Answer: 0.87 g of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a 5.5% w/w gel.
Explanation:
According to the dilution law,
[tex]C_1w_1+C_2w_2=C_3w_3[/tex]
where,
[tex]C_1[/tex] = concentration of ist progesterone gel = 8%w/w
[tex]w_1[/tex] = weight of ist progesterone gel = x g
[tex]C_2[/tex] = concentration of another progesterone gel = 4% w/w
[tex]w_2[/tex] = weight of another progesterone gel = 1.45 g
[tex]C_3[/tex] = concentration of resulting progesterone gel = 5.5%w/w
[tex]w_3[/tex] = weight of resulting progesterone gel = (x+1.45) g
[tex]8\times x+4\times 1.45=5.5\times (x+1.45)[/tex]
[tex]x=0.87[/tex]
Thus 0.87 g of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a 5.5% w/w gel.
0.87 grams of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a 2.32 g of 5.5% w/w gel
Let x represent the number of grams of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a (x + 1.45)g of 5.5% w/w gel
Hence:
(x * 8%) + (1.45 * 4%) = (x + 1.45) * 5.5%
0.08x + 0.058 = 0.055x + 0.07975
0.025x = 0.02175
x = 0.87 g
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