How does cooler air move in relation to warmer air?

A solution is prepared by dissolving 0.5636 g oxalic acid (H2C2O4) in enough water to make 100.0 mL of solution. A 10.00-mL aliq

zepelin [54]

Answer:

The final molarity of the diluted oxalic acid solution is 0.002504 M

Explanation:

Step 1: Data given

Mass of oxalic acid (H2C2O4) = 0.5636 grams

Volume of the solution = 100.0 mL = 0.100 L

A 10 ml of this solution diluted in 250 ml of solution.

Molecular weight of H2C2O4 = 90.03 g/mol

Step 2: Calculate initial moles of H2C2O4

Moles H2C2O4 = mass / molar mass

Moles H2C2O4 = 0.5636 grams / 90.03 g/mol

Moles H2C2O4 = 0.00626 moles

Step 3: Calculate molarity of the solution

Molarity = moles / volume

Molarity = 0.00626 moles / 0.100 L

Molarity = 0.0626 M

Step 4: Calculate moles of a 10.00 mL aliquot

Moles = 0.0626 M * 0.010 L

Moles = 0.000626 moles

Step 5: Calculate the new molarity

Molarity = 0.000626 moles / 0.250 L

Molarity = 0.002504 M

The final molarity of the diluted oxalic acid solution is 0.002504 M

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How does the atomic radius increase?

Helen [10]

Answer:

increases as you move down a group as the number of electrons increases. Therefore, the atomic radius increases as the group and energy levels increase

Explanation:

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Sav [38]

Answer:

The atomic number represents the number of protons

Explanation:

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icang [17]

A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed . (ans: second choice)

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At 25C the density of water is 0.997044 g/mL. Use this value to determine the percent error for the two density measurements

Gnom [1K]

Given that:

At 25C the density of water is 0.997044 g/mL.

From the information attached below, we have the following parameters.

The density of water calculation using a bottle.

Initial volume of Final volume of Mass of water Density (g/mL)

burette (mL) burette (mL) dispensed (g)

Sample 1 2.33 7.34 5.000 -----

Sample 2 7.34 12.37 5.025 -----

Sample 3 12.37 18.50 6.112 -----

Sample 4 18.50 24.57 6.064 -----

Sample 5 24.57 31.31 6.720 -----

The first thing we need to do is to determine the change in the volume of the burette in each sample from the above information.

The change in the volume of the burette = (final volume - the initial volume) mL

Sample 1:

= (7.34 - 2.33) mL

= 5.01 mL

Sample 2:

= (12.37 - 7.34) mL

= 5.03 mL

Sample 3:

= (18.50 - 12.37) mL

= 6.03 mL

Sample 4:

= (24.57 - 18.50) mL

= 6.07 mL

Sample 5:

= (31.31 - 24.57) mL

= 6.74 mL

The mass of the water dispersed in sample 1 is given as = 5.000 g

Using the relation for calculating the density of each, we have:

Sample 1

$\mathbf{density = \dfrac{mass}{volume}}$

$\mathbf{density = \dfrac{5.01 g}{5.000 ml}}$

density = 0.998004 g/ml

Sample 2:

$\mathbf{density = \dfrac{5.025 g}{5.03ml}}$

density = 0.999006 g/ml

Sample 3:

$\mathbf{density = \dfrac{6.112 g}{6.13ml}}$

density = 0.997064 g/ml

Sample 4:

$\mathbf{density = \dfrac{6.064 \ g}{6.07 \ ml}}$

density = 0.999012 g/ml

Sample 5:

$\mathbf{density = \dfrac{6.720 \ g}{6.74 \ ml}}$

density = 0.997033 g/ml

Thus, the average density for all the samples is:

$\mathbf{= \dfrac{( 0.998004 + 0.999006 + 0.997064 + 0.999012 + 0.997033 )}{5}}$

= 0.998024

∴

The percentage error for the two densities measurement is:

$=\dfrac{ (experimental \ value -theoretical \ value)\times 100 }{theoretical \ value}$

Given that the theoretical value = 0.997044 g/ml

Then;

$\mathbf{= \dfrac{(0.998024 - 0.997044)100}{0.997044}}$

= 0.0983%

Therefore, we can conclude that the percent error for the two density measurements is 0.0983%

Learn more about density here:

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3 years ago