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8 days ago

If angle ABE = 2n + 7 and angle EBF=4n-13,find angle ABE.

Chemistry

1 answer:

VMariaS [1K]8 days ago

4 0

Answer:

Angle ABE measures 27°.

Explanation:

Refer to the attached diagram related to this question.

The given values are ∠ABE=2n+7 and ∠EBF=4n-13.

Clearly seen in the diagram, ∠ABE and ∠EBF are equal in measure.

$m\angle ABE=m\angle EBF$

$2n+7=4n-13$

Move variable components to one side of the equation.

$7+13=4n-2n$

$20=2n$

Split both sides by 2.

$10=n$

The solution for n arrives at 10.

The next step is to calculate ∠ABE.

$\angle ABE=2(10)+7=20+7=27$

Consequently, the measurement of angle ABE is 27°.

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Result: A total of 78,500 dye molecules will be required to visualize a single egg.

Explanation: Each egg cell has an approximate diameter of 100 μm.

We can apply this formula to determine the area of the cell membrane;

A = π $(100)^{2} / 4$ ;

Using π as 3.14, we calculate;

A = 3.14 X $(100)^{2} / 4$

Upon solving, we arrive at;

A = 7,850 μ $m^{2}$

Next, we want to compute the quantity of dye molecules needed for 10 fluorescent molecules per μ $m^{2}$ but;

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2 days ago

Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Ca2+, V5+, Br-

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Answer:

CaS, CaBr₂, VBr₅, and V₂S₅.

Explanation:

The ionic compound must exhibit neutrality; its total charge should equal zero.

A binary ionic compound is formed from two distinct ions.

Ca²⁺ combines with either Br⁻ or S²⁻ to create binary ionic compounds.

CaS is created when Ca²⁺ pairs with S²⁻ resulting in the neutral binary ionic compound CaS.

CaBr₂ results from the combination of one mole of Ca²⁺ with two moles of Br⁻ to form the neutral binary ionic compound CaBr₂.

V⁵⁺ can also unite with either Br⁻ or S²⁻ to produce binary ionic compounds.

V₂S₅ is formed when two moles of V⁵⁺ bond with five moles of S²⁻ yielding the neutral binary ionic compound V₂S₅.

VBr₅ is produced by combining one mole of V⁵⁺ with five moles of Br⁻ to form the neutral binary ionic compound VBr₅.

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12 days ago

Help on part "c": The forensic technician at a crime scene has just prepared a luminol stock solution by adding 19.0g of luminol

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1. The luminol stock solution has a molarity of 1.431 M.

2. In 2.00 L of the diluted spray, there are 0.12 moles of luminol.

3. The volume of the stock solution from Part A that contains the same number of moles present in the diluted solution from Part B is 83.86 ml.

Additional Information

Stoichiometry in Chemistry focuses on the quantitative aspects of chemical reactions, which includes calculations related to volume, mass, and the count of ions, molecules, and elements.

Key concepts in stoichiometry include:

1. Relative atomic mass
2. Relative molecular mass

This refers to the relative atomic mass of a molecule.

3. Mole

A mole represents the number of particles in a substance equivalent to the number of atoms in 12 grams of carbon-12.

1 mole = 6.02 × 10²³ particles.

The quantity of moles can also be derived by dividing mass (in grams) by either the relative mass of an element or the relative mass of a molecule.

$\large{\boxed{\bold{mol\:=\:\frac{grams}{ relative\:mass} }}}$

Luminol (C₈H₇N₃O₂) is utilized for detecting blood traces at crime scenes, due to its reaction with iron found in blood.

To prepare a luminol stock solution, 19.0 g of luminol is mixed into a total volume of 75.0 mL of water.

Thus, the molarity is calculated as:

1. Moles of Luminol

- the relative molecular mass of Luminol:

= 8.C + 7.H + 3.N + 2.16

= 8.12 + 7.1 + 3.14 + 2.16

= 177 grams/mol.

Thus, we have:

moles = grams / relative molecular mass.

$mole=\frac{19}{177}$

moles = 0.1073.

2. Molarity (M)

M = moles / volume

$M\:=\:{\frac{ 0.1703 }{75.10^{-3} L}$

M = 1.431.

b. The concentration of luminol in the spray bottle is 6.00 × 10⁻² M. Therefore, in a 2 L solution, the number of moles is:

moles = M × volume

moles = 6 × 10⁻² × 2

moles = 0.12.

c. The molarity of the stock solution (Part A) is 1.431 M.

The diluted solution (Part B) contains 0.12 moles of luminol.

To find the volume of the stock solution (Part A) that has the same moles as the diluted solution (Part B):

volume = moles / M

$volume\:=\:\frac{0.12}{1.431}$

volume = 0.08386 L = 83.86 mL.

Further Learning

moles of water you can generate

the amount of each atom in the chemical's formula

the proportion of hydrogen to oxygen atoms in 2 L of water

Keywords: mole, volume, molarity, Luminol, relative molecular mass

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7 days ago

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A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T

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Answer:

$\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}$

Explanation:

1. Molar concentration

Designate chloroform as C and acetone as A.

The molar concentration for C is derived from Moles of C per Litres of solution.

(a) Moles of C

We are assuming there are 0.187 moles of C.

This resolves that step.

(b) Litres of solution

Next, identify 0.813 moles of A.

(i) Mass of each component

$\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}$

(ii) Volume of each component

$\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}$

(iii) Volume of solution

Assuming mixing doesn't alter the total volume.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

$c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}$

2. Molal concentration of C

Molal concentration is calculated as moles of solute per kilograms of solvent.

Total moles of C = 0.187 mol.

Mass of A = 47.22 g = 0.047 22 kg.

$\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}$

4 0

1 day ago

The [H3O+] in a solution is increased to twice the original concentration. Which change could occur in the pH? 2.0 to 4.0 1.7 to

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Answer: second option: 1.70 to 1.40

Explanation:

1) pH is defined using the formula pH = - log [H₃O⁺]

2) Given that the initial concentration is x and after doubling it becomes 2x, we calculate:

pHi = - logx
pHf = - log 2x = - log 2 - logx

Thus, pHf - pHi = - log2 - logx - (- logx) = - log2 ≈ - 0.30

⇒ pHi - pHf = 0.30, indicating that the final pH (with twice the hydronium ions) is 0.30 lower than the starting pH.

3) The only option that indicates a 0.30 decline in pH is the second one: from 1.70 to 1.40. Therefore, that is the correct choice.

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11 days ago

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If angle ABE = 2n + 7 and angle EBF=4n-13,find angle ABE.​

Additional Information

Further Learning

If angle ABE = 2n + 7 and angle EBF=4n-13,find angle ABE.