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
Answer: 2 (2 neutrons are produced).
Explanation:
1) In the left side of the transmutation equationa appears:
²³⁵U + ¹n →
I am omitting the atomic number (subscript to the leff) because the question does not show them as it is focused on number of neutrons.
2) The right side of the transmutation equation has:
→ ¹⁴⁴Ce + ⁹⁰Sr + ?
3) The total mass number of the left side is 235 + 1 = 236
4) The total mass number of Ce and Sr on the right side is 144 + 90 = 234
5) Then, you are lacking 236 - 234 = 2 unit masses on the right side which are the 2 neutrons that are produced along with the Ce and Sr.
The complete final equation is:
²³⁵U + ¹n → ¹⁴⁴Ce + ⁹⁰Sr + 2 ¹n
Where you have the two neutrons produced.
Answer:
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