How much energy (in Joules) is needed to increase the temperature of 25 grams of glass by 5.0C. Aluminum Iron Glass Copper Water

Chemistry

1 answer:

vitfil [10]3 years ago

4 0

Answer:

25 grams of glass by 5.oc aluminium iron glass copper water

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A chemical reaction occurs when 2 white solids are stirred together they turn into a slushy white solid and the glass jar feels

Anestetic [448]

Answer:

Absorbed, endothermic

Explanation:

Based on the heat of reaction, chemical reactions may be classified as endothermic or exothermic.

In an endothermic reaction, heat is absorbed from the surrounding. As heat is absorbed from the surrounding, the reaction vessel feels cool when touched.

On the other hand, in an exothermic reaction, heat is given out by the system to the surrounding and the reaction vessel feels hot when touched.

Hence the reaction is endothermic and energy is absorbed during the reaction.

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3 years ago

a friend is trying to describe a material to you. which of the following is not a description of a physical property of the subs

Masja [62]

The answer for it is C. when you mix it with water it fizzes

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3 years ago

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

Initial concentration of $\rm HA$ without any dissociation:

$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .