Using accurate measurements, using pure chemicals and performing the reaction under the most ideal conditions is important to get a valuable percent yield.
<h3>How we calculate the percent yield?</h3>
Percent yield of any chemical reaction is define as the ratios of the actual yield to the theoretical yield of the product and multiply by the 100.
To get the high percent yield or actual yield of any reaction, we have to perform the reaction under ideal condition because if we not use the standard condition then we get the low rate of reaction. Reactants should be present in the pure form as impurity make unwanted products and reduce the productivity of main product and accurate amount of reactants also important for the spontaneous reaction.
Hence, options (a), (b) & (c) are correct.
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Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
The answer is homogeneous
<span> d. Reduced visibility
Acid rain doesn't reduce visibility, smog does
</span>
Answer:
F. 2NO + 02 —> 2NO
H. 4NH3 + 502 —> 4NO + 6H20
Explanation:
The law of conservation of mass states that matter can neither be created nor destroyed during a chemical reaction but can be convert from one form to another.
2NO + 02 —> 2NO
From the above, the total number of N on the left balance the total number on the right i.e 2 atoms of N on both side of the equation.
The total number of O on the left balance the total number on the right i.e 2 atoms of O on both side of the equation. This is certified by the law of conservation of mass.
4NH3 + 502 —> 4NO + 6H20
From the above, the total number of N on the left balance the total number on the right i.e 4 atoms of N on both side of the equation.
The total number of O on the left balance the total number on the right i.e 10 atoms of O on both side of the equation.
The total number of H on the left balance the total number on the right i.e 12 atoms of O on both side of the equation.
This is certified by the law of conservation of mass.
The rest equation did not conform to the law of conservation of mass as the atoms on the left side did not balance those on the right side
