Answer:
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Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
Answer:
Can you tell me the question in the comments on this answer or like how you do it then ill answer you in the comments under this answer
Answer:
∇T = 51.68°C
Explanation:
Mass = 150g
Heat Energy (Q) = 1.0*10³J
Change in temperature ∇T = ?
Q = mc∇T
Q = heat energy
M = mass
C = specific heat capacity of the gold = 0.129j/g°C
∇T = change in temperature
Q = Mc∇T
1.0*10³ = 150 * 0.129 * ∇T
1000 = 19.35∇T
Solve for ∇T
∇T = 1000 / 19.35
∇T = 51.679°C = 51.68°C
The change in temperature of gold was 51.68°C
Explanation:
workdone = force x distance
force = mass x acceleration
30 x 10 = 300N
300N x 1m
workdone= 300J
