Answer:
Sr or Strontium
Explanation:
sr: strontium has atomic radius of 255pm
cu: copper has an atomic radius of 128pm
they teach you in chem how to do it based off the chart but I don't remember that method
The initial temperature of the copper metal was 27.38 degrees.
Explanation:
Data given:
mass of the copper metal sample = 215 gram
mass of water = 26.6 grams
Initial temperature of water = 22.22 Degrees
Final temperature of water = 24.44 degrees
Specific heat capacity of water = 0.385 J/g°C
initial temperature of copper material , Ti=?
specific heat capacity of water = 4.186 joule/gram °C
from the principle of:
heat lost = heat gained
heat gained by water is given by:
q water = mcΔT
Putting the values in the equation:
qwater = 26.6 x 4.186 x (2.22)
qwater = 247.19 J
qcopper = 215 x 0.385 x (Ti-24.4)
= 82.77Ti - 2019.71
Now heat lost by metal = heat gained by water
82.77Ti - 2019.71 = 247.19
Ti = 27.38 degrees
The reactant is Mercury (II) Oxide while the products are Mercury and Oxygen separately.
This is because the reactants are typically always on the left side of the yields symbol. In this decomposition reaction, it would still be the same as at the end of the reaction, there were to products produced: Mercury and Oxygen.
Products tend to always be on the right side of the yields symbol, they're what comes out of a reaction no matter what type.
Hope this helps!
Answer:
See explanation
Explanation:
Qualitative analysis in chemistry is a method used to determine the ions present in a solution chiefly by means of chemical reactions.
In this case, I suspect the presence of silver ions and/or barium ions. The first step is to add dilute HCl. This will lead to the precipitation of the silver ion as AgCl. If a white precipitate is formed upon addition of HCl then Ag^+ is present in the solution.
Secondly, I add a carbonate such as NH4CO3(aq). This will cause the barium ions to become precipitated as barium carbonate. Hence, the formation of a white precipitate when NH4CO3(aq) is added to the solution indicates the presence of barium ion in the solution.
Answer:
I hope this will help you and Please mark me as Brilliant
Explanation:
Approximately 2 mL of Solution A (on the left) is added to a sample of Solution B (on the right) with a dropping pipet. If a precipitate forms, the resulting precipitate is suspended in the mixture. The mixture is then stirred with a glass stirring rod and the precipitate is allowed to settle for about a minute.
Solution A: 0.5 M sodium hydroxide, colorless
Solution B: 0.2 M nickel(II) nitrate, green
Precipitate: light green
Ni(NO3)2(aq) + 2 NaOH(aq) —> Ni(OH)2(s) + 2 NaNO3(aq)
Credits:
Design
Kenneth R. Magnell Central Michigan University, Mt. Pleasant, MI 48859
John W. Moore University of Wisconsin - Madison, Madison, WI 53706
Video
Jerrold J. Jacobsen University of Wisconsin - Madison, Madison, WI 53706
Text
Kenneth R. Magnell Central Michigan University, Mt. Pleasant, MI 48859
