Calcite can be either dissolved by groundwater or precipitated by groundwater, depending on several factors including the water temperature, pH, and dissolved ion concentrations. Although calcite is fairly insoluble in cold water, acidity can cause dissolution of calcite and release of carbon dioxide gas.
The first example of diffusion is smoking a cigarette when u light it it spreads through the air.
The second example is lighting a candle in a room it the smoke spreads through the air.
Smoke can be very bad for the air and the people around because it can cause many problems with your breathing.
Answer:
- 13.56 g of sodium chloride are theoretically yielded.
- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.
- 0.50 g of sodium nitrate remain when the reaction stops.
- 92.9 % is the percent yield.
Explanation:
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In this case, according to the question, it is possible to set up the following chemical reaction:
Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:
Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:
Therefore, the leftover of sodium nitrate is:
Finally, the percent yield is computed via:
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Answer:
Sodium chloride and hydrogen gas are the products of this reaction. If the gas collected has a pressure of 1.05 atm at 298 K, then what volume is the hydrogen
Explanation:
Answer:
0.50 mol
Explanation:
The half-life is <em>the time required for the amount of a radioactive isotope to decay to half that amount</em>.
Initially, there are 8.0 moles.
- After 1 half-life, there remain 1/2 × 8.0 mol = 4.0 mol.
- After 2 half-lives, there remain 1/2 × 4.0 mol = 2.0 mol.
- After 3 half-lives, there remain 1/2 × 2.0 mol = 1.0 mol.
- After 4 half-lives, there remain 1/2 × 1.0 mol = 0.50 mol.