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nata0808 [166]
2 years ago
3

Mark measures the volume of water in a beaker marked in 50-ml intervals. He reports the volume is 43.927 ml. What is misleading

about this value?
Chemistry
2 answers:
Triss [41]2 years ago
7 0

The scale to determine laboratory level any liquid is generally milliliter (mL). Now in a beaker which is marked in 50 mL intervals there reporting the volume 43.927 mL of a certain liquid from the beaker is not feasible. And if done there will be serious error. The value of the liquid volume can be measured from the beaker preciously must be in terms of 50 mL that is 50, 100, 150, 200 mL etc. Thus the misleading about the value are-

1. The value is measured by any experiment or calculation, because if its through experiment by 50 mL marked beaker it is not possible.

2.    The volume is been measured by the same beaker or not.

3. The accuracy and precision of the result is questionable.  

Marysya12 [62]2 years ago
4 0

Answer:

The beaker doesn't allow to measure values different than 50 mL, 100 mL, 150 mL and so on, not even with decimals.

Explanation:

Hello,

Such types of beakers just report an approximate measure every 50 mL, thus, what is misleading is that the beaker doesn't allow to measure values different than 50 mL, 100 mL, 150 mL and so on, so the reporting of the decimals and the number below 50 is completely wrong due the beaker's scale and the significant figures it provides.

Best regards.

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Answer:

6.72M of HNO3

Explanation:

In the problem you are diluting the original HNO3 solution by the addition of some water. The final volume is:

290.7mL + 350.0mL = 640.7mL

And you are diluting the solution:

640.7mL / 350.0mL = 1.8306 times

As the original concentration was 12.3M, the final concentration will be:

12.3M / 1.8306 =

<h3>6.72M of HNO3</h3>
5 0
2 years ago
A white powder is known to be a mixture of magnesium oxide and aluminum oxide. 100cm3 of 2moldm-3 NaOH(aq) is just sufficient to
Effectus [21]

Answer: D.Aluminium Oxide 0.10, Magnesium Oxide 0.50

Explanation:

Number of moles of NaOH= number of moles × volume

Number of moles= 100/1000 × 2 = 0.2 moles

Since;

2 moles of NaOH yield 1 mole of Al2O3

0.2 moles of NaOH will yield 0.2 × 1/2 = 0.1 moles of Al2O3.

Number of moles of HCl= 800/1000 × 2 = 1.6 moles

If 1 mole of Al2O3 requires 6 moles of HCl

0.1 moles of Al2O3 requires 0.1 × 6 = 0.6 moles of HCl.

Number of moles of HCl left after reaction with Al2O3 = 1.6- 0.6 = 1 mole

This leftover reacts with MgO

But;

1 mole of MgO reacts with 2 moles of HCl

x moles of MgO reacts with 1 mole of HCl

Thus; x= 0.5 moles of MgO

8 0
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How many molecules of water are needed to completely hydrolyze a polymer that is 11 monomers long?
arlik [135]
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A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is
e-lub [12.9K]

Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

mass sln = 750.0 g = mass solute + mass solvent

∴ solute (b) unknown nonelectrolyte compound

∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

  • ΔT = K*mb = 100.220°C ≅ 373.22 K

∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

⇒ mb = (373.22 K) / (1.86 K.Kg/mol)

⇒ mb = 200.656 mol/Kg

∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg

moles solute:

⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute

molecular weight:

⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol

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The answer would be zinc
3 0
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