Answer:
0.51M
Explanation:
Given parameters:
Initial volume of NaBr = 340mL
Initial molarity = 1.5M
Final volume = 1000mL
Unknown:
Final molarity = ?
Solution;
This is a dilution problem whereas the concentration of a compound changes from one to another.
In this kind of problem, we must establish that the number of moles still remains the same.
number of moles initially before diluting = number of moles after dilution
Number of moles = Molarity x volume
Let us find the number of moles;
Number of moles = initial volume x initial molarity
Convert mL to dm³;
1000mL = 1dm³
340mL gives = 0.34dm³
Number of moles = initial volume x initial molarity = 0.34 x 1.5 = 0.51moles
Now to find the new molarity/concentration;
Final molarity = = = 0.51M
We can see a massive drop in molarity this is due to dilution of the initial concentration.
Answer:
Condensation 212F or 100C, Freezing 32F or 0C
Explanation:
Condensation 212 degrees Fahrenheit or 100 degrees Celsius.
Freezing point 32 degrees Fahrenheit or 0 degrees Celsius.
Answer:
In explanation :P
Explanation:
Cu was detected in CuS with X-ray absorption spectroscopy in the high-energy-resolution fluorescence detection (HERFD) mode. The S L2,3 emission spectrum of CuS was found to be significantly broader than Cu2S due to contributions from inequivalent S sites. <em><u>This difference in the spectral width was used to distinguish between sulfide species formed on the Cu foil exposed to Na2S solutions</u></em>
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Answer:
52.2g of KCl would be left
The mass of KCl will remain despite the solution is been heated
Explanation:
When you are heating a solution, just the solvent (In this case, water), will be evaporated and, in theory, the mass of KCl will remain despite the solution is been heated.
Now, the mass of KCl that you can obtain from 350mL of a 2.0M solution will be:
<em>Moles KCl:</em>
350mL = 0.350L * (2.0mol / L) = 0.700 moles
<em>Mass KCl -Molar mass: 74.55g/mol-:</em>
0.700mol * (74.55g/mol) = 52.2g of KCl would be left
Explanation:
1.
Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)
2.
Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)
A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.
3.
Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)
2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)
Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)
2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)
4.
The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.
Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.