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2 years ago

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Is the atom indicated with an arrow nucleophilic, electrophilic, acidic, more than one of these choices, or none of these choice

s? (For purposes of this question, acidic is defined as pKa ≤ 25.)

1 answer:

9966 [12]2 years ago

7 0

Answer:

Acidic

Explanation:

Please kindly check attachment for the detailed step by step solution of the given problem.

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Calculate the wavelength in meters of electromagnetic radiation that has a frequency of 1.09 × 10⁸ s⁻¹. (c = 3.00 X 10⁸ m/s)

fenix001 [56]

Answer:

2.75X10^16

Explanation:

C=FXwavelength

wavelength=C/F

7 0

1 year ago

a) (1 point) Build anthracene, optimize its geometry and examine its structure. Describe its shape. b) (1 point) Measure the C-C

oksano4ka [1.4K]

Answer:

a) The structure of anthracene is planar with all the pi electrons delocalized in the structure to maintain aromaticity.

b) The C-C bond length in anthracene is about 140 pm with all the bond lengths being similar to each other.

The standard C-C bond length is 154 pm while standard C=C bond is about 134 pm. Therefore the bond length in anthracene is smaller than standard C-C bond length and longer than standard C=C bond length. This can be explained from the fact that the C-C bonds in anthracene has be mixed characteristics of single and double bond because of the delocalization of pi electrons over the whole structure. As a result, they are neither fully single nor fully double bond in nature. Hence the observed bond lengths.

c) This molecule is not flat. The N-atom is sp3 hybridized here and the H-atom attached to N will remain out of plane.

Explanation:

8 0

2 years ago

How many known elements have atoms, in their ground-state, with valence electrons whose quantum numbers are n = 5 and l = 1?a. 3

Goshia [24]

Answer:

c. 6.

Explanation:

Looking at the description given in the question, the elements involved must belong to the p- block of the periodic table and must be in period 5. They also must possess valence electrons in the 5p- orbital.

Now if we look at the p- block of period 5, the following elements satisfy these requirements; Sr, In, Sn, Sb, Te and I.

Hence there are six of such elements.

7 0

1 year ago

An old sample of concentrated sulfuric acid to be used in the laboratory is approximately 98.1 percent h2so4 by mass. calculate

Airida [17]

Basis: 100 mL solution

From the given density, we calculate for the mass of the solution.

density = mass / volume
mass = density x volume

mass = (1.83 g/mL) x (100 mL) = 183 grams

Then, we calculate for the mass H2SO4 given the percentage.

mass of H2SO4 = (183 grams) x (0.981) = 179.523 grams

Calculate for the number of moles of H2SO4,
moles H2SO4 = (179.523 grams) / (98.079 g/mol)
moles H2SO4 = 1.83 moles

Molarity:
M = moles H2SO4 / volume solution (in L)
= 1.83 moles / (0.1L ) = 18.3 M

Molality:
m = moles of H2SO4 / kg of solvent
= 1.83 moles / (183 g)(1-0.983)(1 kg/ 1000 g) = 588.24 m

8 0

2 years ago

A solution is prepared by adding 6.24 g of benzene (C 6H 6, 78.11 g/mol) to 80.74 g of cyclohexane (C 6H 12, 84.16 g/mol). Calcu

tekilochka [14]

Answer:

$x_B=0.0769$

$m=0.990m$

Explanation:

Hello,

In this case, we can compute the mole fraction of benzene by using the following formula:

$x_B=\frac{n_B}{n_B+n_C}$

Whereas n accounts for the moles of each substance, thus, we compute them by using molar mass of benzene and cyclohexane:

$n_B=6.24g*\frac{1mol}{78.11g}=0.0799mol\\ \\n_C=80.74g*\frac{1mol}{84.16g} =0.959mol$

Thus, we compute the mole fraction:

$x_B=\frac{0.0799mol}{0.0799mol+0.959mol}\\ \\x_B=0.0769$

Next, for the molality, we define it as:

$m=\frac{n_B}{m_C}$

Whereas we also use the moles of benzene but rather than the moles of cyclohexane, its mass in kilograms (0.08074 kg), thus, we obtain:

$m=\frac{0.0799mol}{0.08074kg}=0.990mol/kg$

Or just 0.990 m in molal units (mol/kg).

Best regards.

7 0

2 years ago