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3 years ago

Write a generic Lewis structure for the halogens.Draw the Lewis dot structure for X

Chemistry

2 answers:

AysviL [449]3 years ago

5 0

Answer : A Lewis structure is a graphical representation of the distribution of electrons around atoms. Mainly it illustrates the lone pairs of electrons present in molecules. They are also known as Lewis dot diagrams and mainly used for simple way to represent the configuration of atoms within a molecule.

The halogens are the five non-metallic elements found in group 7A of the modern periodic table. The term "halogen" stands for "salt-former" and compounds which contains halogens are called as "salts" as they produce salts when it reacts with metals.

All halogens are supposed to have 7 electrons in their outermost shells.

When a diatomic structure is formed by two elements of halogens, one electron from each halogen combines to form bonds between them.

The diagram is attached here in the answer.

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Alexxx [7]3 years ago

4 0

Halogens exist is diatomic molecules to have stable octet structure

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An excess of mg(s) is added to 100.ml of 0.400 m hcl. at 0c and 1 atm pressure, what volume of h 2 (g) can be obtained?

ra1l [238]

The balanced equation for the reaction between Mg and HCl is as follows
Mg + 2HCl --> MgCl₂ + H₂
stoichiometry of HCl to H₂ is 2:1

number of HCl moles reacted - 0.400 mol/L x 0.100 L = 0.04 mol of HCl
since Mg is in excess HCl is the limiting reactant
number of H₂ moles formed - 0.04/2 = 0.02 mol of H₂

we can use ideal gas law equation to find the volume of H₂
PV = nRT
where
P - pressure - 1 atm x 101 325 Pa/atm = 101 325 Pa
V - volume
n - number of moles - 0.02 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 0 °C + 273 = 273 K
substituting these values in the equation

101 325 Pa x V = 0.02 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 448 x 10⁻⁶ m³
V = 448 mL
therefore answer is
c. 448 mL

5 0

3 years ago

When a solvent can dissolve no more of a solid solute at a specific temperature, we say the solution is ________.

asambeis [7]

The answer is B. Saturated.

7 0

3 years ago

Read 2 more answers

How many molecules are in the substance formula of 2C6H1206 (use coefficients)

FrozenT [24]

Answer: The number of formula units present in 2 moles of $C_6H_{12}O_6$ are $1.2044\times 10^{24}$

Explanation:

Formula units are defined as the number of molecules or atoms present in 1 mole of a compound or element respectively.

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of formula units

Here, 2 represents the number of moles of $C_6H_{12}O_6$

We are given:

Moles of $C_6H_{12}O_6$ (glucose) = 2 moles

Number of formula units of $C_6H_{12}O_6=(2\times 6.022\times 10^{23})=1.2044\times 10^{24}$

Hence, the number of formula units present in 2 moles of $C_6H_{12}O_6$ are $1.2044\times 10^{24}$

6 0

3 years ago

PH is a logarithmic scale used to indicate the hydrogen ion concentration, [H+], of a solution: pH=−log[H+] Due to the autoioniz

ruslelena [56]

Answer:

A) pH = 2.42
B) pH = 12.00

Explanation:

The dissolution of HCl is HCl → H⁺ + Cl⁻

To solve part A) we need to calculate the concentration of H⁺, to do that we need the moles of H⁺ and the volume.

The problem gives us V=2.5 L, and the moles can be calculated using the molecular weight of HCl, 36.46 g/mol:

$0.35g_{HCl}*\frac{1mol_{HCl}}{36.46g_{HCl}} *\frac{1molH^{+}}{1mol HCl}$ = 9.60*10⁻³ mol H⁺

So the concentration of H⁺ is

[H⁺] = 9.60*10⁻³ mol / 2.5 L = 3.84 * 10⁻³ M

pH = -log [H⁺] = -log (3.84 * 10⁻³) = 2.42

The dissolution of NaOH is NaOH → Na⁺ + OH⁻
Now we calculate [OH⁻], we already know that V = 2.0 L, and a similar process is used to calculate the moles of OH⁻, keeping in mind the molecular weight of NaOH, 40 g/mol:

$0.80g_{NaOH}*\frac{1mol_{NaOH}}{40g_{NaOH}} *\frac{1molOH^{-}}{1mol NaOH}$ = 0.02 mol OH⁻

[OH⁻] = 0.02 mol / 2.0 L = 0.01

pOH = -log [OH⁻] = -log (0.01) = 2.00

With the pOH, we can calculate the pH:

pH + pOH = 14.00

pH + 2.00 = 14.00

pH = 12.00

5 0

3 years ago

How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that

tia_tia [17]

Answer:

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,

= 2.50 L × 0.800 mol/L

= 2 mol

The given pH of a buffer is 8.53

pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,

= -log (1.8 ×10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73= 5.37

[NH₄⁺ = 5.37 × 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,

mass of NH₄Cl = 10.74 mol × 53.5 g/mol

= 574.59 grams.

8 0

3 years ago