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Ainat [17]
3 years ago
9

Which of these is an isoelectronic series? question 7 options: 1) na+, k+, rb+, cs+ 2) k+, ca2+, ar, s2– 3) na+, mg2+, s2–, cl–

4) li, be, b, c 5) none of these (a-d) save?
Chemistry
1 answer:
-BARSIC- [3]3 years ago
6 0
An isoelectronic series is where all of the ions listed have the same number of electrons in their atoms. When an atom has net charge of zero or neutral, it has equal number of protons and electrons. Hence, it means that the atomic number = no. of protons = no. of electrons. If these atoms become ions, they gain a net charge of + or -. Positive ions are cations. This means that they readily GIVE UP electrons, whereas negative ions (anions) readily ACCEPT electrons. So, to know which of these are isoelectronic, let's establish first the number of electron in a neutral atom from the periodic table:

Na=11; K=19; Rb=37; Cs = 55; Ca=20; S=16; Mg=12; Li=3; Be=4; B=5; C=6, Ar = 18

A. Na⁺: 11-1 = 10 electrons
     K⁺: 19 - 1 = 18 electrons
     Rb⁺: 37-1 = 36 electrons

B. K⁺: 19 - 1 = 18 electrons
    Ca²⁺: 20 - 2 = 18 electrons
    Ar:  18 electrons
    S²⁻:  16 +2 = 18 electrons

C. Na⁺: 11-1 = 10 electrons
    Mg²⁺: 12 - 2 = 10 electrons
     S²⁻:  16 +2 = 18 electrons

D. Li=3 electrons
    Be=4 electrons
    B=5 electrons
    C=6 electrons

The answer is letter B.
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A bomb calorimeter has a heat capacity of 783 J/oC and contains 254 g of water whose specific heat capacity is 4.184 J/goC. How
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Answer : The amount of heat evolved by a reaction is, 4.81 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

q=[q_1+q_2]

q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]

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q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the water

c_1 = specific heat of calorimeter = 783J/^oC

c_2 = specific heat of water = 4.184J/g^oC

m_2 = mass of water = 254 g

\Delta T = change in temperature = T_2-T_1=(23.73-26.01)=-2.28^oC

Now put all the given values in the above formula, we get:

q=[(783J/^oC\times -2.28^oC)+(254g\times 4.184J/g^oC\times -2.28^oC)]

q=-4208.28J=-4.81kJ

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A sample of gallium Bromide GaBr2,weighing 0.165 g was dissolved in water and treated with silver nitrate AgNO3, and resulting t
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<u>Answer:</u> The percent gallium in gallium bromide is 30.30 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of gallium bromide = 0.165 g

Molar mass of titanium gallium bromide = 229.53 g/mol

Putting values in equation 1, we get:

\text{Moles of gallium bromide}=\frac{0.165g}{229.53g/mol}=0.00072mol

  • The chemical equation for the reaction of gallium bromide and silver nitrate follows:

GaBr_2+2AgNO_3\rightarrow 2AgBr(s)+Ga(NO_3)_2

By Stoichiometry of the reaction:

1 moles of gallium bromide produces 1 mole of gallium nitrate

So, 0.00072 moles of gallium bromide will produce = \frac{1}{1}\times 0.00072=0.00072moles of gallium nitrate

  • Now, calculating the mass of gallium nitrate from equation 1, we get:

Molar mass of gallium nitrate = 193.73 g/mol

Moles of gallium nitrate = 0.00072 moles

Putting values in equation 1, we get:

0.00072mol=\frac{\text{Mass of gallium nitrate}}{193.73g/mol}\\\\\text{Mass of gallium nitrate}=0.139g

Calculating the mass of gallium in the reaction, we use unitary method:

In 1 mole of gallium nitrate, 1 mole of gallium atom is present.

In 193.73 grams of gallium nitrate, 69.72 g of gallium atom is present.

So, in 0.139 grams of gallium nitrate, the mass of gallium present will be = \frac{69.72}{193.73}\times 0.139=g

  • To calculate the percentage composition of gallium in gallium bromide, we use the equation:

\%\text{ composition of gallium}=\frac{\text{Mass of gallium}}{\text{Mass of gallium bromide}}\times 100

Mass of gallium bromide = 0.165 g

Mass of gallium = 0.050 g

Putting values in above equation, we get:

\%\text{ composition of gallium}=\frac{0.050g}{0.165g}\times 100=30.30\%

Hence, the percent gallium in gallium bromide is 30.30 %.

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3 years ago
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