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3 years ago

What two forces limit the height to which capillary action will take water up a tube?

1 answer:

6 0

Capillary action occurs when the adhesion to the walls is stronger than the cohesive forces between the liquid molecules. The height to which capillary action will take water in a uniform circular tube is limited by surface tension and, of course, gravity.

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PLZ HELP

riadik2000 [5.3K]

Too many to know in the world.

8 0

3 years ago

4. How is the coordination number determined?

balandron [24]

A coordination number can be determined by the usage of an atom towards a molecule from seeing how many numbers of atoms would have to be combined together in an atom.

3 0

2 years ago

What is the mass of 3.0 x 1023 atoms of neon

Xelga [282]

1 mole =6.0 * 10^23 atoms
3.0*10^23 atoms = 0.5 moles
the molar mass of neon is 20.18g/mole
0.5 moles = 10.09 grams

8 0

2 years ago

Read 2 more answers

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago

If 45 mL of water are added to 250 mL of a 0.75 M K2SO4 solution, what will the molarity of the diluted solution be?

krok68 [10]

Answer:

$\large\boxed{\large\boxed{0.64M}}$

Explanation:

When you form a diluted solution from a mother (concentrated) solution, the moles of solute are determined by the mother solution.

The main equation is:

$Molarity=\dfrac{\text{moles of solute}}{\text{volume of the solution in liters}}$

Then, since the moles of solute is the same for both the mother solution and the diluted solution:

$\text{Molarity mother solution }\times\text{ volume mother solution}=\\\\=\text{Molarity diluted solution }\times\text{ volume diluted solution}$

Substitute and solve for the molarity of the diluted solution:

$250mL\times 0.75M=(45mL+250mL)\times M\\\\\\M=\dfrac{250mL\times 0.75M}{295mL}=0.64M$

8 0

3 years ago