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3 years ago

Which is the correct Lewis structure for OBr?

Chemistry

1 answer:

____ [38]3 years ago

3 0

Answer:Covalent compound (I,C,W,H are nonmetals)

Valence of I: 7

Valence of C (C must form an octet): 4

Valence of W: 6

Valence of H: 1

Total valence electrons: 19

Molar mass: 324.77 g/mol

Explanation:

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Can a molecule have polar bonds, but still be a no polar molecule?

My name is Ann [436]

A molecule can possess polar bonds and still be nonpolar.

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3 years ago

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What's the Balanced equation for C5H11OH+O2

Galina-37 [17]

The balanced equation is 2 C5H11OH + 15 O2 = 10 CO2 + 12 H2O

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3 years ago

If the valency of oxygen is 2, derive the valencies of the other elements present in the following oxides. Na2O, ZnO , Al2O3 , M

11Alexandr11 [23.1K]

Answer:

1)Na2O

let the valency of Na is x

2(x)+(2)=0

2x+2=0

2x=-2

x=-1

2)ZnO

let the valency of Zn is x

x+2=0

x=-2

3)Al2O3

let the valency of Al is x

2(x)+3(2)=0

2x+6=0

2x=-6

x=-3

4)MgO

let the valency of Mg is x

x+2=0

x=-2

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3 years ago

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An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If t

topjm [15]

Answer:

$\boxed{3}$

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

$\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )$

where

R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

$\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}$

Data:

$n_{f} = 2$

λ = 657 nm

Calculation:

$\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}$

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3 years ago

What is the hydrogen ion concentration, [H+], of a solution with a pH of 5.43?

grigory [225]

Answer: 3.7 x10−6 Mole per dm^3

Explanation:

pH is the negative logarithm of hydrogen ion concentration in a solution.

So, pH = - log(H+)

Since the solution has a pH of 5.43

5.43 = -log(H+)

To get hydrogen ion concentration, find the Antilog of 5.43

(H+) = Antilog (-5.43)

(H+) = 0.000003715

Then, 0.000003715 in standard form becomes 3.7 x10−6 M

Thus, the concentration of hydrogen ion in the solution is 3.7 x10−6 Mole per dm^3

6 0

2 years ago