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3 years ago

What have six valence electrons

2 answers:

6 0

Any element in group 18 has eight valence electrons (except for helium, which has a total of just two electrons). Examples include neon (Ne), argon (Ar), and krypton (Kr). Oxygen, like all the other elements in group 16, has six valence electrons.

Alinara [238K]3 years ago

4 0

You might be interested in

Draw all possible structures for a compound with molecular formula C4H8O that exhibits a broad signal between 3200 and 3600 cm-1

MrRissso [65]

Answer:

Cyclic butanol , 1 cyclopropyl methanol, 1-methyl cyclo propan-1-ol.

Explanation:

IR wave-number of O-H bond = $3600 - 3200 cm^{-1}$

The intensity of the signal is broad and strong.

IR wave number of C=C bond = $1680 - 1600 cm^{-1}$

The intensity of the signal is weak.But for the given chemical formula it didn't show any signal in its IR spectrum.This means that no unsaturation is present.

IR wave number of C=O bond = $1780 - 1650 cm^{-1}$

The intensity of the signal is broad and strong.But for the given chemical formula it didn't show any signal in its IR spectrum. This means that no C=O is present in the compound

For the given formula the possible structures wil be:

$C_4H_8O: C_4H_7OH$

As we know from the I.R. data that no unsaturation is present which indicates that the given compound has cyclic structure.

The possible structures are attached in an image.

4 0

2 years ago

Copper has two naturally occurring isotopes, ⁶³Cu (isotopic mass = 62.9296 amu) and ⁶⁵Cu (isotopic mass = 64.9278 amu). If coppe

Stells [14]

Copper has two naturally occurring isotopes, ⁶³Cu (isotopic mass = 62.9296 amu) and ⁶⁵Cu (isotopic mass = 64.9278 amu). If copper has an atomic mass of 63.546 amu,

Abundance of ⁶³Cu and ⁶⁵Cu ; 70% , 30.848%

<h3>Calculation : </h3>

Let the natural abundance of 63, Cu isotope be x%.

The natural abundance of

Cu isotope will be 100−x.

The average atomic weight is 63.546 g.

Hence, 100×63.546=63x+65(100−x)

6354.6=63x+6500−65x=6500−2x

2x=145.4

x=72.7≃70

Hence, the natural abundance of the 63Cu isotope must be approximately 70%.

For the ⁶⁵Cu calculation is same and the answer is 30.848%

<h3>What is isotope ?</h3>

Isotopes are members of a family of elements that all have the same number of protons and different numbers of neutrons. The number of protons in the nucleus determines the atomic number of an element on the periodic table.

Example : Magnesium has three natural isotopes : ²⁴Mg, ²⁵Mg, and ²⁶Mg.

<h3>What is Atomic mass ?</h3>

Atomic mass is the mass of an atom. The SI unit of mass is the kilogram, while atomic mass is often expressed in the non-SI unit dalton (synonymous with uniform atomic mass unit). 1 Da is defined as 1/12 the mass of a free carbon-12 atom at rest in the ground state.

To know more about Atomic mass please click here : brainly.com/question/338808

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3 0

1 year ago

Convert 15cm^3 to liters

Vinvika [58]

Answer:

0.015 is the answer

Explanation:

6 0

2 years ago

Read 2 more answers

Choose the products that complete the reaction. The chemical equations may not be balanced. HCl + NaOH ? NaCl + H2O NaCl + CO2 +

Aliun [14]

Answer:

a- NaCl + H20

3 0

3 years ago

(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g

Anarel [89]

Answer:

Part A

The volume of the gaseous product is $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

The temperature is $T = 350^oC = 350 +273 =623K$

The pressure is $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

The of $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

$2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

The number of moles of $C_8 H_{18}$ that reacted is mathematically represented as

$n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of $C_8 H_{18}$ is constant value which is

$M = 114.23 \ g/mole$

So $n = \frac{100 }{114.23} }$

$n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

2 moles of $C_8 H_{18}$ will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

1 mole of $C_8 H_{18}$ will react with 12.5 moles of $O_2$ to give 17 moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

This implies that

0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give (17 * 0.8754) of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So the no of moles of gaseous product is

$N_g = 17 * 0.8754$

$N_g = 14.88 \ moles$

From the ideal gas law

$PV = N_gRT$

making V the subject

$V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

$V = \frac{14.88* 0.08206 *623}{0.967}$

$V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

$N_o = 0.8754 * 12.5$

$N_o = 10.94 \ moles$

The volume is

$V_o = \frac{10.94 * 0.08206 *623}{0.967}$

$V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

$V_e = V_o * \frac{0.79}{0.21}$

Substituting values

$V_e = 579 * \frac{0.79}{0.21}$

$V_e = 2178 \ L$

3 0

3 years ago