Why is butanone achiral?

When describing metallic bonding, scientists note that the metal ions seem to be floating in a sea of electrons. This arrangemen

horrorfan [7]

True

Explanation:

Metallic bonds joins atoms of metals and alloys together. The formation of this bond type is favored by large atomic radius, low ionization energy and a large number of electrons in the valence shell.

This bond type is actually an attraction between the positive nuclei of all closely packed atoms in the lattice and the electron cloud jointly formed by all atoms.
Some of the physical properties of metals such as malleability, ductility, electrical and thermal conductivity, luster, high melting point e.t.c stems from this.

Learn more:

Bond types brainly.com/question/6071838

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3 0

3 years ago

How many atoms are in KNO3?

Artyom0805 [142]

Answer:

5

Explanation:

There is 1 K, 1 N, and 3 O.

1 + 1 + 3 = 5

3 0

3 years ago

Identify the anode and cathode please

Nuetrik [128]

Answer:

anode is negative. cathode is positive.

6 0

3 years ago

A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.

Alexxandr [17]

Answer: The empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

Explanation:

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.078 moles.

For Carbon = $\frac{0.313}{0.078}=4.01\approx 4$

For Hydrogen = $\frac{0.316}{0.078}=4.05\approx 4$

For Oxygen = $\frac{0.078}{0.078}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is $C_4H_4O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

$n=\frac{130g/mol}{68g/mol}=1.9\approx 2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2$

Hence, the empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

4 0

3 years ago

__ SnCl4 → __ Sn + __ Cl2

Kisachek [45]

Answer:

Sn (s) + 2 Cl2 (g) → SnCl4 (l)

This is an oxidation-reduction (redox) reaction:

4 Cl0 + 4 e- → 4 Cl-I

(reduction)

Sn0 - 4 e- → SnIV

(oxidation)

Cl2 is an oxidizing agent, Sn is a reducing agent.

Reactants:

Sn

Names: Tin source: wikidata, accessed: 2019-09-07source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02, Sn source: wikidata, accessed: 2019-09-07, Element 50 source: wikidata, accessed: 2019-09-07

Appearance: White crystalline powder source: ICSC, accessed: 2019-09-04; Gray to almost silver-white, ductile, malleable, lustrous solid. source: NIOSH NPG, accessed: 2019-09-02

Cl2

Names: Chlorine source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02, Molecular chlorine source: NIOSH NPG, accessed: 2019-09-02

Appearance: Greenish-yellow compressed liquefied gas with pungent odour source: ICSC, accessed: 2019-09-04; Greenish-yellow gas with a pungent, irritating odor. [Note: Shipped as a liquefied compressed gas.] source: NIOSH NPG, accessed: 2019-09-02

Products:

SnCl4 – Tetrachlorostannane source: wikipedia, accessed: 2019-09-28source: wikidata, accessed: 2019-09-02, Tin tetrachloride source: wikipedia, accessed: 2019-09-28source: wikidata, accessed: 2019-09-02source: ICSC, accessed: 2019-09-04, Tin(IV) chloride source: wikipedia, accessed: 2019-09-28

Other names: Stannic chloride source: wikipedia, accessed: 2019-09-28source: wikidata, accessed: 2019-09-02source: ICSC, accessed: 2019-09-04, Tin(iv) chloride (anhydrous) source: ICSC, accessed: 2019-09-04

Appearance: Colorless to slightly yellow fuming liquid source: wikipedia, accessed: 2019-09-28; Colourless or slightly yellow fuming liquid with pungent odour source: ICSC, accessed: 2019-09-04

6 0

3 years ago