Transition metals
Most transition metals differ from the metals of Groups 1, 2, and 13 in that they are capable of forming more than one cation with different ionic charges. As an example, iron commonly forms two different ions
The balanced chemical reaction is written as:
Sb2S3 + 6HCl = 6SbCl<span>3 + 3H2S
We are given the amount of </span><span>antimony(III) sulfide to be used in the reaction. This is amount will be used for the calculations. We do as follows:
2.85 g Sb2S3 ( 1 mol / </span><span>339.715 g ) ( 6 mol SbCl3 / 1 mol Sb2S3 ) (</span> 228.13 g / mol ) = 11.48 g SbCl3
1. Ca(HCO3)2
2.Ca(HCOO)2
3. Ca(OH)2
4.NaOH
5.KCI
6.MgSO4
7.PbO
8.HCl
9.HNO3
10.H2SO4
11.NH3
12.(NH4)3PO4
13.NaOH
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Answer:
Explanation:
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In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:
Thus, we insert mass, specific heat and temperatures to obtain:
In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:
Now, we plug in to obtain:
NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.
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Answer:
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