Answer: % error of observation is 4.77%
Explanation:
Given:
Observation value = 415nm
theoretical value= 435.8nm
Percent error of observation = theoretical value- observation value/ theoretical value x 100 %
= 435.8-415/435.8= 0.04772 x 100 = 4.77%
therefore % error of observation is 4.77%
this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.
c1v1 = c2v2
where c1 is concentration and v1 is volume of the concentrated solution
and c2 is concentration and v2 is volume of the diluted solution to be prepared
50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL
substituting these values in the formula
1.50 M x 50.0 mL = C x 250 mL
C = 0.300 M
concentration of the final solution is A) 0.300 M
Answer:
9.9652g of water
Explanation:
The establishment of the liquid-vapor equilibrium occurs when the vapour of water is equal to vapour pressurem 26.7 mmHg. Using gas law it is possible to know how many moles exert that pressure, thus:
n = PV / RT
Where P is pressure 26,7 mmHg (0.0351atm), V is volume (1.350L), R is gas constant (0.082 atmL/molK) and T is temperature (27°C + 273,15 = 300.15K)
Replacing:
n = 0.0351atm×1.350L / 0.082atmL/molK×300.15K
n = 1.93x10⁻³ moles of water are in gaseous phase. In grams:
1.93x10⁻³ moles × (18.01g / 1mol) = <u><em>0.0348g of water</em></u>
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As the initial mass of water was 10g, the mass of water that remains in liquid phase is:
10g - 0.0348g = <em>9.9652g of water</em>
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I hope it helps!
Answer:
The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.
Explanation:
Where:
Q = heat absorbed or heat lost
c = specific heat of substance
m = Mass of the substance
ΔT = change in temperature of the substance
We have mass of copper = m = 25.3 g
Specific heat of copper = c = 0.385 J/g°C
ΔT = 39°C - 22°C = 17°C
Heat absorbed by the copper :
The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.
Answer:
glucose
Explanation:
There are two types of respiration:
1. Aerobic respiration
2. Anaerobic respiration
Aerobic respiration:
It is the breakdown of glucose molecule in the presence of oxygen to yield large amount of energy. Water and carbon dioxide are also produced as a byproduct.
Glucose + oxygen → carbon dioxide + water + 38ATP
Anaerobic Respiration:
It is the breakdown of glucose molecule in the absence of oxygen and produce small amount of energy. Alcohol or lactic acid and carbon dioxide are also produced as byproducts.
Glucose→ lactic acid/alcohol + 2ATP + carbon dioxide
This process use respiratory electron transport chain as electron acceptor instead of oxygen. It is mostly occur in prokaryotes. Its main advantage is that it produce energy (ATP) very quickly as compared to aerobic respiration.
Steps involve in anaerobic respiration are:
Glycolysis:
Glycolysis is the first step of both aerobic and anaerobic respiration. It involve the breakdown of one glucose molecule into pyruvate and 2ATP.
Fermentation:
The second step of anaerobic respiration is fermentation. It involve the fermentation of pyruvate into lactic acid or alcohol depending upon the organism in which it is taking place. There is no ATP produced in this step, however carbon dioxide is released.
