Methane, propane, ethane, and butane are four alkanes.
Answer:
Concentration of hydrogen ion, ![[H^+]=5.0118*10^{-6} M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D5.0118%2A10%5E%7B-6%7D%20M)
Explanation:
pH is defined as the negative logarithm of hydrogen ion's concentration.
The lower the value of pH, the higher the acidic the solution is.
The formula for pH can be written as:
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
Given,
pH of the saliva of Marco = 5.3
To calculate: Hydrogen ion concentration in the saliva
Thus, applying in the formula as:
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![5.3=-log[H^+]](https://tex.z-dn.net/?f=5.3%3D-log%5BH%5E%2B%5D)
So,
![log[H^+]=-5.3](https://tex.z-dn.net/?f=log%5BH%5E%2B%5D%3D-5.3)
![[H^+]=10^{(-5.3)}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B%28-5.3%29%7D)
![[H^+]=5.0118*10^{-6} M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D5.0118%2A10%5E%7B-6%7D%20M)
Answer is: freezing point is -0,226°C.
Answer is: the molal concentration of glucose in this solution is 1,478 m.
m(KCl) = 15 g.
n(KCl) = m(KCl) ÷ M(KCl).
n(KCl) = 15 g ÷ 74,55 g/mol.
n(KCl) = 0,2 mol
m(H₂O) = 1650 g ÷ 1000 g/kg = 1,65 kg.
b = n(KCl) ÷ m(H₂O).
b = 0,2 mol ÷ 1,65 kg = 0,122 m.
Kf(water) = 1,86°C/m.
ΔT = Kf(water) · b(solution).
ΔT = 1,86°C/m · 0,122 m.
ΔT = 0,226°C.
<span> aluminum is an element. All elements are pure substances, so that means they are homogenous.
please mark as brainliest
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