Answer:
625 mL
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V₁) = 250 mL
Molarity of stock solution (M₁) = 5 M
Molarity of diluted solution (M₂) = 2 M
Volume of diluted solution (V₂) =?
The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5 × 250 = 2 × V₂
1250 = 2 × V₂
Divide both side by 2
V₂ = 1250 / 2
V₂ = 625 mL
Therefore, the volume of the diluted solution is 625 mL.
<span>As temperature increases, the amount of solute that a solvent can dissolve increases.</span>
<span>Answer: False
</span>
Antibodies can destroy pathogens by (i) binding to and blocking the pathogen's receptors, thus causing neutralization of the pathogen, (ii) binding to the pathogen and activating complement, and (iii) binding to the pathogen and facilitating its opsonization and uptake by macrophages, which utilize their Fc receptors ...
Can you take picture of the whole question? I can’t see.
<span>BaCl2+Na2SO4---->BaSO4+2NaCl
There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent?
"First convert grams into moles"
1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2
1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4
(7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2
"From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl"
The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
