Answer:
0.52 g of KNO₃ are contained in 19.7 mL of diluted solution.
Explanation:
We can work on this problem in Molarity cause it is more easy.
Molarity (mol/L) → moles of solute in 1L of solution.
100 mL of solution = 0.1 L
We determine moles of solute: 44.7 g . 1mol /101.1 g = 0.442 mol of KNO₃
Our main solution is 0.442 mol /0.1L = 4.42 M
We dilute: 4.42 M . (11.9mL / 200mL) = 0.263 M
That's concentration for the diluted solution.
M can be also read as mmol/mmL, so let's find out the mmoles
0.263 M . 19.7mL = 5.18 mmol
We convert the mmol to mg → 5.18 mmol . 101.1 mg / mmol = 523.7 mg
Let's convert mg to g → 523.7 mg . 1 g / 1000 mg = 0.52 g
Answer:
Cr 6+ & SO4 2-
Explanation:
Sulfate is a polyatomic ion that has a charge of 2-. There are three of them in the chemical formula so it equates to a 6- total charge. Thus, chromium must have a 6+ charge to give the compound a neutral charge.
Answer:
Extensibility is a software engineering and systems design principle that provides for future growth.
Explanation:
hope ot helps good day
It would be 4 because the graph is showing the impact of heat on the enzyme not the pH. The pH in this instant is the constant and the graph does not indicate what a change in pH would do, only what a change in temperature would do.
The reaction between the reactants would be:
CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻
Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.
CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I 0.11 0 0
C -x +x +x
E 0.11 - x x x
Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
Since the given information is Kb, let's find Ka in terms of Kb.
Ka = Kw/Kb, where Kw = 10⁻¹⁴
So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]
Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
<em>pH = 5.83</em>
