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Brrunno [24]
3 years ago
14

A reaction occurs when solid X is placed into solution Y. As a result, the temperature of the new solution increases by 3°C.

Chemistry
1 answer:
zzz [600]3 years ago
6 0
It increases the number of molecular collisions.
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I have a 4 amino acid peptide of unknown sequence with a curious titration profile around pH = 6. If Iknow that the peptide N-te
netineya [11]

Answer:

c. Histidine

Explanation:

Histidine is a compound that is normally used for the generation of protein. Three amino acids commonly have basic side chain when the pH is neutral. The conjugate acid in histidine has a pKa of approximately 6. Based on the description of the experimental analysis provided in the statement, the right option is option c.

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3 years ago
4. The picture shows a full moon.
Anuta_ua [19.1K]

Answer:

A full moon occurs when the Moon appears as a complete circle in the sky.

Explanation:

- We see it as a full orb because the whole of the side of the Moon facing the Earth is lit up by the Sun's rays.

- The moon shows no visible light of its own, so we only see the parts of the moon that are lit up by other objects.

6 0
2 years ago
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Sodium bromate chemical formula
Wewaii [24]
NaBrO3 is the chemical formula for Sodium Bromate.
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3 years ago
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What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb
Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

(\text{CH}_3)_3\text{N} in this question acts as a weak base. As seen in the equation in the question, (\text{CH}_3)_3\text{N} produces \text{OH}^{-} rather than \text{H}^{+} when it dissolves in water. The concentration of \text{OH}^{-} will likely be more useful than that of \text{H}^{+} for the calculations here.

Finding the value of [\text{OH}^{-}] from pH:

Assume that \text{pK}_w = 14,

\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}.

[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}.

Solve for [(\text{CH}_3)_3\text{N}]_\text{initial}:

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}

Note that water isn't part of this expression.

The value of Kb is quite small. The change in (\text{CH}_3)_3\text{N} is nearly negligible once it dissolves. In other words,

[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}.

Also, for each mole of \text{OH}^{-} produced, one mole of (\text{CH}_3)_3\text{NH}^{+} was also produced. The solution started with a small amount of either species. As a result,

[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}.

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3},

[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b},

[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}.

8 0
3 years ago
How many liters of dinitrogen tetoxide are formed from 2.5 L of nitrogen?
Black_prince [1.1K]

Answer:

2.5 L

Explanation:

The stoichiometric ratio of N2 to N2O4 is 1:1

7 0
2 years ago
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