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4 years ago

The last answer choice us D. Adding more of the chemical will hear the water but only up to a certain temperature

Chemistry

1 answer:

enot [183]4 years ago

4 0

D. Adding more of the chemical will heat the water but only up to a certain temperature

<h3>Further explanation</h3>

Chemical changes are changes in the process that produce new substances that are different from the initial/ forming substances. Chemical changes are usually known as chemical reactions. So there are products/reactions and reactants and there is a change in the molecule/particle

The chemical reaction is usually followed by the release of heat, gas, change in color, precipitate substance etc

The heat that occurs in the addition of chemicals is caused by the reaction with water. But there is a constant rise in temperature when water reaches <em>its maximum capacity to react with these chemicals.</em>

Evidenced by the addition of 15 g and 20 g of chemicals, the temperature changes are constant

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An object at a distance of 30 cm from a concave mirror gets its image at the same point. The focal length of the mirror is

Nimfa-mama [501]

The focal length of the mirror is 15 cm

From the question given above, the following data were obtained:

Object distance (u) = 30 cm

Image distance (v) = 30 cm

<h3>Focal length (f) =? </h3>

The focal length of the mirror can be obtained as follow:

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\\\\frac{1}{f} = \frac{1}{30} + \frac{1}{30}\\\\\frac{1}{f} = \frac{2}{30} \\\\\frac{1}{f} = \frac{1}{15} \\\\$

<h3>Invert </h3><h3>f = 15 cm</h3>

Therefore, the focal length of the mirror is 15 cm

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3 years ago

A solution is prepared by dissolving 0.26 mol of hydrazoic acid and 0.26 mol of sodium azide in water sufficient to yield 1.00 L

Pavel [41]

Answer:

The pH does not decrease drastically because the HCl reacts with the <u>sodium azide (NaN₃)</u> present in the buffer solution.

Explanation:

The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:

HN₃(aq) + H₂O(l) ⇄ N₃⁻(aq) + H₃O⁺(aq)

The pH of this solution is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60$

When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:

H₃O⁺(aq) + N₃⁻(aq) ⇄ HN₃(aq) + H₂O(l)

The number of moles of NaN₃ after the reaction with HCl is:

$\eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles$

Now, the number of moles of HN₃ is:

$\eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles$

Then, the pH of the buffer solution after the addition of HCl is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43$

The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.

Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

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3 years ago

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Answer:

Your answer is B, Electrochemistry!

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