Answer:
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Explanation:
Answer:
16.89g of PbBr2
Explanation:
First, let us calculate the number of mole of Pb(NO3)2. This is illustrated below:
Molarity of Pb(NO3)2 = 0.595M
Volume = 77mL = 77/1000 = 0.077L
Mole =?
Molarity = mole/Volume
Mole = Molarity x Volume
Mole of Pb(NO3)2 = 0.595x0.077
Mole of Pb(NO3)2 = 0.046mol
Convert 0.046mol of Pb(NO3)2 to grams as shown below:
Molar Mass of Pb(NO3)2 =
207 + 2[ 14 + (16x3)]
= 207 + 2[14 + 48]
= 207 + 2[62] = 207 +124 = 331g/mol
Mass of Pb(NO3)2 = number of mole x molar Mass = 0.046 x 331 = 15.23g
Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol
Equation for the reaction is given below:
Pb(NO3)2 + CuBr2 —> PbBr2 + Cu(NO3)2
From the equation above,
331g of Pb(NO3)2 precipitated 367g of PbBr2
Therefore, 15.23g of Pb(NO3)2 will precipitate = (15.23x367)/331 = 16.89g of PbBr2
Answer:
Atoms He (Avogadro’s number) → Moles of He (molar mass of He) → Mass of He
• molar mass of He (from the periodic table) = 4.003 g/mol
• Avogadro’s Number: Avogadro’s number gives us the number of entities present in 1 mole: 6.022 × 1023 He atoms in 1 mole of He
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I’m don’t kill me if I’m wrong but I think it’s high melting point
Answer:
The correct answer is b.
Explanation:
The quantum number n specifies the energetic level of the orbital, the first level being the one with the least energy. As n increases, the probability of finding the electron near the nucleus decreases and the orbital energy increases.
In the case of atoms with more than one electron, the quantum number l also determines the sublevel of energy in which an orbital is found, within a certain energy level. The value of l is designated by the letters s, p, d, and f.
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