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Bingel [31]
3 years ago
13

Na+ and Cl- __________ ___________________________ Na+ and PO4 3- __________ ___________________________ Na+ and SO4 2- ________

__ ___________________________ Na+ and CO3 2- __________ ___________________________ K+ and Cl- __________ ___________________________ K+ and PO4 3- __________ ___________________________ K+ and SO4 2- __________ ___________________________ K+ and CO3 2- __________ ___________________________ Ca2+and Cl- __________ ___________________________ Ca2+ and PO4 3- __________ ___________________________ Ca2+ and SO4 2- __________ ___________________________ Ca2+ and CO3 2- __________ ___________________________ NH4 + and Cl- __________ ___________________________ NH4 + and PO4 3- __________ ___________________________ NH4 + and SO4 2- __________ ___________________________ NH4 + and CO3 2- __________ ___________________________ Fe3+ and Cl- __________ ___________________________ Fe3+ and PO4 3- __________ ___________________________ Fe3+ and SO4 2- __________ ___________________________ Fe3+ and CO3 2-
Chemistry
1 answer:
bagirrra123 [75]3 years ago
7 0

Answer:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

The cations and anions being oppositely charged attract each other through strong coloumbic forces and form an ionic bond.

(1) Sodium is carrying +1 charge called as Na^{+1} cation and chloride Cl^{-1} is an anion carrying -1 charge. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral NaCl.

(2) Sodium is carrying +1 charge called as Na^{+1} cation and phosphate PO_4^{-3} is an anion carrying -3 charge. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral Na_3PO_4.

(3) Sodium is carrying +1 charge called as Na^{+1} cation and sulfate SO_4^{-2} is an anion carrying -2 charge. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral Na_2SO_4.

(4) Sodium is carrying +1 charge called as Na^{+1} cation and carbonate CO_3^{-2} is an anion carrying -2 charge. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral Na_2CO_3.

(5) Potassium is carrying +1 charge called as K^{+1} cation and chloride Cl^{-1} is an anion carrying -1 charge. They form KCl.

(6) Potassium is carrying +1 charge called as K^{+1} cation and phosphate PO_4^{-3} is an anion carrying -3 charge. They form K_3PO_4.

(7) Potassium is carrying +1 charge called as K^{+1} cation and sulfate SO_4^{-2} is an anion carrying -2 charge. They form K_2SO_4.

(8) Potassium is carrying +1 charge called as K^{+1} cation and carbonate CO_3^{-2} is an anion carrying -2 charge. They form K_2CO_3.

(9) Calcium is carrying +2 charge called as Ca^{+2} cation and chloride Cl^{-1} is an anion carrying -1 charge. They form CaCl_2.

(10) Calcium is carrying +2 charge called as Ca^{+2} cation and phosphate PO_4^{-3} is an anion carrying -3 charge. They form Ca_3(PO_4)_2.

(11) Calcium is carrying +2 charge called as Ca^{+2} cation and sulfate SO_4^{-2} is an anion carrying -2 charge. They form CaSO_4.

(12) Calcium is carrying +2 charge called as Ca^{+2} cation and carbonate CO_3^{-2} is an anion carrying -2 charge. They form CaCO_3.

(13) Ammonium ion is carrying +1 charge called as NH_4^{+1} cation and chloride Cl^{-1} is an anion carrying -1 charge. They form NH_4Cl.

(14) Ammonium ion is carrying +1 charge called as NH_4^{+1} cation and phosphate PO_4^{-3} is an anion carrying -3 charge. They form NH_4_3PO_4.

(15) Ammonium ion is carrying +1 charge called as NH_4^{+1} cation and sulfate SO_4^{-2} is an anion carrying -2 charge. They form NH_4_2SO_4.

(16) Ammonium ion is carrying +1 charge called as NH_4^{+1} cation and carbonate CO_3^{-2} is an anion carrying -2 charge. They form NH_4_2CO_3.

(17) Iron is carrying +3 charge called as Fe^{+3} cation and chloride Cl^{-1} is an anion carrying -1 charge. They form FeCl_3.

(18) Iron is carrying +3 charge called as Fe^{+3} cation and phosphate PO_4^{-3} is an anion carrying -3 charge. They form FePO_4.

(19) Iron is carrying +3 charge called as Fe^{+3} cation and sulfate SO_4^{-2} is an anion carrying -2 charge. They form Fe_2(SO_4)_3.

(20) Iron is carrying +3 charge called as Fe^{+3} cation and carbonate CO_3^{-2} is an anion carrying -2 charge. They form Fe_2(CO_3)_3.

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Elis [28]

Answer:

-low specific heat

3 0
3 years ago
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If the relative activities of two metals are known which metal is more easily oxidized
Ksivusya [100]
Answer: The metal that has a greater reactivity is more easily oxidized.

Explanation:

Oxidation is when the elements lose electrons and increase their oxidation state.

The metals tend to react by losing electrons and form the corresponding cation.

For expample, sodium (an alkalyne metal) loses one elecron and form the cation Na¹⁺ , then this cation combine with an anion and form compounds like NaCl, NaOH. The same do the other alkalyne metals.

Magnesium (an alkalyne earth metal) loses two electrons and form the cation Mg²⁺, then it combines with some anions to form compounds, like MgSO₄, Mg(OH)₂.

So, the easier the metal gets oxidized the greater its reactivity.




6 0
3 years ago
A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200
Sedbober [7]

Answer:

81°C.

Explanation:

To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat released from water (Q = - 1200 J).

m is the mass of the water (m = 20.0 g).

c is the specific heat capacity of water (c of water = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).

∵ Q = m.c.ΔT

∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).

(- 1200 J) = 83.72 final T - 7953.

∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.

<em>So, the right choice is: 81°C.</em>

7 0
3 years ago
47.0ml of a HBr solution were titrated with 37.5ml of a 0.215M LiOH solution to reach the equivalence point. what is the molarit
faltersainse [42]

Hello!

The molarity of the HBr solution is 0,172 M.

Why?

The neutralization reaction between LiOH and HBr is the following:

HBr(aq) + LiOH(aq) → LiBr(aq) + H₂O(l)

To solve this exercise, we are going to apply the common titration equation:

M1*V1=M2*V2

M1=\frac{M2*V2}{V1}= \frac{0,215 M * 37,5 mL}{47 mL}=0,172 M

Have a nice day!

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