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3 years ago

Actual value= -273 Experimental value= -274

Chemistry

1 answer:

Maurinko [17]3 years ago

4 0

Answer:

<h2><u><em>When measuring data, the result often varies from the true value. Error can arise due to many different reasons that are often related to human error, but can also be due to estimations and limitations of devices used in measurement. Regardless, in cases such as these, it can be valuable to calculate the percentage error. The computation of percentage error involves the use of the absolute error, which is simply the difference between the observed and the true value.</em></u></h2>

Explanation:

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If you have 10cm of snow with a volume of 40mL and a density of 0.5 g/mL how many inches of rain is this?

Lelu [443]

Density is the mass of compound divided by its volume can be shown as follows:

$d = 0.5 g /mL = m /V = m /40 = 0.5$

$m = 20 g$

40 mL of snow having 20 g of mass calculated from density.

Now, 10 cm of snow = 3.93 inches = 20 g

As, 10 inches of rain will produce 11 inches of ice as the volume of ice is bigger than rain water.

10 inches rain = 11 inches snow

3.93 inches of snow produced by

$= (3.93 * 10) /11 = 3.57 inches rain.$

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A saturated solution is made by dissolving 0.327 g of a polypeptide (a substance formed by joining together in a chainlike fashi

faust18 [17]

Answer: The approximate molecular mass of the polypeptide is 856 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution in L)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 4.19 torr

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (polypeptide) = 0.327 g

Volume of solution = 1.70 L

R = Gas constant = $62.364\text{ L.torr }mol^{-1}K^{-1}$

T = temperature of the solution = $26^oC=[273+26]K=299K$

Putting values in above equation, we get:

$4.19torr=1\times \frac{0.327}{\text{Molar mass of solute}\times 1.70}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 299K\\\\\text{molar mass of solute}=856g/mol$