Answer:
waxing is in the evening and waning is in the morning if you cant see the moon in the evening it is waning if you can see it in the evening it is waxing
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:
Hello!
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:

Best regards!
Phosphorus has 16 neutrons. Phosphorous is 15 on the periodic table, which means that the atomic number (number of protons) of phosphorous is 15.
For more info: https://study.com/academy/answer/what-is-the-number-of-neutrons-in-phosphorus.html
Answer:
True
Explanation:
Soluble substances dissolve.
Therefore they wouldn't be called soluble if they can't dissolve in solvents
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Answer:
Equilibrium concentration of
is 12.5 M
Explanation:
Given reaction: 
Here, ![K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5BH_%7B2%7DO%5D%7D)
where
represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations
Here,
,
and 
So, ![[H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M](https://tex.z-dn.net/?f=%5BH_%7B2%7DO%5D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5Ctimes%20K_%7Bc%7D%7D%3D%5Cfrac%7B1.69%7D%7B0.015%5Ctimes%209.0%7D%3D12.5M)
Hence equilibrium concentration of
is 12.5 M