Answer: HCI + KOH → KCI + H20
Explanation:
HCI(aq) + KOH(aq) → KCI(aq) + H20(l)
Acid + base → Salt + Water.
The above is a neutralization reaction in which an acid, aqeous HCl reacts completely with an appropriate amount of a base, aqueous KOH to produce salt, aqueous KCl and water, liquid H2O only.
This is a neutralization reaction since, the hydrogen ion, H+, from the HCl is neutralized by the hydroxide ion, OH-, from the KOH to form the water molecule, H2O and salt, KCl only.
A and B are experiencing winter. The picture which isn't available here in this question is attached below.
Option C.
<h3><u>Explanation:</u></h3>
The earth is tilt by an angle of 23.2° to the vertical plane. This makes the seasonal variation of earth, because in some time of the year, the northern hemisphere faces the sun directly, experiencing summer and then southern hemisphere is away from summer experiencing winter and vice versa. The summer occurs when the place directly faces the sun. And the winter happens when the place obliquely faces the sun or doesn't face the sun at all.
Here in this diagram, we can see that the points A and B are the north pole and the part in northern hemisphere respectively which aren't facing the sun directly, whereas C and D are facing the sun. Thus the southern hemisphere is experiencing summer and the northern hemisphere the winter.
The patient needs 1000 ml of 5% (w/v) glucose solution
i.e 1000 ml x 5 g/ 100 ml
where the stock solution is 55% (w/v) = 55 g / 100 ml
So, 1000 ml x 5 g / 100 ml = V (ml) x 55 g / 100 ml
V = 1000 x (5 / 100) / (55 / 100) = 5000 / 55 = 90.9 ml
∴ the patient needs 90.9 ml of 55% (w/v) glucose solution
Answer:
2.5×10⁶ s
Explanation:
From the question given above, the following data were obtained:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
The half-life of a first order reaction is given by:
Half-life (t½) = 0.693 / Rate constant (K)
t½ = 0.693 / K
With the above formula, we can obtain the half-life of the reaction as follow:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
t½ = 0.693 / K
t½ = 0.693 / 2.8×10¯⁷
t½ = 2.5×10⁶ s
Therefore, the half-life of the reaction is 2.5×10⁶ s
