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3 years ago

Plz answer all 3, I will give 5 stars and brainiest!!! :) THANKS

Chemistry

1 answer:

Cerrena [4.2K]3 years ago

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Answer:

B. Positive charge is caused by lack of electrons. A positive ion is formed by the loss of negatively charged electrons. Although the number of protons does not change in the ion, there is an excess number of protons over electrons which produces the positive charge. All electrons in the outer energy level are lost.

C. If there are more electrons than protons, then the element is a negative ion. The amount of neutrons does not play a factor into making a difference between an atom or an ion.

Explanation:

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How does your body obtain energy from the carbohydrates you eat?

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Our bodies digest the food we eat by mixing it with fluids (acids and enzymes) in the stomach. When the stomach digests food, the carbohydrate (sugars and starches) in the food breaks down into another type of sugar, called glucose.

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3 years ago

Briefly describe any example of an endothermic reaction and also mention the word equation

emmainna [20.7K]

Explanation:

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Endothermic reactions are chemical reactions in which the reactants absorb heat energy from the surroundings to form products. These reactions lower the temperature of their surrounding area, thereby creating a cooling effect. Physical processes can be endothermic as well – Ice cubes absorb heat energy from their surroundings and melt to form liquid water (no chemical bonds are broken or formed).

When a chemical bond is broken, it is usually accompanied by a release of energy. Similarly, the formation of chemical bonds requires an input of energy. The energy supplied/released can be of various forms (such as heat, light, and electricity). Endothermic reactions generally involve the formation of chemical bonds through the absorption of heat from the surroundings. On the other hand, exothermic reactions involve the release of heat energy generated from bond-breakage.

Endothermic Reaction Examples

Ammonium nitrate (NH4NO3), an important component in instant cold packs, dissociates into the ammonium cation (NH4+) and the nitrate anion (NO3–) when dissolved in water

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3 years ago

Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the <em>acid</em> dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :