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3 years ago

Why do electrons flow through a wire?

Chemistry

1 answer:

I am Lyosha [343]3 years ago

4 0

Answer:

When electrons flow through a conductor such as a wire, it is called, "Electricity".

Explanation:

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Indicate whether the following represents a Chemical or Physical change: Milk sours

marusya05 [52]

Answer:

Chemical Change

Explanation:

Physical change normally mean that the change can revert back to its orginal state, which in this case that is not possible therfore it is a chemical change.

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3 years ago

A solid cylinder having a diameter of 1.50 cm and a height of 5.15 cm has a mass of 95.56 g. Show the equations needed to calcul

Georgia [21]

Answer:

you can solve the rest of the equation. I only reduced it to that much to show you how to derive it

4 0

2 years ago

Complexes containing metals with d10 electron configurations are typically colorless because ________. Complexes containing meta

algol [13]

Answer:

there is no d electron that can be promoted via the absorption of visible light

Explanation:

One of the properties of transition elements is the possession of incompletely filled d orbitals. This property accounts for their unique colours.

The colours of transition metal compounds stem from d-d transition of electrons due to the presence of vacant d orbitals of appropriate energy to which electrons could be promoted.

For elements whose atoms have a d10 configuration, such vacant orbitals does not exist hence their compounds are not colored.

Sometimes, the colour of transition metal compounds stem from ligand to metal charge transfer(LMCT) for instance in KMnO4.

8 0

2 years ago

snow_lady [41]

Plutons are large chambers of magma under grown

pegmatites generally form in pluton so it cools slow enough to make the crystals big enough to be classified as pegmatite and not just granite

7 0

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 $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

6 0

3 years ago