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azamat
3 years ago
6

Write the symbol for each of the following ions:

Chemistry
1 answer:
kap26 [50]3 years ago
3 0

Answer:

a) _{31}^{71}\textrm {P^{3+}}

b) _{35}^{80}\textrm {Br^{-}}

c) _{90}^{232}\textrm {Th^{4+}}

d) _{38}^{87}\textrm {Sr^{2+}}

Explanation:

For neutral atoms:

Atomic Number (Z)= number of protons = number of electrons

Mass number (A) = number of protons + number of neutrons

For ions with positive net charge:

Number of protons = Z + net charge

For ions with negative net charge:

Number of protons= Z - net charge

a) A = 71, Charge = +3

Number of electrons = 28

Number of protons = 28 +3 =31

_{31}^{71}\textrm {P^{3+}}

b)  A = 35, Z = 45+35=80, Charge = -1

Number of protons =35

Number of neutrons = 45

Number of electrons = 36

Charge = Number of protons- Number of electrons =35-36 = -1

_{35}^{80}\textrm {Br^{-}}

c)  Charge = +4

Number of electrons = 86

Number of protons = Z = 86+4 = 90

mass number = A = 90+142 = 232

_{90}^{232}\textrm {Th^{4+}}

d)  Charge = +2

Atomic number = Number of protons = Z = 38

mass number = A = 87

_{38}^{87}\textrm {Sr^{2+}}

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<u>Answer:</u> The enthalpy of the reaction is 5.30\times 10^{-4}kJ

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

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Given mass of metal = 0.295 g

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Putting values in above equation, we get:

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We are given:

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Molarity of the solution = 1 moles/ L

Putting values in above equation, we get:

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M(s)+2HCl(aq.)\rightarrow MCl_2(aq.)+H_2(g)

By Stoichiometry of the reaction:

1 mole of metal reacts with 2 moles of hydrochloric acid..

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As, given amount of hydrochloric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, metal is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

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Then, 0.0051 moles of metal will absorb = \frac{104J}{1mol}\times 0.0051mol=0.530J of heat.

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The lewis structure is as shown in the diagram.

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The perspective drawing of NH3 is given here

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To know more about ammonia, you can refer to:

brainly.com/question/13960908

#SPJ4

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