Answer:
Chlorine gas.
Explanation:
Hello!
In this case, the undergoing chemical reaction is:

Thus, given the moles of reacting both sodium and chlorine, we compute the moles of sodium chloride yielded by each reactant by considering the 2:2 and 1:2 mole ratios:

Thus, since chlorine yields less moles of sodium chloride, we infer it is the limiting reactant.
Best regards!
Answer:
Explanation:
H ₂ S O ₄ + 2 N a O H ⟶ 2 H ₂ O + N a ₂ S O ₄
29.09 mL of 0.639 M N a O H is mixed with 213.8 mL of H ₂ S O ₄
Let the concentration of H ₂ S O ₄ be S₂ .
In terms of normal or equivalent solution is will be 2 N solution
From the formula S₁ V₁ = S₂ V₂
= 29.09 x .639 = 213.8 x S₂
S₂ = .087 N solution
In terms of molar solution it will be .087 / 2 M
= .0435 M
Answer:
- <u><em>Option a. 6200 K</em></u>
Explanation:
<u>1) Data:</u>
- V₁ = 0.66 liter
- P₁ = 42.9 mmHg
- T = 261.2 K
- T₂ = ?
- V₂ = 7.63 liter
- P₂ = 872.15 mmHg
<u>2) Formula:</u>
Combined law of gases:
<u>3) Solution:</u>
T₂ = P₂ V₂ T₁ / (P₁ V₁)
T₂ = 872.15 mmHg × 7.63 liter × 261.2 K / ( 424.9 mmHg × 0.66 liter)
T₂ = 6198 K
- Rounding to 2 significant figures, that is 6200 K, which is the first choice.
Answer:
Supply All Kinds of Springs
Weak base: [OH⁻] = √Kb.C
pKb = 4.2

c = concentration
MM Amphetamine (C9H13N) = 135.21 g/mol
c = 215 mg/L = (0.215 g : 135,21 g/mol) / L = 0.00159 mol/L = 1.59 x 10⁻³ mol/L
![\tt [OH^-]=\sqrt{10^{-4.2}\times 1.59\times 10^{-3}}=3.17\times 10^{-4}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7B10%5E%7B-4.2%7D%5Ctimes%201.59%5Ctimes%2010%5E%7B-3%7D%7D%3D3.17%5Ctimes%2010%5E%7B-4%7D)
pOH = 4 - log 3.17
pH = 14 - (4 - log 3.17)
pH = 10 + log 3.17 = 10.50