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SCORPION-xisa [38]
3 years ago
11

When an ionic compound dissolves in water:_________.

Chemistry
1 answer:
Katarina [22]3 years ago
5 0

Answer:

a. the negative ends of water molecules surround the positive ions.

Explanation:

Ionization reaction is defined as the reaction in which an ionic compound dissociates into its ions when dissolved in aqueous solution.

Thus,

NaCl_{(s)}\rightarrow Na^+_{(aq)}+Cl^-_{(aq)}

When the ionic compound dissociates, the cation which is positively charged gets attracted to the oxygen (Negative ends) which is present in the water. The anion which is negatively charged gets attracted to the hydrogen (Positive ends) of the water.

Hence, the correct option is:- a. the negative ends of water molecules surround the positive ions.

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Which one of the following is not a fossil fuel?
Finger [1]

The answer is C, Uranium.

Natural gas, coal, and oil are all naturally made from fossils. Uranium is an element which can be found on the Periodic Table of Elements.

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2 years ago
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Identify the three types of metamorphism.
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Answer:

Three types of metamorphism exist: contact, dynamic, and regional. Metamorphism produced with increasing pressure and temperature conditions is known as prograde metamorphism.

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3 years ago
What are the two different ions present in the compound al(no3)3?
Aleks [24]

Al^{3+} and NO_3^{-}

Aluminum nitrate is an ionic compound, meaning that it is made up cations and anions. By convention when writing the formula for an ionic compounds the cation part shall be placed in front the anions. Thus the cation would contain aluminum and the anion shall be a polyatomic nitrate ion consisting of nitrogen and oxygen. The bracket with subscript 3 attached resembles the ratio between the two species, aluminum ions and nitrate ions. For each aluminum ion in the compound there shall be three corresponding nitrate ions.

Aluminum form ions of charge +3 whereas nitrate ions are of charge -1. Write the charge of the ion as the superscript in reverse (i.e., write 3+ as in Al^{3+} and - in NO_3^{-}, omitting any 1.)

5 0
3 years ago
D 6) Select True or False.
Mama L [17]

Answer:

true

Explanation:

6 0
3 years ago
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A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of t
Verdich [7]

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield  = 87.72%

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

  • moles of PbI₂
  • Limiting reactant
  • % yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

=  1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

7 0
3 years ago
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