5.0 mols Li × 6.94 g Li/ 1 mol Li = 34.7 grams of Li
sig figs: 35 grams of Li
0.038 g He
From the question we are given;
Volume of the flask, V = 250 mL
Pressure of He, P = 700 torr
Temperature, T = 25°C, but K = 273 + °C
= 298 K
We are required to calculate the mass of Helium present in the flask.
We are going to use the ideal gas equation to first get the number of moles of He present in the flask.
Ideal gas equation, PV = nRT, where n is the number of moles and R is the ideal gas constant.
In this case, R will be 62.364 L.Torr/mol.K, since the pressure is in torr.
Rearranging the equation;
n = PV ÷ RT
= (0.25 L × 700 torr) ÷ (62.364 × 298 K)
= 0.00942 moles
With the number of moles, we can calculate the mass of He.
Mass = Number of moles × molar mass
Molar mass of He = 4.0 g/mol
Therefore,
Mass = 0.00942 moles × 4.0 g/mol
= 0.03768 g
= 0.038 g He
B. Classifying
C. Inductive reasoning
D. Procedure
Answer;
A. Bias
If the personal opinion of a scientist affects the way that the experimental results are reported, that is called bias.
Explanation;
-An experimenter bias, is a process where the scientists performing the research influence the results, in order to portray a certain outcome.
-Bias is among the factors that makes a qualitative research much more dependent upon experience and judgment than quantitative research.
-Bias in a research may arise from experimental errors and failure to consider all the possible variables and also when the researchers select the subjects that are more likely to generate the desired results.
If the substance entering the cell was in higher concentration inside the cell than outside the cell, the type of transport that would be required here is known as Active transport. Thus, the correct option for this question is A.
Active transport may be defined as a type of transport that occurs against the concentration gradient and therefore it is mediated by carrier proteins. In this type of transport, metabolic energy is utilized in order to move ions or molecules against the concentration gradient. It is directly opposite to passive transport.
Active transport results in the accumulation of solute on one side of the membrane. It is different from the carrier proteins mediated facilitated diffusion. This transport decreases the entropy of the reaction.
Therefore, active transport would be required in order to migrate any substance from a lower concentration to a higher concentration. Thus, the correct option for this question is A.
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B) two D) four
Answer: c
Explanation:brainliest
Grams are the unit of the mass that is used to calculate the moles. From 37.5 gms of iron, 53.6 gms of ferric oxide are produced.
Mass is the measurement of the moles of the substance and the molar mass.
Moles of iron from the mass is calculated as:
Moles of iron = 37.5 gms ÷ 55.84 = 0.671 moles
The balanced chemical reaction:
4Fe + 3O2 → 2Fe2O3
From the above it is deduced that 4 moles of iron produce 2 moles of ferric oxide so, 0.671 moles of iron will produce,
(0.671 × 2) ÷ 4 = 0.3375 moles
Mass of ferric oxide, from moles, is calculated as:
Mass = 0.33 moles × 159.687
= 52.696 gms
Therefore, 53.6 gms of ferric oxide will be produced from 37.5 gms of iron.
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The mass of Fe2O3 that can be produced from 37.5g of iron (Fe) is approximately 53.65g. This is achieved by converting mass of iron to moles, using stoichiometry from the balanced chemical equation to convert moles of iron to moles of Fe2O3, and then converting moles of Fe2O3 back to grams.
First, we need to figure out the molar mass of iron (Fe) which is approximately 55.85 g/mol and the molar mass of iron(III) oxide (Fe2O3) which is approximately 159.69 g/mol. We find this using the atomic masses of Iron (Fe) and Oxygen (O) from the periodic table and add them appropriately.
Next, to find the number of moles of iron we use the provided mass of Fe and its molar mass. We calculate this as (37.5 g Fe / 55.85 g/mol Fe) = 0.671 moles of Fe. Now, the balanced chemical equation for the formation of iron(III) oxide is: 4Fe + 3O2 --> 2Fe2O3. From this balanced equation, we know that it takes 4 moles of iron (Fe) to produce 2 moles of Fe2O3. Therefore, the moles of Fe2O3 formed from 0.671 moles of Fe would be (0.671 moles Fe * 2 moles Fe2O3/4 moles Fe) = 0.336 moles of Fe2O3.
Finally, to find the mass of Fe2O3 produced, we multiply the moles of Fe2O3 by its molar mass. We calculate this as (0.336 moles Fe2O3 * 159.69 g/mol Fe2O3) = 53.657 g of Fe2O3.
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