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
C = 4 mol/l
v = 0.5 l
n(NaCl)=cv
n(NaCl) = 4 mol/l · 0.5 l = 2 mol
2 moles of NaCl must be dissolved
Answer:
Energy is released when bonds are broken, and energy is absorbed when bonds are formed.
Answer:
E
Explanation:
This is because all steps from A-D are important to obtain an accurate result
The molecular weight of a given compound would simply the
sum of the molar weights of each component.
The molar masses of the elements are:
C = 12 amu
H = 1 amu
N = 14 amu
O = 16 amu
where 1 amu = 1 g / mol
Since there are 6 C, 5 H, 1 N and 2 O, therefore the
total molecular weight is:
molecular weight = 6 (12 amu) + 5 (1 amu) + 1 (14 amu) +
2 (16 amu)
molecular weight = 123 amu
Therefore the molecular weight of nitrobenzene is 123 amu
or which is exactly equivalent to 123 g / mol.
