Answer:
About 7.0 × 10³ J or 7.0 kJ
Explanation:
We want to determine the amount of energy needed to raise the temperature of 67 grams of water from 20°C to 45°C.
We can use the heat equation:
Where <em>C</em> is the specific heat of water.
Substitute and evaluate:
Recall that there are 1000 J in a kJ. Hence:
In conclusion, it will take about 7.0 × 10³ J or 7.0 kJ of energy to raise the temperature of 67 grams of water from 20 °C to 45 °C.
I believe A is the is the right answer
Answer:
Graduated cylinder
Explanation:
A graduated cylinder is used routinely for measuring volume and is considered more accurate than a beaker because of the permanently-marked incremental graduations incorporated in the clear cylinder.
<u><em>PLS MARK BRAINLIEST</em></u>
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>
