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pshichka [43]
3 years ago
13

In an ionic bond:

Chemistry
2 answers:
Aleks04 [339]3 years ago
6 0
A. Both atoms share their electrons
Fynjy0 [20]3 years ago
4 0

In an ionic bond :

=》B. one atom accepts electrons from another.

in this bond an atom ( <em><u>metallic</u></em> ) loses its electrons and another atom ( <em><u>non- metallic</u></em> ) accepts the electrons, and since there isn't the equal positive and negative charges in the atoms, they forms <em><u>cations</u></em> ( +ve charge ) and <em><u>anions </u></em>( -ve charge )

and get stacked or <em><u>attracted</u></em> to each other by strong <em><u>electrostatic force</u></em>.

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What is the threshold frequency ν0 of cesium?
Montano1993 [528]
 I think this is what you're after:

 Cs(g) → Cs^+ + e⁻ ΔHIP = 375.7 kJ mol^-1 [1] 


Convert to J and divide by the Avogadro Const to give E in J per photon 


E = 375700/6.022×10^23 = 6.239×10^-19 J 


Plank relationship E = h×ν E in J ν = frequency (Hz s-1) 


Planck constant h = 6.626×10^-34 J s 


6.239×10^-19 = (6.626×10^-34)ν 


ν = 9.42×10^14 s^-1 (Hz) 


IP are usually given in ev Cs 3.894 eV 


<span>E = 3.894×1.60×10^-19 = 6.230×10^-19 J per photon </span>

4 0
3 years ago
A container was found in the home of the victim that contained 120 g of ethylene glycol in 550 g of liquid. How many drinks, eac
Andrei [34K]

Answer:

0.432 drinks are toxic

Explanation:

The toxic dose of ethylene glycol is 0.1 mL per kg body weight (mL/kg). In grams (Density ethylene glycol = 1.11g/mL):

1.11g/mL * (0.1mL / kg) =  0.111g/kg

If the victim weighs 85kg, its letal dose is:

85kg * (0.111g/kg) = 9.435g of ethylene glycol

Using the concentration of ethylene glycol in the liquid:

9.435g of ethylene glycol * (550g liquid / 120g ethylene glycol) = 43.2g of liquid are toxic.

The drinks are:

43.2g of liquid * (1 drink / 100 g) =

<h3>0.432 drinks are toxic</h3>
5 0
3 years ago
(a) ions in a certain volume of .20M NaCl (aq) are represented in the box above on the left. in the box above on the right, draw
mamaluj [8]

Answer:

  • The first picture attached is the diagram that accompanies the question.

  • The<u> second picture attached</u> is the diagram with the answer.

Explanation:

In the box on the left there are 8  Cl⁻ ions and 8 Na⁺ ions.

The dissociaton equation for NaCl(aq) is:

  • NaCl (aq) → Na⁺ (aq) + Cl⁻(aq)

The dissociation equation for CaCl₂ (aq)  is:

  • CaCl₂ (aq) → Ca²⁺ (aq) + 2Cl⁻(aq)

A 0.10MCaCl₂ (aq) solution will have half the number of CaCl₂ units as the number of NaCl units in a 0.20M NaCl (aq) solution.

Thus, while the 0.20M NaCl (aq) solution yields 8 ions of Na⁺ and 8 ions of Cl⁻, the 0.10MCaCl₂ (aq) solution will yield 4 ions of Ca²⁺ (half because the concentration if half)  and 8 ions of Cl⁻ (first take half and then multiply by 2 because the dissociation reaction).

Thus, your drawing must show 4 dots representing Ca²⁺ ions and 8 dots representing Cl⁻ ions in the box on the right.

4 0
3 years ago
A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.692 kg of water
S_A_V [24]

Answer:

494.49 g of NaCl.

Explanation:

Data obtained from the question include the following:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mass of NaCl =.?

Next, we shall determine the number of mole of NaCl in the solution.

Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as

Molality = mole of solute /Kg of solvent

With the above formula, we can obtain the number of mole NaCl in the solution as follow:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mole of NaCl =..?

Molality = mole of solute /Kg of solvent

3.140 = mole of NaCl /2.692

Cross multiply

Mole of NaCl = 3.140 x 2.692

Mole of NaCl = 8.45288 moles

Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:

Mole of NaCl = 8.45288 moles

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =.?

Mole = mass /Molar mass

8.45288 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 8.45288 × 58.5

Mass of NaCl = 494.49 g.

Therefore, 494.49 g of NaCl are present in the solution.

8 0
3 years ago
Calculate the mass of 3.5 mol C6H6
Lilit [14]
(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6
4 0
3 years ago
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