Answer:
H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)
Explanation:
H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq) + SO²⁻₄(aq) + H₂O(l)
A careful observation of the equation above, shows that the equation is already balanced.
To obtain the net ionic equation, we simply cancel Mg²⁺ from both side of the equation as shown below:
H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)
For this problem, we use Graham's Effusion Law to find out the rate of effusion of chlorine gas. The formula is as follows:
R₁/R₂ = √(M₂/M₁)
Let 1 be N₂ while 2 be Cl₂
255/R₂ = √(28/70.8)
Solving for R₂,
R₂ = 405.5 s
<em>Thus, it would take 405.5 s to effuse chlorine gas.</em>
Answer:
Explanation:
Take a random sample of nuts from the jar. Let's take two handfuls, after shaking the jar and mixing the nuts thoroughly. Separate the nuts into almonds and cashews. Count each pile, then do the following calculation (these numbers are random, for example only).
<u> Count</u> <u>Percentage %</u>
Almonds 38 (38)/(87)x100
Cashews <u> 49</u> 49/87x100
87 87/87 = 100%
Ratio of Almonds to Cashews: <u>38/49</u>
Answer:
A. 0.143 M
B. 0.0523 M
Explanation:
A.
Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).
KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄
The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:
1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol
The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.
5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:
M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M
B.
Let's consider the neutralization of potassium hydroxide and perchloric acid.
KOH + HClO₄ → KClO₄ + H₂O
When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.
