Which one is the odd one and why?

Look at the image shown

klemol [59]

Answer:

b) Linear molecule with two domains

Explanation:

The image represents a linear molecule with two domains.

The compound shown here is carbon dioxide in which a central carbon atom is surrounded by two oxygen atoms.

This bond geometry is of the type AX₂
In this linear molecule the bond angle is a perfect 180°
The lone pairs are perfectly balanced out and will not cause distortion of the central carbon.

5 0

3 years ago

A certain kind of light has a wavelength of 4.6 micrometers what is the frequency of this light and gigahertz? Use c= 2.998 × 10

Leno4ka [110]

Answer:

6.52×10⁴ GHz

Explanation:

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

Wavelength (λ) = 4.6 μm

Velocity of light (v) = 2.998×10⁸ m/s

Frequency (f) =?

Next we shall convert 4.6 μm to metre (m). This can be obtained as follow:

1 μm = 1×10¯⁶ m

Therefore,

4.6 μm = 4.6 μm × 1×10¯⁶ m / 1 μm

4.6 μm = 4.6×10¯⁶ m

Next, we shall determine frequency of the light. This can be obtained as follow:

Wavelength (λ) = 4.6×10¯⁶ m

Velocity of light (v) = 2.998×10⁸ m/s

Frequency (f) =?

v = λf

2.998×10⁸ = 4.6×10¯⁶ × f

Divide both side by 4.6×10¯⁶

f = 2.998×10⁸ / 4.6×10¯⁶

f = 6.52×10¹³ Hz

Finally, we shall convert 6.52×10¹³ Hz to gigahertz. This can be obtained as follow:

1 Hz = 1×10¯⁹ GHz

Therefore,

6.52×10¹³ Hz = 6.52×10¹³ Hz × 1×10¯⁹ GHz / 1Hz

6.52×10¹³ Hz = 6.52×10⁴ GHz

Thus, the frequency of the light is 6.52×10⁴ GHz

5 0

3 years ago

In the reaction, 2Na (s) + 2H2O (l) → 2NaOH (aq) + H2 (g), which of the reactants or products is in the gas phase?

lys-0071 [83]

Clearly H2 is in gaseous state as could be seen from (g) written with it which tells state of the product

5 0

3 years ago

The process of converting a liquid into a solid is called

xeze [42]

The answer to this question is C. Freezing, hope this helps :)

4 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