Answer:
36g of H2O.
Explanation:
We'll begin by writing the balanced equation for the reaction:
2NaOH + H2SO4 —> Na2SO4 + 2H2O
Next, we shall determine the mass of NaOH that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:
Molar mass of NaOH = 23 + 16 + 1 = 40g/mol
Mass of NaOH from the balanced equation = 2 x 40 = 80g
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H2O from the balanced equation = 2 x 18 = 36g.
From the balanced equation above, we can see evidently that:
80g of NaOH reacted to produce 36g of H2O.
Answer:
2812.6 g of H₂SO₄
Explanation:
From the question given above, the following data were obtained:
Mole of H₂SO₄ = 28.7 moles
Mass of H₂SO₄ =?
Next, we shall determine the molar mass of H₂SO₄. This can be obtained as follow:
Molar mass of H₂SO₄ = (1×2) + 32 + (16×4)
= 2 + 32 + 64
= 98 g/mol
Finally, we shall determine the mass of H₂SO₄. This can be obtained as follow:
Mole of H₂SO₄ = 28.7 moles
Molar mass of H₂SO₄ =
Mass of H₂SO₄ =?
Mole = mass / Molar mass
28.7 = Mass of H₂SO₄ / 98
Cross multiply
Mass of H₂SO₄ = 28.7 × 98
Mass of H₂SO₄ = 2812.6 g
Thus, 28.7 mole of H₂SO₄ is equivalent to 2812.6 g of H₂SO₄
For a neutralization reaction, the value of q(heat of neutralization) is doubled when the concentration of only the acid is doubled.
A neutralization reaction is a reaction in which an acid reacts with a base to yield salt and water. Ionically, a neutralization reaction goes as follows; H^+(aq) + OH^-(aq) ------> H20(l).
The heat of neutralization (Q) of the system depends on the concentration of the solutions. Since Q is dependent on concentration, if the concentration of any of the reactants is doubled, more heat is evolved hence Q is doubled.
Learn more: brainly.com/question/10323185
Answer:
ºC
Explanation:
We have to start with the variables of the problem:
Mass of water = 60 g
Mass of gold = 13.5 g
Initial temperature of water= 19 ºC
Final temperature of water= 20 ºC
<u>Initial temperature of gold= Unknow</u>
Final temperature of gold= 20 ºC
Specific heat of gold = 0.13J/gºC
Specific heat of water = 4.186 J/g°C
Now if we remember the <u>heat equation</u>:
We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:
Now we can <u>put the values into the equation</u>:
Now we can <u>solve for the initial temperature of gold</u>, so:
ºC
I hope it helps!
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