The mole ratio of NaOH to Al(OH)3 is
(1) 1:1 (2) 3:1
(3) 1:3 (4) 3:7
Answer:
0.60 mol·L⁻¹
Explanation:
Data:
LiBr: c = 0.50 mol/L; V =300 mL
RbBr: c = 0.70 mol/L; V =300 mL
1. Calculate the moles of Br⁻ in each solution
(a) LiBr
(b) RbBr
2. Calculate the molar concentration of Br⁻
(a) Moles of Br⁻
n = 0.150 mol + 0.210 mol = 0.360 mol
(b) Volume of solution
V = 300 mL + 300 mL = 600 mL = 0.600 L
(c) Molar concentration
The concentration of Br ions in the resulting solution of LiBr and RbBr has been 0.6 M.
The addition of LiBr and RbBr has been dissociated into the equal moles of Li, Rb, and Br.
Thus 1 mole of LiBr = 1 mole Br
1 mole RbBr = 1 mole Br.
The moles of LiBr in 0.5 M solution:
Molarity =
0.5 =
Moles of LiBr = 0.15 mol
The moles of Br from LiBr = 0.15 mol.
The moles of RbBr in 0.7 M solution:
Molarity =
0.7 =
Moles of RbBr = 0.21 mol
The moles of Br from RbBr = 0.21 mol.
The total moles of Br ions from LiBr and RbBr has been :
= 0.15 + 0.21
= 0.36 mol.
The total volume of the solution will be:
= 300 + 300 ml
= 600 ml.
The concentration of the Br ion has been:
Molarity =
Molarity of Br ions =
Molarity of Br ions = 0.6 M.
The concentration of Br ions in the resulting solution of LiBr and RbBr has been 0.6 M.
For more information about the concentration of the sample, refer to the link:
Answer:
Concentration of NaOH= 0.0036 M
Explanation:
Given data:
Volume of HCl = 25 mL
Concentration of HCl = 0.05 M
Volume of NaOH = 345 mL
Concentration of NaOH = ?
Solution:
Formula:
C₁V₁ = C₂V₂
C₁ = Concentration of HCl
V₁ = Volume of HCl
C₂ = Concentration of NaOH
V₂ = Volume of NaOH
Now we will put the values in formula.
C₁V₁ = C₂V₂
0.05 M × 25 mL = C₂ × 345 mL
1.25 M.mL = C₂ × 345 mL
C₂ = 1.25 M.mL/345 mL
C₂ = 0.0036 M
To find the concentration of the NaOH solution, we can use the concept of titration. By using the equation Moles = Concentration * Volume, we can calculate the moles of HCl used and then use the ratio of moles between HCl and NaOH to find the concentration of the NaOH solution.
To find the concentration of the NaOH solution, we need to use the concept of titration. From the given information, it takes 25 mL of 0.05 M HCl to neutralize 345 mL of NaOH solution. We can use the equation Moles = Concentration * Volume to find the amount in moles of HCl used. Then, we can use this information to calculate the concentration of the NaOH solution.
First, let's calculate the moles of HCl used:
Moles of HCl = (0.05 M) x (0.025 L) = 0.00125 mol
Next, we can use the ratio of moles between HCl and NaOH, which is 1:1, to find the moles of NaOH in the solution:
Moles of NaOH = 0.00125 mol
Finally, we can calculate the concentration using the formula:
The concentration of NaOH = (0.00125 mol) / (0.345 L) = 0.00362 M
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Answer:
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Explanation: