Answer:
We have to add 9.82 grams of calcium acetate
Explanation:
Step 1: Data given
Molarity of the calcium acetate solution = 0.207 M
Volume = 300 mL = 0.300 L
Molar mass calcium acetate = 158.17 g/mol
Step 2: Calculate moles calcium acetate
Moles calcium acetate = molarity * volume
Moles calcium acetate = 0.207 M * 0.300 L
Moles calcium acetate = 0.0621 moles
Step 3: Calculate mass calcium acetate
Mass calcium acetate = moles * molar mass
Mass calcium acetate = 0.0621 moles * 158.17 g/mol
Mass calcium acetate = 9.82 grams
We have to add 9.82 grams of calcium acetate
Answer:
my mass turned into gasses and other substances besides ash
Explanation:
burning a log products smoke and ash. the smoke takes some mass and the ash takes the rest.
Empirical formula is the simplest ratio of whole numbers of components in a compound
molecular formula is the actual ratio of components in a compound .
the molecular formula for the compounds given are as follows
ethyne - C₂H₂
ethene - C₂H₄
ethane - C₂H₆
methane - CH₄
the actual ratios of the elements simplified ratio
C : H C : H
ethyne 2:2 1:1
ethene 2:4 1:2
ethane 2:6 1:3
methane 1:4 1:4
the only compound where the actual ratio is equal to the simplified ratio is methane
therefore in methane molecular formula CH₄ is the same as empirical formula CH₄
Answer:
Explanation:
A pressure of 1 atm and a temperature of 0 °C is the old definition of STP. Under these conditions, 1 mol of a gas occupies 22.4 L.
1. Calculate the moles of hydrogen.
2. Calculate the number of molecules
