the accepted density of helium at stp is 0.18 g/L. during an expirement, 0.10g of helium with a volume of 0.54L at stp was collected. what is the percent error in this expirement?
(accepted-expiremental)/accepted X 100
Please explain because I don't understand. You have the data and the formula. What's left? Plug and chug.
To calculate the percent error in the experiment, use the formula:
Percent Error = ((Accepted Value - Experimental Value) / Accepted Value) * 100
Given:
Accepted Density of Helium at STP = 0.18 g/L
Experimental mass of Helium collected = 0.10g
Experimental volume of Helium collected = 0.54L
First, calculate the experimental density using the collected mass and volume:
Experimental Density = Experimental mass / Experimental volume
Experimental Density = 0.10g / 0.54L
Now, substitute the values into the formula to calculate the percent error:
Percent Error = ((0.18 g/L - Experimental Density) / 0.18 g/L) * 100
Percent Error = ((0.18 g/L - (0.10g / 0.54L)) / 0.18 g/L) * 100
Simplify the expression inside the parentheses:
Percent Error = ((0.18 - 0.10/0.54) / 0.18) * 100
Perform the division:
Percent Error = ((0.18 - 0.185) / 0.18) * 100
Calculate the subtraction:
Percent Error = (-0.005 / 0.18) * 100
Perform the division:
Percent Error = -0.0277778 * 100
Calculate the multiplication:
Percent Error = -2.77778
Therefore, the percent error in this experiment is approximately -2.78%. Note that the negative sign indicates an underestimation of experimental value compared to the accepted value.