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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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The molecular formula of compound is Al(C_{2}H_{3}O_{2})_{3}.

Since, in 1 mole of  Al(C_{2}H_{3}O_{2})_{3} there are 6.023\times 10^{23} atoms of  Al(C_{2}H_{3}O_{2})_{3}.

Thus, according to molecular formula, in 1 mole of  Al(C_{2}H_{3}O_{2})_{3} there are 6\times 6.023\times 10^{23}=3.61\times 10^{24} atoms of oxygen atoms.

1 atom of oxygen will be present in \frac{1}{3.61\times 10^{24}} moles of Al(C_{2}H_{3}O_{2})_{3} . Thus,

2.63\times 10^{24} atoms of oxygen \rightarrow \frac{2.63\times 10^{23}}{3.61\times 10^{24}}= 0.7285 moles of Al(C_{2}H_{3}O_{2})_{3}.

Molar mass of Al(C_{2}H_{3}O_{2})_{3} is 236 g/mol, mass can be calculated as follows:

m=n\times M=0.7285 mol\times 236 g/mol=172 g

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C is the correct answer
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Calculate the amount in grams of Na2CO3 needed to react with HCL to produce 120g NaCl
worty [1.4K]
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Na_2CO_{3_{(s)}} + \textbf2HCl_{_{(aq)}}\rightarrow CO_{2_{(g)}} + H_2O_{_{(l)}} + \textbf2NaCl_{_{(aq)}}

Assuming the sodium carbonate is the limiting reagent, look at the coefficients of sodium carbonate and sodium chloride and use those as a ratio of sodium carbonate to sodium chloride: 1:2.

Since you have the required mass of NaCl, convert this to moles.

Assuming you know how to find the molar mass of NaCl:

M_{NaCl} = 58.44g/mol
n = \frac{m}{M}
n_{NaCl} = \frac{120g}{58.44g/mol}
n_{NaCl} = 2.053mol

Using the ratio, since 1 mole of sodium carbonate is required to produce 2 moles of sodium chloride, cross-multiply the ratios:

1:2 = x:2.053mol
2x = 2.053mol
x = 1.027mol

Therefore 1.027 moles of sodium carbonate is required to produce the required amount of sodium chloride. Convert to mass for your final answer:

n_{Na_2CO_3} = 1.027mol
M_{Na_2CO_3} = 105.99g/mol
m = nM
m = (1.027mol)(105.99g/mol)
m_{Na_2CO_3} = 108.85g

Hope this helps :)
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