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2 years ago

When Z-4,5-dimethyloct-4-ene is treated with hydrogen chloride, HCl, the result is:________.

Chemistry

1 answer:

Vika [28.1K]2 years ago

6 0

Answer:

The correct option is c

Explanation:

The chemical equation for the reaction of Z-4,5-dimethyloct-4-ene and HCl is shown on the first uploaded image

Now looking at the product we see that there are two who has four different groups attached to them this carbon are known as chiral carbons hence the product formed is a pair of diastereomers

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5.00 g of hydrogen gas and 50.0g of oxygen gas are introduced into an otherwise empty 9.00L steel cylinder, and the hydrogen is

GenaCL600 [577]

1) Balanced chemical reaction:

2H2 + O2 -> 2H20

Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O

2) Reactant quantities converted to moles

H2: 5.00 g / 2 g/mol = 2.5 mol

O2: 50.0 g / 32 g/mol = 1.5625 mol

Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).

3) Products

H2 totally consumed -> 0 mol at the end

O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end

H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.

Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol

4) Pressure

Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K

p = nRT/V = 7.9 atm

3 0

1 year ago

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A 23.74-mL volume of 0.0981 M NaOH was used to titrate 25.0 mL of a weak monoprotic acid solution to the stoi- chiometric point.

Dmitrij [34]

Answer:

Molar concentration of the weak acid solution is 0.0932

Explanation:

Using the formula: $\frac{C_aV_a}{C_bV+b} = \frac{n_a}{n_b}$

Where Ca = molarity of acid

Cb = molarity of base = 0.0981 M

Va = volume of acid = 25.0 mL

Vb = volume of base = 23.74 mL

na = mole of acid

nb = mole of base

Since the acid is monopromatic, 1 mole of the acid will require 1 mole of NaOH. Hence, na = nb = 1

Therefore, $C_a = \frac{C_bV_b}{V_a}$

Ca = 0.0981 x 23.74/25.0

= 0.093155 M

To 4 significant figure = 0.0932 M

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1 year ago

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A balance measures mass to 0.001 g. If you determine the mass of an object that weighs about 30 g, would you record the mass as

solong [7]

Answer:

The mass is recorded as 32.075 g

Explanation:

"The first digit of uncertainty is taken as the last significant digit", this is the rule for significant figures in the analysis. The balance measures the mass up to three decimal places, so it makes the most sense to note the whole figure.

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1 year ago

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Add electron dots and charges as necessary to show the reaction of potassium and bromine to form an ionic compound

S_A_V [24]

Explanation: Electron dot structures are the lewis dot structures which represent the number of valence electrons around an atom in a molecule.

The electronic configuration of potassium is $[Ar]4s^1$

Valence electrons of potassium are 1.

The electronic configuration of Bromine is $[Ar]4s^24p^5$

Valence electrons of bromine are 7.

These two elements form ionic compound.

Ionic compound is defined as the compound which is formed from the complete transfer of electrons from one element to another element.

Here, one electron is released by potassium which is accepted by bromine element. In this process, Potassium becomes cation having +1 charge and Bromine become anion having (-1) charge.

The ionic equation follows:

$K^++Br^-\rightarrow KBr$

The electron dot structure is provided in the image below.

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1 year ago

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HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.

Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, $C_A$ = 6.0mol/dm³

volume of HA, $V_A$ = 25.0cm³, = 0.025dm³

Concentration of HB, $C_B$ = 3.0mol/dm³

volume of HB, $V_B$ = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

$C_iVi = C_fV_f$

where;

$C_i$ is initial concentration

$V_i$ is initial volume

$C_f$ is final concentration of the solution

$V_f$ is final volume of the solution

$C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3$

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0

1 year ago