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2 years ago

A large sport utility vehicle has a mass of 2600 kg.

Chemistry

1 answer:

Pepsi [2]2 years ago

4 0

Mass of CO2 = 225.632g

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Value of an extensive property

FinnZ [79.3K]

Answer:

The value of an intensive property may vary with time and its position within the system. Examples of intensive properties include temperature, velocity, mass density, specific volume, and specific energy. An extensive property does not have a value at a point, and its value depends on the extent or size of the system.

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3 years ago

After 1911, most scientists accepted the theory that nucleus of an atom was very dense and very small and has a positive charge.

Yanka [14]

After 1911 most scientists accepted the theory that the nucleus of an atom was very dense and very small and has a positive charge.

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3 years ago

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Resources found in lithosphere: gold and iron etc

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2 years ago

9. According to an "alternative theory", H2O is

Jlenok [28]

Answer:(4) ----accepts a proton

Explanation:

H2O water can produce both hydrogen and hydroxide ions

H2O --> H+ + OH-

According to the Bronsted-Lowry theory, it can be a proton donor and a proton acceptor.this means that It can donate a hydrogen ion to become its conjugate base, or can accept a hydrogen ion to form its conjugate acid,

When , a water molecule, H2O accepts a proton it will act as a Brønsted-Lowry base especially when dissolved in a strong acidic medium. for eg

HCl + H2O(l) → H3O+(aq) + Cl−(aq)

Here, Hydrochloric acid is a strong acid and ionizes completely in water, since it is more acidic than water, the water will act as a base.

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2 years ago

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,

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

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

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

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

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3 years ago