Answer:
70 mL of 5% HCl and 30 mL of 15% HCl
Explanation:
We will designate x to be the fraction of the final solution that is composed of 5% HCl, and y to be the fraction of the final solution that is composed of 15% HCl. Since the percentage of the final solution is 8%, we can write the following expression:
5x + 15y = 8
Since x and y are fractions of a total, they must equal one:
x + y = 1
This is a system of two equations with two unknowns. We will proceed to solve for x. First, an expression for y is found:
y = 1 - x
This expression is substituted into the first equation and we solve for x.
5x + 15(1 - x) = 8
5x+ 15 - 15x = 8
-10x = -7
x = 7/10 = 0.7
We then calculate the value of y:
y = 1 - x = 1 - 0.7 = 0.3
Thus 0.7 of the 100 mL will be the 5% HCl solution, so the volume of 5% HCl we need is:
(100 mL)(0.7) = 70 mL
Similarly, the volume of 15% HCl we need is:
(100 mL)(0.3) = 30 mL
The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.
Given, T = 25°C.
<h3>Chemical equation:</h3>
Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2
PbCl2 in aqueous solution split into following ions
PbCl2 ------ Pb(+2) + 2Cl-
Q = [Pb(+2)] [Cl-]^2
The Concentration of Pb(+2) ions and Cl- ions can be calculated as
[Pb(+2)] = 0.06 × 125/200
= 0.0375
[Cl-] = 0.02 × 75/200
= 0.0075
By substituting all the values, we get
[0.0375] [0.0075]^2
= 2.11 × 10^(-6).
Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
learn more about Ions:
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He did experiments with combustion and gas's
The pressure in the flask is 3.4 atm.
<em>pV</em> = <em>nRT
</em>
<em>T</em> = (20 + 273.15) K = 293.15 K
<em>p</em> = (<em>nRT</em>)/<em>V</em> = (1.4 mol × 0.082 06 L·atm·K⁻¹mol⁻¹ × 293.15 K)/10 L = 3.4 atm
