Which of the following has dispersion forces as its only intermolecular force? A) CH4 B) HCl C) C6H13NH2 D) NaCl E) CH3Cl

Which step of the scientific method involves obtaining resources for your question?

djyliett [7]

I would say it is constructing a hypothesis, that is because a hypothesis is basically gathering information or resources to support what your question was. so i say it is B.

3 0

3 years ago

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Sodium hydroxide (a base) reacts with hydrochloric acid. After the reaction is complete, a pH indicator is added to the solution

fomenos

Answer:

hydrochloric acid

Explanation:

The hydrochloric acid in this reaction is the limiting reactant. A limiting reactant is the reactant that is used up in a chemical reaction. It determines the extent of the reaction.

Since the solution indicates a basic one after the end of the reaction, this suggests that more of the sodium hydroxide is still left unreacted with.

The reactant in excess supply here is the sodium hydroxide and the bulk of it is till left in solution.

7 0

3 years ago

A 15.00-ml sample of a naoh solution of unknown concentration requires 17.88 ml of a 0.1053 m h2so4 solution to reach the equiva

Sergeu [11.5K]

Answer:

0.1255 M

Explanation:

1) Data:

Base: NaOH

Vb = 15.00 ml = 15.00 / 1,000 liter

Mb = ?

Acid: H₂SO₄

Va = 17.88 ml = 17.88 / 1,000 liter

Ma = 0.1053

2) Chemical reaction:

The titration is an acid-base (neutralization) reaction to yield a salt and water:

Acid + Base → Salt + Water

H₂SO₄ (aq) + NaOH(aq) → Na₂SO₄ (aq) + H₂O (l)

3) Balanced chemical equation:

H₂SO₄ (aq) + 2 NaOH(aq) → Na₂SO₄ (aq) + 2H₂O (l)

Placing coefficient 2 in front of NaOH and H₂O balances the equation

4) Stoichiometric mole ratio:

The coefficients of the balanced chemical equation show that 1 mole of H₂SO₄ react with 2 moles of NaOH. Hence, the mole ratio is:

1 mole H₂SO₄ : 2 mole NaOH

5) Calculations:

a) Molarity formula: M = n / V (in liter)

⇒ n = M × V

b) Nunber of moles of acid:

nₐ = Ma × Va = 0.1053 (17.88 / 1,000)

c) Number of moles of base, nb:

nb = Mb × Vb = Mb × (15.00 / 1,000)

d) At equivalence point number of moles of acid = number of moles of base

nₐ = nb

0.1053 × (17.88 / 1,000) = Mb × (15.00 / 1,000)

Mb = 0.1053 × 17.88 / 15.00 = 0.1255 mole/liter = 0.1255 M

3 0

3 years ago

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How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

8 0

3 years ago

Which rules are violated in the following electronic configuration 1s2 2s3

aalyn [17]

There is no element in 2s3

5 0

3 years ago