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2 years ago

6 Which element requires the least amount of

Chemistry

2 answers:

quester [9]2 years ago

6 0

Answer: 1) Na because it requires least amount of energy.

DaniilM [7]2 years ago

3 0

Which element requires the least amount of

energy to remove the most loosely held electron

from a gaseous atom in the ground state?

<h3>Answer-</h3><h3>Na</h3>

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Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

pentagon [3]

Answer: molecular formula = C12H16O8

Explanation:

NB Mm CO2= 44g/mol

Mm H2O= 18g/mol

Moles of CO2 = 36.86/44=0.84mol

0.84mole of CO2 has 0.84 mol of C

Moles of H2O = 10.06/18= 0.56mol

1mol of H20 contains 1mol of O and 2 mol H,

Hence there are 0.56mol O and (0.56×2)mol H

Hence the compound contains

C= 0.84 mol H= 1.12mol O=0.56mol

Divide through by smallest number

C= 0.83/0.56= 1.5mol

H= 1.12/0.55= 2mol

O= 0.56/0.56= 1mol

Multiply all by 2 to have whole number of moles = 3:4:2

Hence empirical formula= C3H4O2

(C3H4O2)n = 288.38

[(12×3) + 4+(16×2)]n= 288.38

72n=288.38

n= 4

:. Molecular formula=(C3H4O2)4= C12H16O8

7 0

2 years ago

Why is a lot of energy needed to break covalent bonds

gulaghasi [49]

Answer:

The strength of a bond between two atoms increases as the number of electron pairs in the bond increases.

Explanation:

6 0

2 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

3 years ago

Consider the second-order reaction:

kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)
$

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

5 0

3 years ago

.Inherited traits are passed from parent to offspring by information coded in

Liono4ka [1.6K]

Answer:

Inherited traits are passed from parent to offspring by information coded in

the DNA or Deoxyribonucleic Acid

8 0

3 years ago