Answer:
Eight foods are responsible for 90 percent of all reactions: peanut, tree nut, milk, egg, fish, shellfish, soy, and wheat.
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Answer:
![[Cu^{2+}]=0.041 M](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%20M)
Explanation:
Hello!
In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:
![[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B1molCu%5E%7B2%2B%7D%7D%7B1molCu%28NH_3%29_4%5E%7B2%2B%7D%7D%20%3D0.041%20%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D)
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Answer:
what are the answer choices?
Explanation:
#48 might be (A) but in not for sure
Answer is: 56 gallons of
70% antifreeze and 84 gallons of 95% antifreeze.
ω₁ = 70% ÷ 100% = 0.7; 70% pure antifreeze.
ω₂ = 95% ÷ 100% = 0.95.
ω₃<span> = 85% ÷ 100% = 0.85.
V</span>₁ = ?; volume of 70% antifreeze.
V₂ = ?; volume of 95% antifreeze.<span>
V</span>₃ = V₁ + V₂<span>.
V</span>₃ = 140 gal.
V₁ = 140 gal - V₂<span>.
ω</span>₁ · V₁ + ω₂ ·V₂ = ω₃ · V₃.
0.70 · (140 gal -
V₂) + 0.95 · V₂ = 0.85 · 140 gal.
98 gal - 0.7V₂ + 0.95V₂ = 119 gal.
0.25V₂ = 21 gal.
V₂ = 21 gal ÷ 0.25.
V₂ = 84 gal.
V₁ = 140 gal - 84 gal.
V₁ = 56 gal.
