Answer:
The products are carbon dioxide and water
Explanation:
Step 1: Data given
Combustion = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O2 as one reactant.
Step 2: The complete combustion of C3H7OH:
For the combustion of 1-propanol, we need O2.
The products of this combustion are CO2 and H2O.
C3H7OH + O2→ CO2 + H2O
On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3
C3H7OH + O2→ 3CO2 + H2O
On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.
C3H7OH + O2→ 3CO2 + 4H2O
On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).
To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.
2C3H7OH + 9O2→ 6CO2 + 8H2O
For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O
The products are carbon dioxide and water
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 pH at equivalence point is 12.46
At equivalence point, number of moles of acid, n equals number of moles of base, n'
So, n = n'
CV = C'V' where
- C = concentration of acid (HCl) = 0.0470 M,
- V = volume of acid = 16.0 mL,
- C' = concentration of base (ammonia solution) and
- V' = volume of base = 26.0 mL.
<h3>Concentration of ammonia solution</h3>
Making C' subject of the formula, we have
C' = CV/V'
Substituting the values of the variables into the equation, we have
C' = CV/V'
C' = 0.0470 M × 16.0 mL/26.0 mL
C' = 0.752 MmL/26.0 mL
C' = 0.0289 M
<h3>The concentration of acid at equivalence point</h3>
We know that the ion-product of water Kw is
Kw = [H⁺][OH⁻] = where
- [H⁺] = concentration of HCl at equivalence point,
- [OH⁻] = C' = concentration of ammonia solution = 0.0289 M and
- Kw = 1.01 × 10⁻¹⁴
Making [H⁺] subject of the formula, we have
[H⁺} = Kw/[OH⁻]
[H⁺] = 1.01 × 10⁻¹⁴/0.0289
[H⁺] = 34.95 × 10⁻¹⁴
[H⁺] = 3.495 × 10⁻¹³
<h3>pH at equivalence point</h3>
Since pH = -㏒[H⁺]
pH = -㏒[3.495 × 10⁻¹³]
pH = -㏒[3.495] + (-㏒10⁻¹³)
pH = -㏒[3.495] + [-13(-㏒10)]
pH = 13 - 0.5434
pH = 12.4566
pH ≅ 12.46
So, the pH at equivalence point is 12.46
Learn more about pH at equivalence point here:
brainly.com/question/25487920
Answer:
There is 54.29 % sample left after 12.6 days
Explanation:
Step 1: Data given
Half life time = 14.3 days
Time left = 12.6 days
Suppose the original amount is 100.00 grams
Step 2: Calculate the percentage left
X = 100 / 2^n
⇒ with X = The amount of sample after 12.6 days
⇒ with n = (time passed / half-life time) = (12.6/14.3)
X = 100 / 2^(12.6/14.3)
X = 54.29
There is 54.29 % sample left after 12.6 days
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