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Goryan [66]
3 years ago
5

If it requires 18.2 mL of 0.45 molar barium hydroxide to neutralize 38.5 mL of nitric acid, solve for the molarity of the nitric

acid solution. Show all the work used to solve this problem
Chemistry
1 answer:
krok68 [10]3 years ago
4 0

The moles of OH- ions from the Ba(OH)2 must equal the moles of H+ ions from the HNO3 in order for them to neutralize. You must multiply volume (in liters) by the molar to get number of moles. There is 0.90 molar of OH- because there is twice as many OH- as there is Ba(OH)2. The molarity of H+ is unknown.

Let X be the unknown molarity


(0.0182 L)(0.90 M) = (0.0385 L)(X M)

X = 0.43 M (2 significant figures)


So the molarity of H+ ions, and therefore HNO3 is 0.43 M

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Find the density of an object that has a mass of 6.13 g and volume of 1.50 cm³.​
storchak [24]
  • <em>Answer:</em>

<em>≈ 4.10 g/cm³</em>

  • <em>Explanation:</em>

<em>Hi there ! </em>

<em><u /></em>

<em><u>Density formula</u></em>

<em>d = m/V</em>

<em>d = 6.13g/1.5cm³</em>

<em>d = 4.08(6) g/cm³ ≈ 4.10 g/cm³</em>

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<em>Good luck !</em>

8 0
3 years ago
Which of the following is the best hypothesis?
torisob [31]

Answer:

b

Explanation:

i think it is b

8 0
3 years ago
Dissolving brass requires an oxidizing acid such as concentrated nitric acid. Nitrogen dioxide is produced as a byproduct in thi
polet [3.4K]

Answer:

                  Cu  +  4 HNO₃    →    Cu(NO₃)₂  +  2 NO₂  +  2 H₂O

Explanation:

Step 1: Write down the chemical formulas of given substances,

                                    Copper Metal  =  Cu

                                    Nitric Acid  =  HNO₃

                                    Copper (II) Nitrate  =  Cu(NO₃)₂

                                    Nitrogen Dioxide  =  NO₂

                                    Water  =  H₂O

Step 2: Write down the unbalance Chemical equation,

                         Cu  +  HNO₃    →    Cu(NO₃)₂  +  NO₂  +  H₂O

Step 3: Balance Cu atoms on both sides;

The number of Cu atoms on both sides are same. Hence, there number will remain the same.

Step 4: Balance N atoms on both sides;

As there is 1 N atom on left hand side and 3 N atoms on right hand side, so we will multiply HNO₃ by 3 to balance N on both sides, hence,

                         Cu  +  3 HNO₃    →    Cu(NO₃)₂  +  NO₂  +  H₂O

Step 5: Balance O atoms on both sides;

As there are 9 O atom on left hand side and 9 O atoms on right hand side, so they are balance.

Step 6: Balance H atoms on both sides;

As there are 3 H atom on left hand side and 2 H atoms on right hand side, so we will multiply H₂O by 2 as,

                         Cu  +  3 HNO₃    →    Cu(NO₃)₂  +  NO₂  +  2 H₂O

By doing so the number of O atoms got imbalanced, so to balance O atoms again we will multiply HNO₃ by 4 as,

                         Cu  +  4 HNO₃    →    Cu(NO₃)₂  +  NO₂  +  2 H₂O

Now, The Cu and H atoms are balanced, and the O atoms are greater on left hand side and the N atoms are greater on right hand side, therefore we will multiply NO₂ by 2 to balance both N and O as,

                         Cu  +  4 HNO₃    →    Cu(NO₃)₂  +  2 NO₂  +  2 H₂O

7 0
3 years ago
Draw the major organic substitution product(s) for (2R,3S)-2-bromo-3-methylpentane reacting with the given nucleophile. Indicate
Andrew [12]

Answer:

(2R,3S)-2-ethoxy-3-methylpentane

and

(2S,3S)-2-ethoxy-3-methylpentane

Explanation:

For this case, we will have  CH_3CH_2O^- as nucleophile. Also, this compound is also in excess. So, we will have as solvent CH_3CH_2OH a protic solvent. Therefore the Sn1 reaction would be favored.

The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).

7 0
2 years ago
What is the concentration of the barium hydroxide solution if 50.0 mL of a 0.425 M HNO3 solution is required to neutralize a 36.
fomenos

The molarity of Barium Hydroxide is 0.289 M.

<u>Explanation:</u>

We have to write the balanced equation as,

Ba(OH)₂ + 2 HNO₃ → Ba(NO₃)₂ + 2 H₂O

We need 2 moles of nitric acid to react with a mole of Barium hydroxide, so we can write the law of volumetric analysis as,

V1M1 = 2 V2M2

Here V1 and M1 are the volume and molarity of nitric acid

V2 and M2 are the volume and molarity of Barium hydroxide.

So the molarity of Ba(OH)₂, can be found as,

$ M2 = \frac{V1 \times M1}{2 \times V2}

   $M2 = \frac{50 \times 0.425 }{2 \times 36.8}

      = 0.289 M

5 0
3 years ago
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