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marishachu [46]
3 years ago
15

What are valence electrons (be specific)?

Chemistry
2 answers:
Helen [10]3 years ago
5 0
In chemistry, a valence electron is an outer shell electron that is associated with an atom, and that can participate in the formation of a chemical bond if the outer shell is not closed; in a single covalent bond, both atoms in the bond contribute one valence electron in order to form a shared pair.
coldgirl [10]3 years ago
5 0
Valence electrons are the electrons present in the outermost shell of an atom. You can easily determine the number of valence electrons an atom can have by looking at its Group in the periodic table. ... Atoms in Groups 13 and 18 have 3 and 8 valence electrons, respectively.
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Why is it, do you think, that so many of the geniuses throughout history were poor
Alexxandr [17]

<em>Hi </em>

<em>I'm</em><em> </em><em>Abraar </em><em>Shaikh</em>

here is your explanation

Explanation:

People with high IQ might have bad grades. It is not necessary that high IQ requires higher grades. Mind of a person with high IQ works at a faster rate so they tend to ignore normal problems almost everywhere even in the examinations.

8 0
3 years ago
Tina's calculations of the tarantula found that the spider was able to cover 20 centimeters in 5 seconds ,what was the average s
12345 [234]

4 centimetres per second

7 0
3 years ago
Consider the mechanism. Step 1: A+B↽−−⇀CA+B↽−−⇀C equilibrium Step 2: C+A⟶DC+A⟶D slow Overall: 2A+B⟶D2A+B⟶D Determine the rate la
tatuchka [14]

Answer:

rate = k[A][B] where k = k₂K

Explanation:

Your mechanism is a slow step with a prior equilibrium:

\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}

(The arrow in Step 1 should be equilibrium arrows).

1. Write the rate equations:

-\dfrac{\text{d[A]}}{\text{d}t} = -\dfrac{\text{d[B]}}{\text{d}t} = -k_{1}[\text{A}][\text{B}] + k_{1}[\text{C}]\\\\\dfrac{\text{d[C]}}{\text{d}t} = k_{1}[\text{A}][\text{B}] - k_{2}[\text{C}]\\\\\dfrac{\text{d[D]}}{\text{d}t} = k_{2}[\text{C}]

2. Derive the rate law

Assume k₋₁ ≫ k₂.  

Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.  

In an equilibrium, the forward and reverse rates are equal:

k₁[A][B] = k₋₁[C]

[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)

rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]

The rate law is  

rate = k[A][B] where k = k₂K

5 0
4 years ago
If i have 340mL of a 1.5 M NaBr solution, What will the concentration be for 1000mL?
Gekata [30.6K]

Answer:

0.51M

Explanation:

Given parameters:

Initial volume of NaBr = 340mL

Initial molarity  = 1.5M

Final volume  = 1000mL

Unknown:

Final molarity = ?

Solution;

This is a dilution problem whereas the concentration of a compound changes from one to another.

In this kind of problem, we must establish that the number of moles still remains the same.

    number of moles initially before diluting = number of moles after dilution

Number of moles  = Molarity x volume

Let us find the number of moles;

          Number of moles  = initial volume x initial molarity

Convert mL to dm³;

                  1000mL  = 1dm³

                     340mL gives \frac{340}{1000}   = 0.34dm³

Number of moles  = initial volume x initial molarity  = 0.34 x 1.5 = 0.51moles

Now to find the new molarity/concentration;

               Final molarity  = \frac{number of moles}{Volume}    = \frac{0.51}{1}    = 0.51M

We can see a massive drop in molarity this is due to dilution of the initial concentration.

6 0
4 years ago
Please Help! (Chemistry Molecules to Grams)
Ivahew [28]

Answer:

20.95 g of caffeine, C₈H₁₀N₄O₂

Explanation:

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

Number of molecules of C₈H₁₀N₄O₂ = 6.5×10²² molecules

Mass of C₈H₁₀N₄O₂ =?

From Avogadro's hypothesis,

1 mole of C₈H₁₀N₄O₂ = 6.02×10²³ molecules

Next, we shall determine the mass of 1 mole of C₈H₁₀N₄O₂. This can be obtained as follow:

1 mole of C₈H₁₀N₄O₂ = (8×12) + (10×1) + (4×14) + (2×16)

= 96 + 10 + 56 + 32

1 mole of C₈H₁₀N₄O₂ = 194 g

Thus,

194 g of C₈H₁₀N₄O₂ = 6.02×10²³ molecules

Finally, we shall determine the mass of caffeine, C₈H₁₀N₄O₂ that contains 6.5×10²² molecules. This can be obtained as follow:

6.02×10²³ molecules = 194 g of C₈H₁₀N₄O₂

Therefore,

6.5×10²² molecules = (6.5×10²² × 194) / 6.02×10²³

6.5×10²² molecules = 20.95 g of C₈H₁₀N₄O₂.

Therefore, 20.95 g of caffeine, C₈H₁₀N₄O₂ contains 6.5×10²² molecules

6 0
3 years ago
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