Calcium sulfate, carbon dioxide, and water are produced when sulfuric acid and calcium carbonate interact.
<h3>What results from the reaction of calcium carbonate and sulfuric acid?</h3>
The insoluble in water free lime (CaCO3) interacts with sulfuric acid. It takes only a few minutes for sulfuric acid and lime to react and produce gypsum. H2SO4 + Lime CaCO3 = CaSO4 (gypsum).
<h3>What occurs when calcium and sulfuric acid are combined?</h3>
Each of these metals forms a layer of insoluble sulfate when exposed to sulfuric acid, which slows or stops the reaction altogether. A white precipitate of calcium sulfate and a small amount of hydrogen are formed in the calcium case.
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Answer:
The enthalpy of vaporization of a given substance is much greater than its enthalpy of fusion because it takes more energy to completely separate molecules (c
<u>Answer:</u> The volume of barium chlorate is 195.65 mL
<u>Explanation:</u>
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:
Given mass of barium chlorate = 25.0 g
Molar mass of barium chlorate = 304.23 g/mol
Molarity of solution = 0.420 mol/L
Volume of solution = ?
Putting values in above equation, we get:
Hence, the volume of barium chlorate is 195.65 mL
Answer:
142.82 g
Explanation:
The following data were obtained from the question:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Density of gol= 19.3 g/cm³
Mass of gold =.?
Next, we shall determine the volume of the gold. This can be obtained as follow:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Volume of gold =.?
Volume of gold = (Volume of water + gold) – (Volume of water)
Volume of gold = 19.4 – 12
Volume of gold = 7.4 mL
Finally, we shall determine the mass of the gold as follow:
Note: 1 mL is equivalent to 1 cm³
Volume of gold = 7.4 mL
Density of gol= 19.3 g/cm³ = 19.3 g/mL
Mass of gold =?
Density = mass /volume
19.3 = mass of gold /7.4
Cross multiply
Mass of gold = 19.3 × 7.4
Mass of gold = 142.82 g
Therefore, the mass of the gold pebble is 142.82 g
Answer:
The answer to your question is Aluminum
Explanation:
We need to warm from 22°C to 85°C
a) 45 g of water
b) 200 g of aluminum Cp = 0.905 J/cal°C
a) Water
Q = mCpΔT
Q = (45)(1)(85 - 22) = 45(63) = 2835 cal
b) Aluminum
Q = (200)(0.905)(85 - 22) = 114030 cal
