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Molodets [167]
3 years ago
11

C6H6NCl On the basis of the molecular formula, 1H NMR data, and IR data provided, propose a consistent structure.

Chemistry
1 answer:
Aliun [14]3 years ago
6 0

Answer:

Please refer to the attachment below.

Explanation:

Please refer to the attachment below for explanation.

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Black holes are the final stage of what star
Andreyy89

Answer:

This:

Explanation:

Black holes occur when a massive star or larger reaches the final stage of it's lifespan. The star implodes and a black hole is the dying star's remains

4 0
3 years ago
Read 2 more answers
How many grams of carbon are contained in one mole of<img src="https://tex.z-dn.net/?f=C_%7B3%7DH_%7B8%7D" id="TexFormula1" titl
NNADVOKAT [17]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \:36 \:\: g

____________________________________

\large \tt Explanation \: :

The given compound has 3 carbon atoms, so in 1 mole of that compound, there will be 3 moles of carbon atoms.

Mass of each mole of carbon atoms = 12 g

For 3 mole carbon atoms, it will be 12 × 3 = 36 grams

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

4 0
2 years ago
What is the concentration of 10.00mL of HBr if it takes 16.73 mL of a 0.253M LiOH solution to neutralize it?
Fudgin [204]
The chemical reaction would be expressed as follows:

HBr + LiOH = LiBr + H2O

We are given the volumes and corresponding concentration to be used for the reaction. We use these values to solve for the concentration of the other reactant. We do as follows:

0.253 mol LiOH / L solution ( 0.01673 L ) ( 1 mol HBr / 1 mol LiOH ) = 0.00423 HBr needed

Concentration of HBr =0.00423mol / .010 L = 0.423 M HBr 
3 0
4 years ago
Given that a chlorine-oxygen bond has an enthalpy of 243 kJ/mol , an oxygen-oxygen bond has an enthalpy of 498 kJ/mol , and the
Alecsey [184]

Explanation:

The chemical equation is as follows.

      \frac{1}{2}Cl_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow ClO(g)

And, the given enthalpy is as follows.

    \frac{1}{2}Cl_{2}(g) + O_{2}(g) \rightarrow ClO_{2}(g);  \Delta H = 102.5 kJ

    Cl-Cl = 243 kJ/mol,      O=O = 498 kJ/mol

Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.

   \Delta H = \sum \text{bond broken in reactants} - \sum \text{bond broken in products}

    102.5 = [(\frac{1}{2})x + 498] - [(2)(243)]

    102.5 = (\frac{1}{2})x + 498 - 486

     102.5 - 12 = \frac{x}{2}

           x = 181 kJ

Now, total bond enthalpy of per mole of ClO is calculated as follows.

       \Delta H = \sum E(\text{bond broken in reactants}) - \sum (\text{bond broken in products})

              x = [(\frac{1}{2})181 + (\frac{1}{2})498] - 243

                 = 339.5 - 243

                 = 96.5 kJ

Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.

8 0
3 years ago
An element crystallizes in a face-centered cubic lattice. If the length of an edge of the unit cell is 0.408 nm, and the density
V125BC [204]

Answer:

Au

Explanation:

For the density of a face-centered cubic:

Density = \dfrac{4 \times M_w}{N_A \times a^3}

where

M_w = molar mass of the compound

N_A= avogadro's constant

a^3 = the volume of a unit cell

Given that:

Density (\rho) = 19.30 g/cm³

a = 0.408 nm

a = 0.408 \times 10^{-9} \times 10^{2} \ cm

a = 4.08 \times 10^ {-8} \ cm

∴

19.3 = \dfrac{4 \times M_w}{(6.023 \tmes 10^{23})\times (4.08 \times 10^{-8})^3}

M_w= \dfrac{19.3\times (6.023 \times 10^{23})\times (4.08 \times 10^{-8})^3}{4}

M_w=197.37 \ g/mol

Thus, the molar mass of 197.37 g/mol element is Gold (Au).

4 0
3 years ago
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