Answer:
4.00 is the pH of the mixture
Explanation:
The ethyl amine reacts with HNO3 as follows:
C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻
To solve this question we need to find the moles of ethyl amine and the moles of HNO3:
<em>Moles C2H5NH2:</em>
0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine
<em>Moles HNO3:</em>
0.201L * (0.025mol/L) = 0.005025 moles HNO3
That means HNO3 is in excess. The moles in excess are:
0.005025 moles HNO3 - 0.00500 moles ethyl amine =
2.5x10⁻⁵ moles HNO₃
In 50 + 201mL = 251mL = 0.251L:
2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]
As pH = -log [H+]
pH = -log 9.96x10⁻⁵M
pH = 4.00 is the pH of the mixture
Answer:
3.01 ·10↑22
Explanation:
First you want to convert the grams of Glucose to moles of Glucose.
Next find the formula units of glucose.
.008326Moles of Glucose · 6.022 · 10↑23Forumula Units*Moles↑-1 =
5.01 ·10↑21 Formula Units of Glucose
Now multiply the formula units of glucose by the amount of each element in the molecule.
So for Carbon:
6carbon · 5.01 · 10↑21 = 3.01 · 10↑22
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Answer:easy
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Answer:
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