Answer:
A₃B₈ + 5C₂ –> 3AC₂ + 4B₂C
Explanation:
A₃B₈ + C₂ –> AC₂ + B₂C
The equation can be balance as illustrated below:
A₃B₈ + C₂ –> AC₂ + B₂C
There are 3 atoms of A on the left side and 1 atom on the right side. It can be balance by writing 3 before AC₂ as shown below:
A₃B₈ + C₂ –> 3AC₂ + B₂C
There are 8 atoms of B on the left side and 2 atoms on the right side. It can be balance by writing 4 before B₂C as shown below:
A₃B₈ + C₂ –> 3AC₂ + 4B₂C
There are 2 atoms of C on the left side and a total of 10 atoms on the right side. It can be balance by writing 5 before C₂ as shown below:
A₃B₈ + 5C₂ –> 3AC₂ + 4B₂C
Now, the equation is balanced.
160 g of SO3 are needed to make 400 g of 49% H2SO4.
<h3>How many grams of SO3 are required to prepare 400 g of 49% H2SO4?</h3>
The equation of the reaction for the formation of H2SO4 from SO3 is given below as follows:
1 mole of SO3 produces 1 mole of H2SO4
Molar mass of SO3 = 80 g/mol
Molar mass of H2SO4 = 98 g/mol
80 g of SO3 are required to produce 98 og 100%H2SO4
mass of SO3 required to produce 400 g of 100 %H2SO4 = 80/98 × 400 = 326.5 g of SO3
Mass of SO3 required to produce 49% of 400 g H2SO4 = 326.5 × 49% = 160 g
Therefore, 160 g of SO3 are needed to make 400 g of 49% H2SO4.
Learn more about mass and moles at: brainly.com/question/15374113
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I believe the answer is B because it donates H+ ions
Answer:
26.2g = Mass of water in the calorimeter
Explanation:
The heat absorbed for the water is equal to the heat released for the metal. Based on the equation:
Q = m*C*ΔT
<em>Where Q is heat, m is the mass of the sample, C is specific heat of the material and ΔT is change in temperature</em>
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Replacing we can write:
13.9g * 0.449J/g°C * (54.2°C-15.6°C) = m(H₂O) * 4.184J/g°C * (15.6°C-13.4°C)
240.9J = m(H₂O) * 9.2J/g
<h3>26.2g = Mass of water in the calorimeter</h3>
Answer:
Butanoic acid present in solution
Explanation:
In this case, we have a buffer solution of butanoic acid and sodium butanoate. In other words a reaction like this:
HC₄H₇O₂ + H₂O <------> C₄H₇O₂⁻ + H₃O⁺ Ka = 1.5x10⁻⁵
The low value of Ka means that this is a weak acid. So, after this, the NaOH is added to the solution.
The NaOH is a really strong base, so we might expect that the pH of the solution increase drastically, however this do not occur.
The reason for this is because the first thing to happen in this reaction is an acid base reaction.
The NaOH react with the butanoic acid still present in solution, because is a weak acid, so in solution, this acid is not completely dissociated into it's respective ions. So the butanoic acid reacts with the NaOH and the products:
HC₄H₇O₂ + NaOH <------> Na⁺C₄H₇O₂⁻ + H₂O
So, because of this, the pH increase but not much.
