Unbalanced it should be 2Zn+2Hcl=2ZnCl2+H2
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
According to the question, the determined melting point of the compound is 112.5-113.0oC. When the solidified compound was retried, the melting point was found to be 133.6-154.5oC. This greater range higher than 112°C is caused by reusing samples leads to errors.
A pure sample is known by its sharp melting point. A pure sample does not melt over a large range. We can see this in the predetermined melting points of the pure sample(112.5-113.0oC).
However, reusing a sample introduces errors because the pure sample may become contaminated leading to a larger and higher range of melting point (133.6-154.5oC) which is far above 112°C.
Learn more: brainly.com/question/5325004
<u>Answer:</u> The entropy change of the ethyl acetate is 133. J/K
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of ethyl acetate = 398 g
Molar mass of ethyl acetate = 88.11 g/mol
Putting values in above equation, we get:
To calculate the entropy change for different phase at same temperature, we use the equation:
where,
= Entropy change = ?
n = moles of ethyl acetate = 4.52 moles
= enthalpy of fusion = 10.5 kJ/mol = 10500 J/mol (Conversion factor: 1 kJ = 1000 J)
T = temperature of the system = ![84.0^oC=[84+273]K=357K](https://tex.z-dn.net/?f=84.0%5EoC%3D%5B84%2B273%5DK%3D357K)
Putting values in above equation, we get:
Hence, the entropy change of the ethyl acetate is 133. J/K
The main <span>hazard </span>is the Radiation and the Gamma rays that are dispersed
