Answer:
The titration process has quite a few real-world uses, including key roles in the food industry and medical community. The titration process is essentially an analytical technique, as it is used to determine a chemical or physical property of a chemical substance, element, or mixture (such as food). Specifically in the food industry, it is used to allow food manufactuers to determine the quantity of a reactant in a sample. To provide an example, it can be used to find the specific amount of stuff that is usually labeled on the nutrition label, such as sugar, salt, protein, calcium, vitamin C, etc. As for the medical world, pharamcists typically use this process to get the proper mix when compounding medicines. It is used to get the necessary proportions in intravenous drips.
3.9L
Explanation:
Initial volume = 25L
Final volume = 28.9L
The initial volume is the volume of the liquid without the metal in originally in the container. It is 25L
After adding the metal, the volume changes to 28.9L
Final volume = Initial volume + volume of metal
To find the volume of metal make it the subject of the expression;
Volume of metal = Final volume - initial volume = 28.9 - 25 = 3.9L
The volume of irregular solids are measured by immersing them in water. The change in volume gives their volume.
Sometimes the volume of water displaced is the volume of the solid.
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Answer:
Scientists must collect accurate information that allows them to make evolutionary connections among organisms. Similar to detective work, scientists must use evidence to uncover the facts. In the case of phylogeny, evolutionary investigations focus on two types of evidence: morphologic (form and function) and genetic.
Explanation:
This what I think
<span>C2Br2
First, we need to determine how many moles of the gas we have. For that, we'll use the Ideal Gas Law which is
PV = nRT
where
P = pressure (1.10 atm = 111458 Pa)
V = volume (10.0 ml = 0.0000100 m^3)
n = number of moles
R = Ideal gas constant (8.3144598 (m^3 Pa)/(K mol) )
T = Absolute temperature
Solving for n, we get
PV/(RT) = n
Now substituting our known values into the formula.
(111458 Pa * 0.0000100 m^3) / (288.5 K * 8.3144598 (m^3 Pa)/(K mol))
= (1.11458/2398.721652) mol
= 0.000464656 mol
Now let's calculate the empirical formula for this compound.
Atomic weight carbon = 12.0107
Atomic weight bromine = 79.904
Relative moles carbon = 13.068 / 12.0107 = 1.08802984
Relative moles bromine = 86.932 / 79.904 = 1.087955547
So the relative number of atoms of the two elements is
1.08802984 : 1.087955547
After dividing all numbers by the smallest, the ratio becomes
1.000068287 : 1
Which is close enough to 1:1 for me to consider the empirical formula to be CBr
Now calculate the molar mass of CBr
12.0107 + 79.904 = 91.9147
Finally, let's determine if the compound is actually CBr, or something like C2Br2, or some other multiple. Using the molar mass of CBr, multiply by the number of moles and see if the result matches the mass of the gas. So
91.9147 g/mol * 0.000464656 mol = 0.042708701 g
0.0427087 g is a lot smaller than 0.08541 g. So the compound isn't exactly CBr. Let's divide them to see what the factor is.
0.08541 / 0.0427087 = 1.99982673
1.99982673 is close enough to 2 to within the number of significant digits we have for me to claim that the formula for the unknown gas isn't CBr, but instead is C2Br2.</span>
Answer:
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