How many grams are 2.50 * 10^23 molecules of CaO?
1 answer:
134.56 grams are 2.50 * 10^23 molecules of CaO.
Explanation:
According to Avagadro hypothesis, 1 mole of a substance contain 6.023 x 10^23 molecules or particles.
Data given:
Number of molecules of CaO =
FORMULA USED IN CALCULATION:
number of molecules = 6.02 x 10^23 x number of moles of CaO
At first number of moles is calculated:
= 2.4 moles of CaO is present
to calculate the mass of CaO in grams formula used is
mass = atomic mass x number of moles
atomic mass of CaO = 56.07 grams/mole
mass = 2.4 x 56.07
= 134.56 grams of calcium oxide is present is 2.50 * 10^23 molecules of CaO.
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Answer:
20 g Ag
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Using Dimensional Analysis
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
Explanation:
<u>Step 1: Define</u>
[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)
[Given] 10 g Cu
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol Cu = 1 mol Ag
Molar Mass of Cu - 63.55 g/mol
Molar Mass of Ag - 197.87 g/mol
<u>Step 3: Stoichiometry</u>
<u /> = 16.974 g Ag
<u>Step 4: Check</u>
<em>We are given 1 sig fig. Follow sig fig rules and round.</em>
16.974 g Ag ≈ 20 g Ag