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Scorpion4ik [409]
2 years ago
14

What volume in milliliters of 0.0180 M Ba(OH)₂ is required to neutralize 55.0 mL of 0.0500 M HCl?

Chemistry
1 answer:
Nataly_w [17]2 years ago
5 0

The volume in milliliters of 0.0180 M Ba(OH)₂ that is required to neutralize 55.0 mL of 0.0500 M HCl is 152.78mL.

<h3>HOW TO CALCULATE VOLUME?</h3>

The volume of a solution can be calculated by using the following expression:

C1V1 = C2V2

Where;

  • C1 = initial concentration
  • C2 = final concentration
  • V1 = initial volume
  • V2 = final volume

The volume can be calculated as follows;

0.0180 × V1 = 55 × 0.05

0.0180V1 = 2.75

V1 = 2.75 ÷ 0.0180

V1 = 152.78mL

Therefore, the volume in milliliters of 0.0180 M Ba(OH)₂ that is required to neutralize 55.0 mL of 0.0500 M HCl is 152.78mL.

Learn more about volume at: brainly.com/question/1578538

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Polonium is a rare element with 33 radioisotopes. The most common one, 210Po, has 84 protons and 126 neutrons. When 210Po decays
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Answer:

Lead Pb

Explanation:

Firstly, we need to know what occurs when a radioisotope emits an alpha particle. An alpha particle is an helium atom. When an isotope emits an alpha particle, it loses an helium atom corresponding to subtracting 4 from its mass number and 2 from its atomic number. This of course coupled with the release of radiation.

Now, we polonium has a proton number of 84 and a mass number of 210. Subtracting 2 and 4 respectively from its proton and mass numbers will yield 82 and 206 proton and mass numbers respectively.

Hence, the decomposition of the Po-210 isotope will yield an element with 82 proton number and 206 mass number. This corresponds to the element Lead.

210Po ——> 206Pb + alpha particle + radiation

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A 2.0% (w/v) solution of sodium hydrogen citrate, Na2C6H6O7, which also contains 2.5% (w/v) of dextrose, C6H12O6, is used as an
tamaranim1 [39]

Answer:

0.0847M is molarity of sodium hydrogen citrate in the solution

Explanation:

The 2.0%(w/v) solution of sodium hydrogen citrate contains 2g of the solute in 100mL of solution. To find the molarity of the solution we need to convert the mass of solute to moles using molar mass and the mL of solution to Liters because molarity is the ratio between moles of sodium hydrogen citrate and liters of solution.

<em>Moles Na2C6H6O7:</em>

<em>Molar Mass:</em>

2Na: 2*22.99g/mol: 45.98g/mol

6C: 6*12.01g/mol: 72.01g/mol

6H: 6*1.008g/mol: 6.048g/mol

7O: 7*16g/mol: 112g/mol

45.98g/mol + 72.01g/mol + 6.048g/mol + 112g/mol = 236.038g/mol

Moles of 2g:

2g * (1mol / 236.038g) = <em>8.473x10⁻³ moles</em>

<em />

<em>Liters solution:</em>

100mL * (1L / 1000mL) = <em>0.100L</em>

<em>Molarity:</em>

8.473x10⁻³ moles / 0.100L =

<h3>0.0847M is molarity of sodium hydrogen citrate in the solution</h3>
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