A catalyst only speed up a chemical reaction
D is correct
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
It isn't a materialistic object.
Answer:
Atoms of sulfur = 9.60⋅g32.06⋅g⋅mol−1×6.022×1023⋅mol−1
Explanation:
because the units all cancel out, the answer is clearly a number, ≅2×1023 as required.
Answer is: Keq expression for this system is Keq = <span>[O</span>₂<span> ]</span> · [H₂<span>]</span>² ÷ [H₂O<span>]</span>².<span>
Chemical reaction: 2H</span>₂O(g) ⇄ O₂(g) + 2H₂(g).
The equilibrium constant<span> (Keq) is a ratio of the concentration of the products (in this reaction oxygen and hydrogen) to the concentration of the reactants (in this reaction water).</span>
