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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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Answer:

allows for you to account for only the weight of the substance being measured and not the vessel it’s being measured in.

Explanation:

What does it mean to tare a balance and why do you think it is important to complete this before you begin measuring mass? Explanation: The term tare is used when weighing chemicals on a balance, using the tare button allows for you to account for only the weight of the substance being measured and not the vessel it’s being measured in.

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A mineral sample is obtained from a region of the country that has high arsenic contamination. An elemental analysis yields the
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Answer:

<em><u>CaAsHO₄</u></em>

Explanation:

The data has a mistake in one of the values there. I believe the mistake is on the hydrogen. So, I'm going to assume the value of Hydrogen is 0.6%, so the total percent composition would be 100.1% (Something better). All you have to do is replace the correct value of H (or the value with the mistaken option) and do the same procedure.

Now, to calculate the empirical formula, we can do this in three steps.

<u>Step 1. Calculate the amount in moles of each element.</u>

In these case, we just divide the percent composition with the molar mass of each one of them:

Ca: 22.3 / 40.078 = 0.5564

As: 41.6 / 74.9216 = 0.5552

O: 35.6 / 15.9994 = 2.2251

H: 0.6 / 1.00794 = 0.5953

Now that we have done this, let's calculate the ratio of mole of each of them. This is doing dividing the smallest number of mole between each of the moles there. In this case, the moles of As are the smallest so:

Ca: 0.5564/0.5552 = 1.0022

As: 0.5552/0.5552 = 1

O: 2.2251/0.5552 = 4.0077

H: 0.5953/0.5552 = 1.0722

Now, we round those numbers, and that will give us the number of atoms of each element in the empirical formula

<u>Step 3. Write the empirical formula with the rounded numbers obtained</u>

In this case we will have:

Ca: 1

As: 1

O: 4

H: 1

The empirical formula would have to be:

<em><u>CaAsHO₄</u></em>

3 0
3 years ago
The reaction of aluminum with bromine is shown here. The equation for the reaction is
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2,3,1 is the stoichiometry

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Step by step exp.
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