It depends on the substance,but for most substance it is in the gaseous state.
Answer:
H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-
Explanation:
Reduction half reaction
2H_2O + CrO_4^2- + 3e -> CrO_2^- + 4OH^-
Oxidation half reaction
2OH^- + SO_3^2- -> SO_4^2- + H_2O + 2e
Balanced overall equation
H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-
Explanation:
Earlier, we located the valence electrons for elements Z < 20 by drawing modified Bohr structures. We can obtain these values quicker by referring to the roman numeral numbers above each family on the periodic table. The total number of valence electrons for an atom can vary between one and eight. If an element is located on the left side of the table (metal) and has less than three valence electrons, it will lose its valence in order to become stable and achieve an octet. In contrast, elements on the right side of the table (nonmetals) will gain up to eight electrons to achieve octet status.
Answer:
226.8 mg of mupirocin powder are required
Explanation:
Given that;
weight of standard pack = 22 g
mupirocin by weight = 2%
so weight of mupirocin = 2% × 22 = 2/100 × 22 = 0.44 g
so by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment
mg of mupirocin powder are required = ?, lets rep this with x
Total weight of ointment = 22 + x g
Amount of mupirocin = 0.44 + x g
percentage of mupirocin in ointment is 3?
so
3/100 = 0.44 + x g / 22 + x g
3( 22 + x g ) = 100( 0.44 + x g )
66 + 3x g = 44 + 100x g
66 - 44 = 100x g - 3x g
97 x g = 22
x g = 22 / 97
x g = 0.2268 g
we know that; 1 gram = 1000 Milligram
so 0.2268 g = x mg
x mg = 0.2268 × 1000
x mg = 226.8 mg
Therefore, 226.8 mg of mupirocin powder are required
