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3 years ago

How do lone pairs of electrons influence the structure of molecules

Chemistry

1 answer:

dybincka [34]3 years ago

6 0

Answer:

Electron pairs repel each other and influence bond angles and molecular shape. The presence of lone pair electrons influences the three-dimensional shape of the molecule.

Explanation:

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The symbol for gold is a. go. b. g. c. au. d. au.

Zolol [24]

Au is the symbol for gold. It comes from Latin word for gold, which is Aurunum. Its atomic number is 79.

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3 years ago

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Rasek [7]

Explanation:

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3 years ago

The reactant concentration in a first-order reaction was 8.10×10−2 M M after 15.0 s s and 1.80×10−3 M M after 90.0 s s . What is

kipiarov [429]

Answer:

The answer to the question is

The rate constant for the reaction is 1.056×10⁻³ M/s

Explanation:

To solve the question, e note that

For a zero order reaction, the rate law is given by

[A] = -k×t + [A]₀

This can be represented by the linear equation y = mx + c

Such that y = [A], m which is the gradient is = -k, and the intercept c = [A]₀

Therefore the rate constant k which is the gradient is given by

Gradient = $\frac{[A]_{2} - [A]_{1} }{t_{2} - t_{1} }$ where [A]₁ = 8.10×10⁻² M and [A]₂ = 1.80×10⁻³ M

= $\frac{1.80*10^{-3} M- 8.10*10^{-2} M}{90 s - 15 s}$ = -0.001056 M/s = -1.056×10⁻³ M/s

Threfore k = 1.056×10⁻³ M/s

3 0

3 years ago

Using the Bohr model, determine the energy, in joules, necessary to ionize a ground-state hydrogen atom. Show your calculations.

lord [1]

Answer:

The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.

Explanation:

To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.

It is given by Bohr's Theory as:

1/λ = Rh (1/n1² - 1/n2²)

where,

λ = wavelength of photon

n1 = initial state = 1 (ground-state of hydrogen)

n2 = final state = ∞ (since, electron goes far away from atom after ionization)

Rh = Rhydberg's Constant = 1.097 x 10^7 /m

Therefore,

1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)

λ = 9.115 x 10^-8 m = 91.15 nm

Now, for energy (E) we know that:

E = hc/λ

where,

h = Plank's Constant = 6.625 x 10^-34 J.s

c = speed of light = 3 x 10^8 m/s

Therefore,

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)

E = 2.18 x 10^-18 J

E = (2.18 x 10^-18 J)(1 eV/1.6 x 10^-19 J)

E = 13.6 eV

5 0

4 years ago

How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction

nekit [7.7K]

Answer : The volume of $NaOH$ required to neutralize is, 340 mL

Explanation :

To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?$

Putting values in above equation, we get:

$2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL$

Hence, the volume of $NaOH$ required to neutralize is, 340 mL

4 0

3 years ago