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4vir4ik [10]
3 years ago
6

Write the molecular formula of the given compounds.

Chemistry
1 answer:
timurjin [86]3 years ago
8 0

Answer:

(i) Al2(SO4)3

(ii) CaCl2

(iii) Fe(OH)2

(iv)AlPO4

(v) Fe(NO3)3

(vi) AuCl3

(vii) AgNO3

Explanation:

The numbers after the element/ion is a subscript meaning its put below the symbol

(i) Al has a charge of 3+ and SO4 has a charge of 2-

So to balance the charges 2 Al is needed to make the charge a 6+ and 3 SO4 is needed to make the charge a 6-. Once you can add the two charge values together to create 0, then you have the correct quantities

(ii) Same applies here, but Ca is 2+ and Cl is 1-, so 2 Cl- is needed to balance out the Ca

(iii) Ferrous is an ion of Iron with a 2+ charge, the other ion of Iron is Ferric which has a 3+ charge. OH has a charge of -1, so 2 OH is needed to balance out one Fe

(iv) Aluminum again has a charge of 3+ while PO4 has a charge of 3-, so they both balance out with one of each.

(v) Ferric is an ion of iron with a charge of 3+ and NO3 has a charge of 1-, so three of NO3 is needed to balance.

(vi) Gold has two charges, 1+ or 3+. However used in Gold chloride it is a 3+ and Cl is a 1-, so 3 of Cl is needed to balance one gold.

(vii) Silver has a charge of 1+ while NO3 has a charge of 1-, so they balance each other at one each.

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5 0
3 years ago
Determine the electron geometry (eg) and molecular geometry (mg) of CO2.
eimsori [14]

Answer:

eg=linear, mg=linear

Explanation:

First of all, it must be stated that most triatomic molecules are either linear or bent. This depends on the electron geometry of the molecule and the number of bonding groups, multiple bonds and lone pairs present.

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6 0
4 years ago
How many milliliters of 0.0050 N KOH are required to neutralize 41 mL of 0.0050 M H2SO4?
Kipish [7]

Answer:

V KOH = 41 mL

Explanation:

for neutralization:

  • ( V×<em>C </em>)acid = ( V×<em>C </em>)base

∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L

∴ V H2SO4 = 41 mL = 0.041 L

∴ <em>C</em> KOH = 0.0050 N = 0.0050  eq-g/L

∴ E KOH = 1 eq-g/mol

⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L

⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH

⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)

⇒ V KOH = 0.041 L

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