Write the name for sn(so4)2. remember that sn forms several ions.

2 answers:

5 0

Answer is: name of Sn(SO₄)₂ is tin(IV) sulfate or stannic sulfate.
Major oxidation numbers for tin are +2 and +4 (group 14 of the periodic table). In this chemical compond tin has oxidation number +4, besause sulfate has oxidation number -2 (2·(-2) = -4). Compound has neutral charge and it is moderately water and acid soluble.

Shalnov [3]3 years ago

4 0

Answer: The name of the given compound is tin (IV) sulfate.

Explanation:

The given compound is formed by the combination of tin ions and sulfate ions. It is an ionic compound.

The nomenclature of ionic compounds is given by:

Positive ion is written first.
The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.
In case of transition metals, the oxidation state are written in roman numerals in bracket in front of positive ions.

In the given compound, tin is present in +4 oxidation state.

Hence, the name of the given compound is tin (IV) sulfate

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The heat of vaporization ΔHv of dichloromethane CH2Cl2 is 28.0 /kJmol . Calculate the change in entropy ΔS when 3.3g of dichloro

ella [17]

Answer:

$\Delta _VS=3.48x10^{-3}kJ=3.48J$

Explanation:

Hello,

In this case, the change in the entropy, by knowing the change in the enthalpy and the temperature for a vaporization process is given by:

$\Delta _VS=\frac{\Delta _VH}{T_V}$

Thus, we start by computing the change in the enthalpy for the 3.3g of dichloromethane as shown below:

$\Delta _VH=3.3gCH_2Cl_2 *\frac{1molCH_2Cl_2}{84.9gCH_2Cl_2}*28\frac{kJ}{molCH_2Cl_2} =1.09kJ$

Finally, the change in the entropy, considering the temperature in kelvins:

$\Delta _VS=\frac{1.09kJ}{(39.8+273.15)K}\\\\\Delta _VS=3.48x10^{-3}kJ=3.48J$

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