How to atoms share more than one pair of electrons?

Answers

Answer 1

Answer:

Answer:

A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.

Explanation:

More than one pair of electrons can be shared between atoms to form double or triple covalent bonds. Unlike ionic bonds, covalent bonds are often formed between atoms where one of the atoms cannot easily attain a noble gas electron shell configuration through the loss or gain of one or two electrons.

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Use the equation editor or "Insert Chemistry - WIRIS editor" to write the balanced molecular chemical equation for the reaction of aqueous 0.13 M lead (II) nitrate, with 0.19 M potassium carbonate. You may need to consult Appendix E to determine the states of each reactant and product. Assume any insoluble products are completely insoluble.

Answers

Answer:

Balanced equation:

$Pb(NO_(3))_(2)(aq)+K_(2)CO_(3)(aq)\rightarrow PbCO_(3)(s)+2KNO_(3)(aq)$

Explanation:

The chemical reaction between Lead(II) Nitrate and potassium carbonate is as follows.

$Lead(II)Nitrate+Potassium\,carbonate \rightarrow Lead(III)\,\,carbonate+Potassium\,nitrate$

$Pb(NO_(3))_(2)(aq)+K_(2)CO_(3)(aq)\rightarrow PbCO_(3)(s)+2KNO_(3)(aq)$

Ionic equation:

$Pb^(2+)(aq)+2NO_(3)^(-)(aq)+2K^(+)(aq)+CO_(3)^(2-)(aq)\Leftrightarrow PbCO_(3)(s)+K^(+)(aq)+2NO_(3)^(-)$

Cancel the same ions on the both sides of the reaction.

The net ionic equation is as follows.

$Pb^(2+)(aq)+CO_(3)^(2-)(aq)\Leftrightarrow PbCO_(3)(s)$

What formula was used to find the answer

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Can you give more information?

Caffeine, a molecule found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 10.0 g of caffeine produces 18.13 g of CO₂, 4.639 g of H₂O, and 2.885 g of N₂. Determine the molar mass of the compound if it is between 150 and 210 g/mol.

Answers

The molar mass of the compound is found by finding the empirical and

molecular formula of the compound.

The molar mass of the compound, MM = 2 × 97 g/mol = 194 g/mol

Reasons:

Molar mass of CO₂ = 44.01 g/mol

Number of moles of CO₂ produced = $(18.13)/(44.01)$ ≈ 0.412 moles

Number of moles of produced C = 0.412 moles

Mass of C = 12 × 0.412 = 4.944 g

Molar mass of H₂O = 18.015 g/mol

Moles of H₂O produced = $(4.639)/(18.015 )$ = 0.2575 moles

Moles of produced H = 2 × 0.2575 = 0.515 moles

Mass of H = 1.00784 × 0.515 ≈ 0.519 g

Molar mass of N₂ = 28.0134 g/mol

Moles of N produced = 2 × $(2.885)/(28.0134 )$ = 2 × 0.103 = 0.206 moles

Mass of oxygen = 10 - 4.944 - 2.885 - 0.519 = 1.652

Moles of oxygen, O = $(1.652 \, g)/(16 \, g/mol)$ ≈ 0.103 moles

Therefore, we get;

Number of moles of produced C = 0.412 moles

Number of moles of produced H = 0.515 moles

Number of moles of oxygen, O ≈ 0.103 moles

Number of moles of N produced = 0.206 moles

Dividing by 0.103 gives;

Mole ratio of C = $(0.412)/(0.103) = 4$

Mole ratio of H = $(0.515)/(0.103) = 5$

Mole ratio of O = 1

Mole ratio of N = $(0.206)/(0.103) = 2$

The empirical formula of the compound is therefore; C₄H₅N₂O

The general molecular formula is of the form (C₄H₅N₂O)ₙ

Molar mass of the compound is between 150 g/mol and 210 g/mol (given)

The molar mass of C₄H₅N₂O = 4×12 + 5×1.00784 + 2×14 + 16 ≈ 97

The molar mass of C₄H₅N₂O ≈ 97 g/mol

Molar mass of the compound is between 150 and 210 g/mol, therefore, n in

(C₄H₅N₂O)ₙ = 2, which gives;

The molar mass of the compound, MM = 2 × 97 g/mol = 194 g/mol

Learn more here:

brainly.com/question/12238420

Answer:

194 g/mol.

Explanation:

Hello,

In this case, one first must compute the mass of each element as shown below:

$C=18.13gCO_2*(12gC)/(44gCO_2) =4.945gC\nH=4.639gH_2O*(2.016gH)/(18.0152gH_2O)=0.519gH\nN=2.885gN_2\nO=10.0g-4.945g-0.519g-2.885g=1.651gO$

Next, the corresponding moles:

$C=4.945gC*(1molC)/(12gC)=0.412mol\nH=0.519gH*(1molH)/(1gH)=0.519mol\nN=2.885gN*(1molN)/(14gN)=0.206molN\nO=1.648gO*(1molO)/(16gO) =0.103molO$

Then, each element's subscripts is found to be:

$C=(0.412)/(0.103)=4\nH=(0.519)/(0.103)=5\nN=(0.206)/(0.103) =2\nO=(0.103)/(0.103)=1$

Therefore, the empirical formula is:

$C_4H_5N_2O$

Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:

$C_8H_10N_4O_2$

Which has a molar mass of 194 g/mol being correctly contained in the given interval.

Best regards.

Why can't 1−methylcyclohexanol be prepared from a carbonyl compound by reduction? select the single best answer?

Answers

1−methylcyclohexanol is a tertiary alcohol. Tertiary Alcohols are synthesized by either reacting Ketone with Organometallic compounds like Grignard reagent or by hydration of substituted alkenes. 1−methylcyclohexanol can not be synthesized by reduction of carbonyl compound because it is not possible to have a starting carbonyl compound having carbonyl group along with three other alkyl groups (as carbon can only form 4 bonds).

Result:
Tertiary alcohols don't contain a hydrogen atom at carbon attached to hydroxyl group that is why it is not possible to synthesize 1−methylcyclohexanol by reduction of carbonyl compound.

How many grams of CO2 and H2O are produced from the combustion of 220. g of propane (C3H8)? C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

Answers

Answer:

(C3H8) produces 660 g of CO2 and 360 g of H2O

Explanation:

The balanced chemical equation for the combustion of propane (C3H8) is:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

This equation tells us that for every molecule of propane (C3H8) that reacts with 5 molecules of oxygen (O2), 3 molecules of carbon dioxide (CO2) and 4 molecules of water (H2O) are produced.

So, if we have 220. g of propane (C3H8), we can find the amount of CO2 and H2O produced by using the mole ratio from the balanced equation:

1 mole C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O

We can find the number of moles of C3H8 by dividing the mass by the molar mass of C3H8 (44 g/mol):

220 g / 44 g/mol = 5 moles C3H8

So, the number of moles of CO2 and H2O produced can be found by multiplying the number of moles of C3H8 by the mole ratio:

3 moles CO2 = 3 moles CO2/1 mole C3H8 * 5 moles C3H8 = 15 moles CO2

4 moles H2O = 4 moles H2O/1 mole C3H8 * 5 moles C3H8 = 20 moles H2O

Finally, we can convert the number of moles of CO2 and H2O to grams by multiplying by their molar masses (44 g/mol for CO2 and 18 g/mol for H2O):

15 moles CO2 * 44 g/mol = 660 g CO2

20 moles H2O * 18 g/mol = 360 g H2O

So, the combustion of 220 g of propane (C3H8) produces 660 g of CO2 and 360 g of H2O.

The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (M) Initial Rate (10–3 M/s) 0.4 0.4 0.2 160 0.2 0.4 0.4 80 0.6 0.1 0.2 15 0.2 0.1 0.2 5 0.2 0.2 0.4 20 Using the initial-rate method, what is the order of the reaction with respect to C? a. zero-order b. first-order c. third-order d. second-order e. impossible to tell from the data given

Answers

The dependence of the power of the reaction rate on the concentration is called the order of the reaction. The order of the reaction is the first order.

What is the initial-rate method?

The initial rate method is the estimation of the order of the reaction by the initial rates of the reactants and products and by performing the reaction several times by measuring the rate.

The reaction is given as,

$\rm A + B + C \rightarrow Products$

The rate of reaction can be given as:

$\rm rate = k[A]^(x)[B]^(y)[C]^(z)$

Here the variables x, y and z are orders respective to the reactant concentration and k is the rate constant.

Value of x with respect to A:

$\begin{aligned} \rm \frac {Rate 3}{Rate 4} &= \rm [([A(3)])/([A(4)])]^(\rm x)\n\n(15)/(5) &= [([0.6])/([0.2])]^(\rm x)\n\n\rm x &= 1\end{aligned}$

Value of y with respect to B:

$\begin{aligned}\rm \frac {Rate 2}{Rate 5} &= \rm [([B(2)])/([B(5)])]^(\rm y)\n\n(80)/(20) &= [([0.4])/([0.2])]^(\rm y)\n\n\rm y &= 2\end{aligned}$

Value of z with respect to C:

$\rm \frac {Rate 1}{Rate 2} &= [([A(1)])/([A(2)])]^(x) [([B(1)])/([B(2)])]^(y) [([C(1)])/([C(2)])]^(z)$

Substituting value of x = 1 and y = 2 in the above equation:

$\begin{aligned}(160)/(80) &= [([0.4])/([0.2])]^(1)[([0.4])/([0.4])]^(2) [([0.2])/([0.4])]^(\rm z)\n\n1 &= (0.5)^(\rm z)\n\n&= 1\end{aligned}$

Therefore option b. with respect to C = 1, the order of the reaction is first-order.

Learn more about the order of reaction here:

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Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

[A]₀ [B]₀ [C]₀ Initial Rate (10⁻³ M/s)

1. 0.4 0.4 0.2 160

2. 0.2 0.4 0.4 80

3. 0.6 0.1 0.2 15

4. 0.2 0.1 0.2 5

5. 0.2 0.2 0.4 20

The rate of the above reaction is given as:

$Rate = k[A]^(x)[B]^(y)[C]^(z)$

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

$(Rate3)/(Rate4)= [([A(3)])/([A(4)])]^(x)$