Current is the rate at which electric charge flows past a point in a circuit. In other words, current is the rate of flow of electric charge. Voltage, also called electromotive force, is the potential difference in charge between two points in an electrical field. ... Current is the effect (voltage being the cause).
Answer:
83.33ml
Explanation:
Firstly, let's notice that 200ml are equal to 0.2L
M=number of moles of the solute ÷ number of liters. Let's call the number of liter by x:
2.5=x÷0.2
x=2.5×0.2
x=0.5
It means you'll need 0.5 mol of HNO. But your solution have 6 moles per liter, so how much do you need? Divide the moles by 12 to have 0.5, so now you need to divide the liter per 12 too. The liter is equal to 1000ml, so:
1000ml÷12=83.33ml
The balanced reaction is: H2O2 -> H2O + (1/2) O2
However, this is a very awkward equation since it has a
coefficient in fraction form.
Another way to write this equation is:
2 H2O2 -> 2 H2O + O2
Hope this helps.
1) Chemical reaction
HCl + NaOH ---> NaCl + H2O
25.0 ml
0.150 M 0.250M
2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution
0.001875 mol HCl => 0.001875 mol H(+)
Volume = Volume of HCl solution + Volumen of NaOH solution added
Volume of HCl solution = 0.0250 l
Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l
Total volume = 0.0250 l + 0.0075 l = 0.0325 l
[H+] = 0.001875 mol / 0.0325 l = 0.05769 M
pH = - log [H+] = - log (0.05769) = 1.23
Answer: 1.23
3) Equivalence point
0.02500 l * 0.150 M = 0.250M * V
=> V = 0.02500 * 0.150 / 0.250 = 0.015 l
4) 1.00 ml NaOH added beyond the equivalence point
1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess
0.00025 mol NaOH = 0.00025 mol OH-
Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l
[OH-] = 0.00025 mol / 0.041 l = 0.00610 M
pOH = - log (0.00610) = 2.21
pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76
Answer: 11.76
Answer:
Explanation:
Let the monoprotic acid be HX
HX ⇄ H⁺ + X⁻
pH = 2.53
Hydrogen ion concentration
Concentration of undissociated acid will remain almost the same as it is a weak acid
So
Ka = concentration of H⁺ x concentration of Cl⁻ / concentration of acid
= [ H⁺] x [Cl⁻ ] / [ HX]
= 5.24 x 10⁻⁴ M .