The percentage of KCl present in the mixture is approximately <span>40%</span><span>If we consider a 100-gram sample of the mixture:ω(K) = 44.20% ÷ 100% = 0.442.
m(K) = 0.442 · 100 g = 44.2 g.</span>n(K) = 44.2 g ÷ 39.1 g/mol.
n(K) = 1.13 mol.
n(KCl) + n(KNO₃) = n(K)
m(KCl) = x.
m(KNO₃) = y.
Two equations:
1) x + y = 100 g.
2) m(KCl)/M(KCl) + m(KNO₃)/M(KNO₃) = 1.13 mol.
x/74.55 g/mol + y/101.1 g/mol = 1.13 mol.
From the first equation, we find x = 100 - y and substitute into the second equation:
(100 - y)/74.55 + y/101.1 = 1.13 /×101.1.
135.61 - 1.356y + y = 114.24.
0.356y = 22.37 g.
y = 62.83 g.
Thus, x = m(KCl) = 100 g - 62.83 g = 37.17 g.
The amount to administer to the child is 2,469 mL.
To convert to kilograms (kg), the child's weight in pounds (lb) is multiplied by 0.45359237: m(child) = 72.6 · 0.045359237 = 32.93 kg.
To find m(Medrol), the child's mass in kilograms is multiplied by 1.5 mg/kg.
Thus, m(Medrol) = 32.93 kg · 1.5 mg/kg = 49.39 mg.
The concentration of Medrol is d(Medrol) = 20.0 mg/mL.
To find the volume of Medrol needed, use V(Medrol) = m(Medrol) ÷ d(Medrol).
So, V(Medrol) = 49.39 mg ÷ 20 mg/mL = 2,469 mL.
A heavier player collides with a lighter player using greater force.
The lighter player sustains more injuries following the impact.
Explanation:
A heavier player impacts a lighter player with greater intensity, resulting in more pronounced injuries to the lighter player post-collision.
Force is defined as mass multiplied by the acceleration of an object;
Force = mass x acceleration
We observe that as mass and acceleration increase, the force exerted rises accordingly.
Clearly, the heavier player's mass surpasses that of the lighter player, leading to a greater force exerted upon collision.
Moreover, the lighter player is likely to be injured more severely after the clash. The momentum generated by the heavier player during the impact is considerably significant. Once they collide, the lighter player will certainly alter their speed and trajectory.
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Momentum
The weight of 16.3 moles of nickel amounts to 956.647 g.
Answer: 1.14
Explanation:
To find the molarity of the acid, we will utilize the equation derived from the neutralization process:
where
are the n-factor, molarity, and volume for the acid and
represent the n-factor, molarity, and volume for NaOH.
We know that:
By substituting the known values into the equation, we get:
To determine the pH of gastric juice:
The molarity amounts to = 0.072
Thus, the pH level of the gastric juice is 1.14
