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2 years ago

Explain why neon is monatomic but chlorine is diatomic.

Chemistry

2 answers:

Elza [17]2 years ago

5 0

Chlorine is a halogen and all halogens and oxygen, nitrogen and hydrogen are diatomics

kirza4 [7]2 years ago

4 0

Explanation:

Neon is in group 8/0, which means that it has a full outer shell. Atoms react to get a full outer shell but as neon already has a full outer shell, it doesn't react and stays monatomic.

However, chlorine is in group 7, so it needs 1 more electron to complete its outer shell, two chlorine atom share one electron each with each other so they have full outer shells. This is why chlorine is diatomic.

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Hell me with this ASAP please

Hatshy [7]

Answer:

Losing 2 valence electrons

Gaining 2 electrons

3 0

3 years ago

Use the definition of molarity to calculate the concentration of 12.34 g of CaSO4 completely dissolved in water, with a solution

Ede4ka [16]

Answer:

[CaSO₄] = 36.26×10⁻² mol/L

Explanation:

Molarity (M) → mol/L → moles of solute in 1L of solution

Let's convert the volume from mL to L

250 mL . 1L/1000 mL = 0.250L

We need to determine the moles of solute. (mass / molar mass)

12.34 g / 136.13 g/mol = 0.0906 mol

M → 0.0906 mol / 0.250L = 36.26×10⁻² mol/L

8 0

3 years ago

If 20 grams of N2 react completely with H2 , how many moles of NH3 are produced? Show Work

Maslowich

This is the balanced eq
N2 + 3H2 -> 2NH3
first you need to find mole of N2 by using
mol = mass ÷ molar mass.
mol N2= 20g ÷ (14.01×2)g/mol
=0.7138mol
then look at the coefficient between H2 and NH3.
it is N2:NH3
1:2
0.7138:0.7138×2
0.7138:1.4276 moles
moles of NH3 = 1.4276 moles

5 0

3 years ago

A smaple of 1cm cuvette gives an absorbance reading of 0.558. if they absorptivity for this sample is 15000l/(mol.cm), what is t

Mazyrski [523]

To solve this problem, we use Beer's Law: A= ε.l.c
A is the absorbance- 0,558
ε is the molar absorptivity- is 15000 L⋅mol-1cm-1
l is the length of the cuvette- 1 cm
c is the molar concentration

Applying the formula,
0,558= 15000 x 1 x c
0,558/15000= c
c= 3.72×10⁻⁵ mol⋅L⁻¹

3 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 acid dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the base 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