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Stella [2.4K]
3 years ago
7

Using the periodic table, determine the ion charges of the following families of elements if valence electrons were removed or a

dded to reach the octet. Group I = Group VI = Group III =
Chemistry
2 answers:
kogti [31]3 years ago
6 0

Group I=1+

Group VI=-2

Group III=3+

igomit [66]3 years ago
4 0

<u>Answer:</u> The ion charge formed by Group I will be +1, by Group VI will be -2 and by Group III will be +3.

<u>Explanation:</u>

Ions are formed when electrons are lost or gained by an atom. When an atom looses electrons, it forms a positive ion known as cations and when an atom gains electrons, it forms a negative ion known as anions.

The electronic configuration of the elements present in

  • <u>Group I:</u> ns^1 where n is the number of period

This configuration will loose 1 electron and thus will form an ion of +1.

  • <u>Group VI:</u> ns^2np^4 where n is the number of period

This configuration will gain 2 electrons and thus will form an ion of -2.

  • <u>Group III:</u> ns^2np^1 where n is the number of period

This configuration will loose 3 electrons and thus will form an ion of +3.

Hence, the ion charge formed by Group I will be +1, by Group VI will be -2 and by Group III will be +3.

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In the reaction, A → Products, the rate constant is 3.6 × 10−4 s−1. If the initial concentration of A is 0.548 M, what will be t
Arada [10]

Answer:

        \large\boxed{\large\boxed{0.529M}}

Explanation:

Since the <em>rate constant</em> has units of <em>s⁻¹</em>, you can tell that the order of the reaction is 1.

Hence, the rate law is:

       r=d[A]/dt=-k[A]

Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:

      [A]=[A]_0e^{-kt}

You know [A]₀, k, and t, thus you can calculate [A].

       [A]=0.548M\times e^{-3.6\cdot 10^{-4}/s\times99.2s}

       [A]=0.529M

7 0
3 years ago
Argon has a pressure of 34.6 atm. It is transferred to a new tank with a volume of 456 L and pressure of 2.94 atm. What was the
NemiM [27]

Answer:

38.75 L

Explanation:

From the question,

Applying Boyles Law,

PV = P'V'....................... Equation 1

Where P = Original pressure of the Argon gas, V = Original Volume of Argon gas, P' = Final pressure of Argon gas, V' =  Final Volume of Argon gas.

make V the subject of the equation

V = P'V'/P.................... Equation 2

Given: P = 34.6 atm, V' = 456 L, P' = 2.94 atm.

Substitute these values into equation 2

V = (456×2.94)/34.6

V = 38.75 L

3 0
3 years ago
A 0.89% (w/v) sodium chloride solution is referred to as physiological saline solution because it has the same concentration of
maks197457 [2]
1) 0.89% m/v = 0.89 grams of NaCl / 100 ml of solution

=> 8.9 grams of NaCl in 1000 ml of solution = 8.9 grams of NaCl in 1 liter of solution

2) Molarity = M = number of moles of solute / liters of solution

=> calculate the number of moles of 8.9 grams of NaCl

3) molar mass of NaCl = 23.0 g /mol + 35.5 g/mol = 58.5 g / mol

4) number of moles of NaCl = mass / molar mass = 8.9 g / 58.5 g / mol = 0.152 mol

5) M = 0.152 mol NaCl / 1 liter solution = 0.152 M

Answer: 0.152 M
4 0
3 years ago
Please help with number 24
vladimir1956 [14]
Layla it is A. that's the only one in standard form.
3 0
3 years ago
Read 2 more answers
At 25 ∘C the reaction CaCrO4(s)←→Ca2+(aq)+CrO2−4(aq) has an equilibrium constant Kc=7.1×10−4. What is the equilibrium concentrat
Nitella [24]

Answer:

2.67 × 10⁻²

Explanation:

Equation for the reaction is expressed as:

CaCrO₄(s)    ⇄      Ca₂⁺(aq)         +        CrO₂⁻⁴(aq)

Given that:

Kc=7.1×10⁻⁴

Kc= [Ca^{2+}][CrO^{2-}_4]

Kc= [x][x]

Kc= [x²]

7.1×10⁻⁴ =  [x²]

x = \sqrt{7.1*10^{-4}}

x = 0.0267

x = 2.67*10^{-2}

6 0
3 years ago
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