Anatomical evidence of evolution focuses on similarities and differences in the body structures of different species. ... Similarities in anatomical structures of different species signify that the two species have a relatively recent common ancestor.
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Answer: The molarity of KBr in the final solution is 1.42M
Explanation:
We can calculate the molarity of the KBr in the final solution by dividing the total number of moles of KBr in the solution by the final volume of the solution.
We will first calculate the number of moles of KBr in the individual sample before mixing together
In the first sample:
Volume (V) = 35.0 mL
Concentration (C) = 1.00M
Number of moles (n) = C × V
n = (35.0mL × 1.00M)
n= 35.0mmol
For the second sample
V = 60.0 mL
C = 0.600 M
n = (60.0 mL × 0.600 M)
n = 36.0mmol
Therefore, we have (35.0 + 36.0)mmol in the final solution
Number of moles of KBr in final solution (n) = 71.0mmol
Now, to get the molarity of the final solution , we will divide the total number of moles of KBr in the solution by the final volume of the solution after evaporation.
Therefore,
Final volume of solution (V) = 50mL
Number of moles of KBr in final solution (n) = 71.0mmol
From
C = n / V
C= 71.0mmol/50mL
C = 1.42M
Therefore, the molarity of KBr in the final solution is 1.42M
Answer:
Q = 1461.6 J
Explanation:
Given data:
Mass of ice = 36 g
Initial temperature = -20°C
Final temperature = 0°C
Amount of heat absorbed = ?
Solution:
specific heat capacity of ice is 2.03 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
ΔT = 0°C - (-20°C)
ΔT = 20°C
Q = 36 g ×2.03 j/g.°C×20°C
Q = 1461.6 J
To convert a mass of a substance to units of moles, we would need the molar mass of the substance since is it describes the mass of one mole of that substance. For CuCl2, the molar mass is 134.45 g/mol. For Al, the atomic mass is 26.98 g/mol.
2.50 g CuCl2 ( 1 mol / 134.45) = 0.019 mol CuCl2
0.25 g Al ( 1 mol / 26.98 g ) = 0.0093 mol Al