During an experiment, a student adds 2.90 g CaO to 400.0 mL of 1.500 M HCl . The student observes a temperature increase of 6.00 °C . Assuming that the solution's final volume is 400.0 mL , the density is 1.00 g/mL , and the heat capacity is 4.184 J/g⋅°C , calculate the heat of the reaction, ΔHrxn .
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Answer:
ΔHrxn = 193107.69 J/mol
Explanation:
ΔHrxn = mcΔT
m = mass
c = heat capacity
ΔT = temperature variation
density = m/V
m = density x V
m = 1.00 g/mL x 400.0 mL
m = 400.0 g
ΔHrxn = mcΔT
ΔHrxn = 400 g x 4.184 J/g°C x 6.00 °C
ΔHrxn = 10041.6 J
CaO + 2HCl → CaCl₂ + H₂O
CaO = 56.0774 g/mol
2.90 g CaO = 0.052 mol
400.0 mL of 1.500 mol/L HCl = 0.6 mol HCl
ΔHrxn = 10041.6 J is for 0.052 mol of CaO
ΔHrxn = 193107.69 J is for 1 mol of CaO
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6.579 rounded to nearest hundreth
Four different methods are described for validating the results of a particular analysis. Indicate for each whether the method primarily checks the accuracy of the analysis or the precision of the analysis.a. Five aliquots of the same sample are injected for gas chromatographic analysis by one person on the same day.i. Accuracyii. Precisionb. A known amount of analyte is added to an aliquot of the sample and analyzed with sample.i. Accuracyii. Precisionc. Aliquots from a blood sample are sent to three separate laboratories for analysis using the same method.i. Accuracyii. Precisiond. Identical standard are analyzed by two different methods.i. Accuracyii. Precision
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Hope this helps!
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Answer:
A polysaccharide (n) can be formed by linking several monosaccharides through glycosidic linkages.
Explanation:
Polysaccharides are carbohydrates or complex carbohydrates, where monosaccharides join with glucosidic bonds to form a more complex structure that would be the polysaccharide.
An example of a polysaccharide is starch, or glycogen.
Starch is found in many foods such as potatoes or rice, and glycogen is a form of energy reserve of our organism housed in muscles and liver to fulfill locomotion, physical activity, and other activities that consist of glycolysis.
Polysaccharides are degraded in our body by different stages, and several enzymes unlike monosoccharides or disaccharides, since they have more unions and a more complex structure to disarm in our body and thus assimilate it.
Polysaccharides are also part of animal structures, such as insect shells or nutritional sources, among others.
B) Jet Fuel
C) steel
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Option-C, STEEL is not a product of the fractional distillation of petroleum.
Explanation:
Petroleum is the mixture of Hydrocarbons *carbon and hydrogen containing compounds) present beneath the Earth's surface. Petroleum is formed from the remains of animals and plants beneath earth's surface in an anaerobic conditions.
Petroleum contains from small hydrocarbons (gases) to medium (liquids) and long chain hydrocarbons (Solids). These hydrocarbons are separated from each other by Fractional Distillation method (separation due to difference in boiling points)
Gasoline is a derivative of one of the fraction of petroleum used in internal combustion engines.
Jet Fuel is also derived from Kerosene and Naphtha fractions of petroleum.
While, Steel is inorganic Alloy (mixture of metals) composed of mainly Iron, Carbon and other elements.
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Answer:
13440 J
Explanation:
c ≈ 4200 J / (kg * °C)
m = 80 g = 0,08 kg
= 10 °C
= 50 °C
The formula is: Q = c * m * ()
Calculating:
Q = 4200 * 0,08 * (50 - 10) = 13440 (J)
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Answer:
Average atomic mass = 19.9 amu
Explanation:
Isotopes can be defined as two or more forms of a chemical element that are made up of equal numbers of protons and electrons but different numbers of neutrons.
Generally, the isotopes of a chemical element have the same chemical properties because of their atomic number but different physical properties due to their atomic weight (mass number).
Given the following data;
Relative abundance of Z-19 = 55%
Relative abundance of Z-21 = 45%
Atomic mass of Z-19 = 19 amu
Atomic mass of Z-21 = 21 amu
To find the average atomic mass;
Average atomic mass = 19 * (55/100) + 21 * (45/100)
Average atomic mass = 19*0.55 + 21*0.45
Average atomic mass = 10.45 + 9.45
Average atomic mass = 19.9 amu
Therefore, the average atomic mass for element Z is 19.9 amu.