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stepan [7]
3 years ago
14

If chlorine gas is bubbles through an aqueous solution of sodium iodide,the result is elemental iodine and aqueous sodium chlori

de solution. What kind of reactions took place?
Chemistry
1 answer:
Mrac [35]3 years ago
3 0
The balanced reaction that describes the reaction of chlorine gas and sodium iodide to produce elemental iodine and sodium chloride in aqueous solution is expressed Cl2+2NaI= I2 + 2NaCl. This kind of reaction is called single replacement reaction where the anion, in this case, is only replaced
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Explanation:

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how much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60% antifreeze?
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1 gallon of antifreeze = 60% of mixture. Total mixture:
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3 years ago
How many moles of AgCl2 are in 5.78x1024 molecules of AgCl2?
Zolol [24]

Answer:

9.6 mol AgCl2

Explanation:

You have to use Avogadro's number: 6.023 x 10^23

5.78 x 10^24 molecules (1 mol AgCl2/ 6.023 x 10^23 molecules) =9.6 mol AgCl2

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3 years ago
What is used to classify galaxies?star typesestimated agecolorshape
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The correct answer among the choices given is the last option. Galaxies are classified according to their shapes or visual morphology. There three main types of galaxies currently. They are the elliptical, spiral and irregular. Elliptical galaxies are like a spheriod or an elongated sphere. Spiral galaxies are characterized and distinguished by having a bulge, a disk and a halo. Lastly, irregular galaxies do not have regular or symmetrical shape.
3 0
3 years ago
Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

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

7 0
3 years ago
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