Answer: 0.0000332mol
Explanation: 1mole of CCl4 contains 6.02x10^23 molecules.
Therefore, X mol of CCl4 will contain 2 x 10^19 molecules i.e
Xmol of CCl4 = 2 x 10^19/ 6.02x10^23 = 0.0000332mol
Answer:
7. ok
please mark me brainlistb
The grams that would be produced from 7.70 g of butanoic acid and excess ethanol is 7.923grams
calculation
Step 1: write the chemical equation for the reaction
CH3CH2CH2COOH + CH3CH2OH → CH3CH2CH2COOCH2CH3 +H2O
step 2: find the moles of butanoic acid
moles= mass/ molar mass
= 7.70 g/ 88 g/mol=0.0875 moles
Step 3: use the mole ratio to determine the moles of ethyl butyrate
moles ratio of CH3CH2CH2COOH :CH3CH2CH2COOCH2CH3 is 1:1 therefore the moles of CH3CH2CH2COOCH2CH3 = 0.0875 x78/100=0.0683moles
step 4: find mass = moles x molar mass
= 0.0683 moles x116 g/mol=7.923grams
The density is calculated as mass per volume, so if we want to solve for mass, we would multiply density by volume.
For Part A: if we have a density of 0.69 g/mL, and a volume of 280 mL, multiplying these will give a mass of: (0.69 g/mL)(280 mL) = 193.2 g. Rounded to 2 significant figures, this is 190 g gasoline.
For Part B: if we have a density of 0.79 g/mL, and a volume of 190 mL, multiplying these will give a mass of: (0.79 g/mL)(190 mL) = 150.1 g. Rounded to 2 significant figures, this is equal to 150 g ethanol.
