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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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Answer:

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Explanation:

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3 years ago
Calcium bicarbonate (Ca(HCO3)2) Ca:H:C:O = 1:2:2 ___ Lithium sulfide (Li:2s) Li:S = 2:__
Anastaziya [24]

Answer: The ratio of atoms in calcium bicarbonate ; Ca : H : C : O = 1:2:2:6.

The ratio of atoms in lithium sulfide; Li : S = 2 : 1

Explanation:

In calcium bicarbonate:Ca(HCO_3)_2

In a molecular formula of calcium carbonate there are:

Number of Calcium atoms = 1

Number of Hydrogen atom = 1 × 2 = 2

Number of Carbon atoms = 1 × 2 = 2

Number of Oxygen atoms = 3 × 2 = 6

So, Ca : H : C : O = 1 : 2 : 2 : 6

In lithium sulfide :Li_2S

In a molecular formula of lithium sulfide there are:

Number of Lithium atoms =  1 × 2 = 2

Number of Sulfur atoms = 1

So, the Li : S = 2 : 1

8 0
3 years ago
What is the acceleration of an object if it goes from a velocity of +25m/s to rest in 5.0s?
kari74 [83]

Vf = Vi + at

0 m/s = 25 m/s + a(5sec)

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5 0
3 years ago
Which Ion has the same electron configuration as an Ar atom
Troyanec [42]
Ar (argon) has 18 electrons

Cl- would give you 18 electrons
F- would give you 10 electrons
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Cl- would be the correct answer
7 0
3 years ago
How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?
xenn [34]

Answer:

5.225x10^{22}atoms\ I

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I

Best regards!

4 0
3 years ago
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