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Ket [755]
3 years ago
13

Water is a polar molecule. What does that mean?

Chemistry
2 answers:
Dimas [21]3 years ago
8 0
Answer: it is a molecule with no charge
aalyn [17]3 years ago
7 0
The answer. Is c I hope this helps
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3 A specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state.
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When a specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state, this emitted energy can
<span>be used to determine the "identity of the element".</span>
5 0
3 years ago
Write and balance molecular equations for the following reactions between aqueous solutions. You will need to decide on the form
musickatia [10]

Answer:

This is the balanced equation:

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

Explanation:

This are the reactants:

PbNO₃

NaI

Iodide can react to Pb²⁺ to make a solid compound.

4 0
3 years ago
How many molecules are in 20,484 grams of H2O?​
alexandr1967 [171]

Answer:

3.4027x10^-22

Explanation:

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3 years ago
Convert a mass defect of 0.115 g to (a) Joules and (b) MeV
timofeeve [1]

Calculate the mass defect and nuclear binding energy of an atom ... To convert to joules per nucleon, simply divide by the number of nucleons.

7 0
3 years ago
n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a ph
olasank [31]

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

<em>where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.</em>

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = <em>0.111M</em>

<em></em>

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = <em>0.2501 g of acetic acid</em>

Now, assuming density of solution as 1.00g/mL, 37.54mL weights <em>37.54g</em>.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = <em>0.666% (w/w)</em>

8 0
2 years ago
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