Answer:
The concentration of hydrogen ion at pH is equal to 2 :![= [H^+]=0.01 mol/L](https://tex.z-dn.net/?f=%3D%20%5BH%5E%2B%5D%3D0.01%20mol%2FL)
The concentration of hydrogen ion at pH is equal to 6 : ![[H^+]'=0.000001 mol/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%27%3D0.000001%20mol%2FL)
There are 0.009999 more moles of
ions in a solution at a pH = 2 than in a solution at a pH = 6.
Explanation:
The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
The hydrogen ion concentration at pH is equal to 2 = [H^+]
![2=-\log [H^+]\\](https://tex.z-dn.net/?f=2%3D-%5Clog%20%5BH%5E%2B%5D%5C%5C)
![[H^+]=10^{-2}M= 0.01 M=0.01 mol/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-2%7DM%3D%200.01%20M%3D0.01%20mol%2FL)
The hydrogen ion concentration at pH is equal to 6 = [H^+]
![6=-\log [H^+]\\\\](https://tex.z-dn.net/?f=6%3D-%5Clog%20%5BH%5E%2B%5D%5C%5C%5C%5C)
![[H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-6%7DM%3D%200.000001%20M%3D%200.000001%20mol%2FL)
Concentration of hydrogen ion at pH is equal to 2 =![[H^+]=0.01 mol/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01%20mol%2FL)
Concentration of hydrogen ion at pH is equal to 6 = ![[H^+]'=0.000001 mol/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%27%3D0.000001%20mol%2FL)
The difference between hydrogen ion concentration at pH 2 and pH 6 :
![= [H^+]-[H^+]' = 0.01 mol/L- 0.000001 mol/L = 0.009999 mol/L](https://tex.z-dn.net/?f=%3D%20%5BH%5E%2B%5D-%5BH%5E%2B%5D%27%20%3D%200.01%20mol%2FL-%200.000001%20mol%2FL%20%3D%200.009999%20mol%2FL)
Moles of hydrogen ion in 0.009999 mol/L solution :

There are 0.009999 more moles of
ions in a solution at a pH = 2 than in a solution at a pH = 6.
Answer:
12 mins
Explanation:
The distance covered is 5 km, divide this by 25 to get the fraction of an hour it takes. Doing this you get .2, times this by 60 min (1 hour) to get how many mins it takes
Answer:
a. I = 0.76 A
b. Z = 150.74
c. RL₁ = 34.41 , RL₂ = 602.58
d. RL₂ = 602.58
Explanation:
V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V , Rc = 473 Ω
a.
Using law of Ohm
V = I * R
I = Vc / Rc = 364 V / 473 Ω
I = 0.76 A
b.
The impedance of the circuit in this case the resistance, capacitance and inductor
V = I * Z
Z = V / I
Z = 116 v / 0.76 A
Z = 150.74
c.
The reactance of the inductor can be find using
Z² = R² + (RL² - Rc²)
Solve to RL'
RL = Rc (+ / -) √ ( Z² - R²)
RL = 473 (+ / -) √ 150.74² 77.0²
RL = 473 (+ / -) (129.58)
RL₁ = 34.41 , RL₂ = 602.58
d.
The higher value have the less angular frequency
RL₂ = 602.58
ω = 1 / √L*C
ω = 1 / √ 602.58 * 473
f = 285.02 Hz
Answer:
The quantitative relationship between heat transfer and temperature change contains all three factors: Q = mcΔT, where Q is the symbol for heat transfer, m is the mass of the substance, and ΔT is the change in temperature. The symbol c stands for specific heat and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00ºC. The specific heat c is a property of the substance; its SI unit is J/(kg ⋅ K) or J/(kg ⋅ ºC). Recall that the temperature change (ΔT) is the same in units of kelvin and degrees Celsius. If heat transfer is measured in kilocalories, then the unit of specific heat is kcal/(kg ⋅ ºC).
Explanation:
Answer:
0.1g to 0.0000001g hope it helps uu