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natulia [17]
2 years ago
5

The chemical formula for the ionic compound aluminum sulfite is Al2(SO3)3. Explain why there are 2 cations for every 3 anions in

this compound.
Chemistry
1 answer:
stiks02 [169]2 years ago
3 0

my mom !!!Explanation:

Musscle man

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What are some examples of homogeneous mixtures
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3 0
3 years ago
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When aqueous solutions of AgNO 3 and KI are mixed, AgI precipitates. The balanced net ionic equation is ________. Ag (aq) I- (aq
dimulka [17.4K]

Answer:

Ag⁺(aq) + Cl⁻(aq) ==> AgCl(s)

Explanation:

The net ionic equation can be described as the equation that contains only those species which would be participated in the chemical reaction. The spectator ions are the type of the ions that are present in both sides of the chemical equation these ions could not be present in the net ionic equation

First, it is easiest if you write the compete molecular equation:

AgNO₃(aq) + KCl(aq) ⇔ AgCl(s) + KNO₃(aq)

we look up which compounds are soluble (aq) and which are not (s). In this case, silver chloride (AgCl) is not soluble.  Thus, the net ionic equation is...  

Ag⁺(aq) + Cl⁻(aq) ==> AgCl(s)

5 0
3 years ago
A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
AysviL [449]

Answer:- Heat lost by the metal is 279.45 cal.

Solution:- This type of problems are solved by using the concept, heat given = - heat taken

Metal temperature is decreasing from 45.00 degree C to 11.08 degree C. It means the heat is lost by the metal and this heat lost by metal is gained by water and the calorimeter to raise their temperature.

the equation we use is, q=mc\Delta T .

where, q is the heat energy, m is mass, c is specific heat and \Delta T is change in temperature.

Combined mass of calorimeter and water is 250.0 g and the specific heat is \frac{1.035cal}{g.^0C} .

\Delta T  for calorimeter and water (combined) = 11.08 - 10.00 = 1.08 degree C

\Delta T  for metal = 11.08 - 45.00 = -33.92 degree C

let's plug in the values in the above equation and calculate heat gained by combined system.

q=250.0g*\frac{1.035cal}{g.^0C}*1.08^0C

q = 279.45 cal

So, the heat lost by the metal is 279.45 cal.


3 0
3 years ago
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