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DaniilM [7]
3 years ago
9

What is the correct name for CaS04? (a) calcium sulfoxide (b) calcium sulfite (c) calcium sulfur oxide (d) calcium sulfate (e) c

alcium sulfide tetroxide
Chemistry
2 answers:
Zielflug [23.3K]3 years ago
5 0

Answer: OPTION, D

Explanation:

It is an example for ionic salt and has Calcium ion and sulfate ion.

Hence it's name is Calcium sulfate.

Calcium sulfoxide does not exist.

Calcium sulfite has CaSO3.

krok68 [10]3 years ago
5 0

<u>Answer:</u> The name of CaSO_4 is calcium sulfate.

<u>Explanation:</u>

An ionic compound is defined as the compound which is formed when electron gets transferred from one atom to another atom. These are usually formed when a metal reacts with a non-metal or a metal reacts with a polyatomic ion or a reaction between two polyatomic ions takes place.

CaSO_4 is an ionic compound because calcium element is a metal and sulfate ion is a polyatomic ion. The bond formed between a metal and a non-metal is always ionic in nature.

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 polyatomic ions, the name of the ion is written as such. <u>For Example:</u> OH^- is named as hydroxide, SO_4^{2-} is named as sulfate and so on..

Hence, the name of CaSO_4 is calcium sulfate.

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Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
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Answer:

Final temperature: 659.8ºC

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Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

U_{2}-U_{1}=Q-W

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

dw=Pdv

The pressure is constant, so:  w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg} (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

U_{2}-U_{1}=500-(3*75)=275kJ

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}  

Subtracting the first from the second:

1.5P_{1}V_{1}=nR(T_{2}-T_{1})

Isolating T_{2}:

T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}

Assuming that it is water steam, n=0.1666 kmol

V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}

T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76 ºC

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