You may feel cold when you touch certain kinds of matter because some matter gives off cold air. I’m NOT SURE
To answer this question, we need to convert the given number of moles of the compound copper (II) cyanide to grams by multiplying the compound's molar mass. The molar mass of copper (II) cyanide is 115.55 g/mol. Multiply 4 by this number, the answer is 462.2 grams.
Answer:
77460 g
Explanation:
To solve this problem first we<u> convert molecules into moles</u>, using <em>Avogadro's number</em>:
3.47x10²⁶ molecules ÷ 6.023x10²³ molecules/mol = 576.12 moles
Then we <u>convert 576.12 CuCl₂ moles into grams</u>, using its<em> molar mass</em>:
576.12 mol * 134.45 g/mol = 77460 g
So the answer is 77460 g, or 77.46 kg.
Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
