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Gre4nikov [31]
3 years ago
15

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes.

Chemistry
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

a. Ca2+  1M ; NO3-  2M

b. Na+  4M ;  SO4-2  2M

c.  NH4+ , Cl-  0,186M  

d. K+ 0,056M ; PO4-3 0,0188 M

Explanation:

Here are the ionization equations

Ca(NO3)2 ----> Ca2+  +  2NO3-

Na2SO4 ------> 2Na+   +  SO4-2

NH4Cl ------> NH4+  +  Cl-

K3PO4 -----> 3K+   +   PO4-3

Concentrations must be in Molarity (moles/L)

Ca(NO3)2 0,1 mole in 100mL

100mL = 0,1 L

Molarity Ca(NO3)2  1M  (0,1 /0,1)

Ca(NO3)2 ----> Ca2+  +  2NO3-

1 M                      1M            2M

Na2SO4 2,5 moles in 1,25L

Molarity Na2SO4 2,5 moles /1,25L = 2M

Na2SO4 ------> 2Na+   +  SO4-2

2M                     4M             2M

5.00 g of NH4Cl in 500mL

We have to resort to the molar mass

Let's make the rule of three

500mL ______ 5g

1000mL _____ (1000mL . 5g) / 500mL = 10 g

(These are the grams in 1L, so with the molar mass, we get the moles)

Mass / Molar mass = moles --> 10g / 53,5 g/m = 0.186 moles

NH4Cl ------> NH4+  +  Cl-

0,186 M      0,186M  +  0,186M

1.00 g K3PO4 in 250.0 mL

(The same as before)

250 mL ________1 g

1000 mL _______ (1000 mL . 1 g) /250mL = 4 g

These are the grams in 1L, so with the molar mass, we get the moles

Mass / Molar mass = moles -->  4g /212,26 g/m = 0,0188 moles

K3PO4 -----> 3K+   +   PO4-3

0,0188 M      0,056M   +  0,0188 M

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<u>Answer:</u>

<u>For a:</u> The isotopic representation of iodine is _{53}^{131}\textrm{I}

<u>For b:</u> The isotopic representation of cesium is _{55}^{137}\textrm{Cs}

<u>For c:</u> The isotopic representation of strontium is _{38}^{52}\textrm{Sr}

<u>Explanation:</u>

The isotopic representation of an atom is: _Z^A\textrm{X}

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

  • <u>For a:</u>

We are given:

Number of neutrons = 78

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Mass number = 53 + 78 = 131

Thus, the isotopic representation of iodine is _{53}^{131}\textrm{I}

  • <u>For b:</u>

We are given:

Number of neutrons = 82

Atomic number of cesium = 55 = Number of protons

Mass number = 55 + 82 = 137

Thus, the isotopic representation of cesium is _{55}^{137}\textrm{Cs}

  • <u>For c:</u>

We are given:

Number of neutrons = 52

Atomic number of strontium = 38 = Number of protons

Mass number = 38 + 52 = 90

Thus, the isotopic representation of strontium is _{38}^{52}\textrm{Sr}

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Answer:

The pressure  increases.

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