8.03 solutions report is described below.
Explanation:
8.03 Solutions Lab Report
In this laboratory activity, you will investigate how temperature, agitation, particle size, and dilution affect the taste of a drink. Fill in each section of this lab report and submit it and your pre-lab answers to your instructor for grading.
Pre-lab Questions:
In this lab, you will make fruit drinks with powdered drink mix. Complete the pre-lab questions to get the values you need for your drink solutions.
Calculate the molar mass of powered fruit drink mix, made from sucrose (C12H22O11).
Using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 M solution of 100 mL.
d. Fe(s) and Al(s)
<h3>Further explanation</h3>
In the redox reaction, it is also known
Reducing agents are substances that experience oxidation
Oxidizing agents are substances that experience reduction
The metal activity series is expressed in voltaic series
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent
The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent
So that the metal located on the left can push the metal on the right in the redox reaction
The electrodes which are easier to reduce than hydrogen (H), have E cells = +
The electrodes which are easier to oxidize than hydrogen have a sign E cell = -
So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)
The metal : d. Fe(s) and Al(s)
Answer:
21.10g of H2O
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2C7H14 + 21O2 —> 14CO2 + 14H2O
From the balanced equation above, 2L of C7H14 produced 14L of H2O.
Therefore, 3.75L of C7H14 will produce = (3.75 x 14)/2 = 26.25L of H2O.
Next, we shall determine the number of mole of H2O that will occupy 26.25L at stp. This is illustrated below:
1 mole of a gas occupy 22.4L at stp
Therefore, Xmol of H2O will occupy
26.25L i.e
Xmol of H2O = 26.25/22.4
Xmol of H2O = 1.172 mole
Therefore, 1.172 mole of H2O is produced from the reaction.
Next, we shall convert 1.172 mole of H2O to grams. This is illustrated below:
Number of mole H2O = 1.172 mole
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H2O =..?
Mass = mole x molar mass
Mass of H2O = 1.172 x 18
Mass of H2O = 21.10g
Therefore, 21.10g of H2O is produced from the reaction.
The question is missing the data sets.
This is the complete question:
A single penny has a mass of 2.5 g. Abbie and James
each measure the mass of a penny multiple times. Which statement about
these data sets is true?
O Abbie's measurements are both more accurate
and more precise than James'.
O Abbie's measurements are more accurate,
but less precise, than James'.
O Abbie's measurements are more precise,
but less accurate, than James'.
O Abbie’s measurements are both less
accurate and less precise than James'.
Penny masses (g)
Abbie’s data
2.5, 2.4, 2.3, 2.4, 2.5, 2.6, 2.6
James’ data
2.4, 3.0, 3.3, 2.2, 2.9, 3.8, 2.9
Answer: first option, Abbie's measurements are both more accurate
and more precise than James'.
Explanation:
1) To answer this question, you first must understand the difference between precision and accuracy.
<span>Accuracy is how close the data are to the true or accepted value.
</span>
<span>Precision is how close are the data among them, this is the reproducibility of the values.</span>
Then, you can measure the accuracy by comparing the means (averages) with the actual mass of a penny 2.5 g.
And you measure the precision by comparing a measure of spread, as it can be the standard deviation.
2) These are the calculations:
Abbie’s data
Average: ∑ of the values / number of values
Average = [2.5 + 2.4 + 2.3 + 2.4 + 2.5 + 2.6 + 2.6 ] / 7 = 2.47 ≈ 2.5
Standard deviation: √ [ ∑ (x - mean)² / (n - 1) ] = 0.11
James’ data
Average = [2.4 + 3.0 + 3.3 + 2.2 + 2.9 + 3.8 + 2.9] / 7 = 2.56 ≈ 2.6
Standard deviation = 0.53
3) Conclusions:
1) The average of Abbie's data are closer to the accepted value 2.5g, so they are more accurate.
2) The standard deviation of Abbie's data is smaller than that of Jame's data, so the Abbie's data are more precise.
The % yield if 500 g of sulfur trioxide reacted with excess water to produce 575 g of sulfuric acid is calculated using the below formula
% yield = actual yield/ theoretical yield x100
actual yield =575 grams
to calculate theoretical yield
find the moles of SO3 used =mass/molar mass
= 500g/ 80 g/mol =6.25 moles
SO3+H2O=H2SO4
by use of mole ratio of SO3 : H2SO4 which is 1:1 the moles of H2SO4 is also= 6.25 moles
the theoretical yield of H2SO4 is therefore = moles /molar mass
= 6.25 x98= 612.5 grams
%yield is therefore= 575 g/612 g x100= 93.9 %
