42.34 g of water could be warmed from 21.4°C to 43.4°C by the pellet dropped inside it
Heat loss by the pellet is equal to the Heat gained by the water.
….(1)
where, 
is the heat gained by water
is the heat loss by pellet
= mCΔT
where m = mass of water
C = specific heat capacity of water = 4.184 J/g-°C
ΔT = Increase in temperature
ΔT for water = 43.4 - 21.4 = 22°C
= m × 4.184 × 22 …. (2)
Now
= 
×ΔT
where = Heat capacity of pellet = 56J/°C
Δ T for pellet = 43.4 - 113 =- 69.6°C
= 56 × -69.6 = -3897.6 J
From equation (1) and (2)
-m× 4.184 × 22 =-3897.6
m= 42.34 g
Hence, 42.34 g of water could be warmed from 21.4 degrees Celsius to 43.4 degrees Celsius by the pellet dropped inside it.
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Answer:
Percent yield = 61.58%
Explanation:
C4H9OH + NaBr + H2SO4 --> C4H9Br + NaHSO4 + H2O
From the reaction;
1 mol of C4H9OH reacts with 1 mol of NaBr and 1 mol of H2SO4 to form 1 mol of C4H9Br
We have to obtain the limiting reagent. We do this by calculatig the mols of each reactant in the reaction;
Number of moles = Mass / Molar mass
C4H9OH; 15 / 74g/mol = 0.2027
NaBr; 22.4 / 102.89 = 0.2177
H2SO4; 32.7 / 98 = 0.3336
In this case, the limiting reagent is C4H9OH, it determines the amount of product formed.
From the stoichometry of the reaction; one mole of C4H9OH forms one mole of C4H9Br. This means 0.2027 mol of C4H9Br was formed.
Mass = Number of moles * Molar mass
Mass of C4H9Br formed = 0.2027 * 137
Mass = 27.77g
Percent yield = (Practical yield / theoretical yeield ) * 100%
Percent Yield = 17.1 / 27.77 * 100
Percent yield = 61.58%
The number of electrons in an atom is directly proportional to the number of protons, which is clearly represented by atomic number.
If you were to notate all of the electron configurations of the atoms in the 2nd row, you would clearly see an increase in the number of valence electrons from left to right.
First: Test the hypothesis
Second: Test or modify
Third: Ask a question
Fourth: Make observations
Fifth: Analyze the results and make a conclusion
The reaction is: Mg + 2HCl -----> MgCl2 + H2
<span>M = mol / L </span>
<span>mol = M x L = 4.5 mol/L x 6.3 L = 28.4 mol HCl </span>
<span>1 mol Mg reacts with 2 mol HCl </span>
<span>28.4 mol HCl x (1 mol Mg / 2 mol HCl) = 14.2 mol Mg </span>
<span>Molar mass of Mg is 24.3 g/mol. </span>
<span>14.2 mol Mg x (24.3 g / 1 mol) = 344 g Mg</span>
