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2 years ago

The valency of sodium metal

Chemistry

1 answer:

Ronch [10]2 years ago

3 0

Answer:

The atomic number of sodium is 11 (Z=11). The electronic configuration of sodium can be written as 2, 8, 1. Therefore, valence electron in sodium is 1 and it needs to lose 1 electron from the outermost orbit to attain octet. Hence, the valency of sodium is 1.

Explanation:

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Using the mmoles of (35)-2,2,-dibromo-3,4-dimethylpentane calculated earlier and the molecular weight of the product (962 g/mol)

Sedbober [7]

Answer:

The yield of the product in gram is $\mathsf{{w_P}=0.26 \ gram}$

Explanation:

Given that:

the molecular mass weight of the product = 96.2 g/mol

the molecular mass of the reagent (3S)-2,2,-dibromo-3,4-dimethylpentane is 257.997 g

given that the millimoles of the reagent = 2,7 millimoles = $2.7 \times 10^{-3} \ moles$

We know that:

Number of moles = mass/molar mass

Then:

$2.7 \times 10^{-3} = \dfrac{ mass}{257.997}$

$mass = 2.7 \times 10^{-3} \times 257.997$

mass = 0.697

Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100

i.e

Theoretical yield = $\dfrac{n_P}{n_R}\times 100\%$

where;

$n_P = \dfrac{w_P}{m_P}$ and $n_R = \dfrac{w_R}{m_R}$

Theoretical yield = $\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%$

Given that the theoretical yield = 100%

Then:

$100\% =\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%$

$\dfrac{w_P}{m_P}=\dfrac{w_R}{m_R}$

${w_P}=\dfrac{w_R \times m_P}{m_R}$

where,

$w_P$ = derived weight of the product

$m_P =$ the molecular mass of the derived product

$m_R =$ the molecular mass of the reagent

$w_R$ = weight in a gram of the reagent

${w_P}=\dfrac{w_R \times m_P}{m_R}$

${w_P}=\dfrac{0.697 \times 96.2}{257.997}$

$\mathsf{{w_P}=0.26 \ gram}$

8 0

2 years ago

CAN SOMEONE HELP PLS!!!

Pachacha [2.7K]

Answer:

Matals tend to LOSE electrons to become POSITIVE ions

6 0

2 years ago

Read 2 more answers

Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit

luda_lava [24]

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3>Further explanation </h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

$K~=~10^{-pK}$

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

$K~=~10^{4.9}$

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

$K`=~10^{1.12]$

Reaction can take place

2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

$K~=~10^{ -3.56}$

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

$K`=~10 ^{-12.93}$

Reaction does not happen

3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

$K~=~10^{-3.26}$

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

$K~=~ 10^{-12.63}$

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3>Learn more </h3>

the lowest ph

brainly.com/question/9875355

the concentrations at equilibrium.

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the ph of a solution

brainly.com/question/9560687

Keywords : acid base reaction, the equilibrium constant

5 0

2 years ago

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What is the empirical formula for c12h24o6? what is the empirical formula for c12h24o6? ch2o cho c2h5o c2h4o cho2?

stira [4]

Empirical formula: The formula consist of proportions of the elements which is present in the compound or the simplest whole number ratios of atoms.

Now, molecular formula is equal to the product of n (ratio) and empirical formula.

Molecular formula = $n\times empirical formula$ (1)

molecular formula = $C_{12}H_{24}O_{6}$ (given)

Since, 6 is the smallest subscript in above molecular formula to get the simpler whole number of atoms. Therefore, divide all the subscripts i.e. number of carbon atoms (12), number of hydrogen atoms (24) and number of oxygen atoms (6) by 6.

empirical formula becomes $C_{2}H_{4}O$