solution its PH value changes very little when a small amount
of strong acid or base is added to it, and here the bicarbonate buffering system is used to regular the PH of the blood that keeping the PH at nearly constant value by maintaining the original acidity or basicity of the solution.</span>
An incandescent bulb becomes hotter than a fluorescent bulb when turned on because in a regular incandescent bulb, there is tungsten wire where electricity is converts into heat. A regular incandescent light bulb requires 4 times more energy than a fluorescent bulb in order to produce the same amount of light. The conversion is such that for a 75-watt bulb, temperature get raised to approximately 2000 K. For such a high temperature, the radiating energy from the wire have some visible light. In such bulbs, 90% of the electricity get consumed in producing heat and only 10% produces light thus, they are not much efficient source of light.
On the other hand, fluorescent bulbs produce light with less amount of heat. In them, 40% of electricity is consumed in producing light and 60% in heat which is very less as compared to heat produced by a incandescent bulb. This is because when it get turned on, mercury atoms inside the bulb collides with electrons and produce UV light which is then converted into visible light using thin layer of phosphor power present inside the bulb. This produces low amount of heat thus, the bulb stays cooler, the bigger size of bulb also helps in dispersing heat.
Therefore, a fluorescent light bulb is not as hot as an incandescent light bulb.
The correct answer is 0.014467 M.
Molarity is defined as the number of moles present in a liter solution, that is, number of moles / liter solution.
The molar mass of sodium (Na) is 23.0 g/mol
Thus, 1 mole of Na contains 23.0 g
Now, x moles of Na contains 0.50 g
Moles = 0.50 × 1 / 23.0
Moles = 0.50 / 23.0
= 0.0217 moles of Na
Molarity = Number of moles / liters of solution
= 0.0217 / 1.5
= 0.014467 M
<span>2Li⁺(aq) + Zn⁰(s) → 2Li⁰(s) + Zn²⁺(aq)
</span>2Li⁺(aq) + 2e⁻ → 2Li⁰(s)
Zn⁰(s) → Zn²⁺(aq) +2e⁻
2 electrons are transferred from atom of Zn⁰ to 2 ions of Li⁺.
