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timama [110]
3 years ago
10

[HELP PLEASE]What is the electronegativity from largest to smallest for Chlorine [Cl], Carbon [C], and Arsenic [As]? (hint: use

your interactive periodic table)
Chemistry
1 answer:
valentina_108 [34]3 years ago
8 0

Answer:

Electronegativity varies in a predictable way across the periodic table. Electronegativity increases from bottom to top in groups, and increases from left to right across periods. Thus, fluorine is the most electronegative element, while francium is one of the least electronegative.Explanation:

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1. Ibuprofen (C13H18O2) is the active ingredient in many nonprescription pain relievers. Each tablet contains 200 mg of ibuprofe
Dmitry_Shevchenko [17]

Answer :

The molar mass of ibuprofen is, 206.29 g/mole.

The number of moles of ibuprofen in a single tablet is, 0.000969 moles

The number of moles of ibuprofen in four doses is, 0.007752 moles

Solution : Given,

Molar mass of carbon = 12.01 g/mole

Molar mass of hydrogen = 1.01 g/mole

Molar mass of oxygen = 15.99 g/mole

1) Now we have to calculate the molar mass of ibuprofen.

Molar mass of ibuprofen, C_{13}H_{18}O_2 = (13\times 12.01)+(18\times 1.01)+(2\times 15.99)=206.29g/mole

The molar mass of ibuprofen = 206.29 g/mole

2) Now we have to calculate the moles of ibuprofen.

Formula used : Moles=\frac{Mass}{\text{ Molar mass}}

Given : Mass of ibuprofen = 200 mg = 0.2 g         (1 mg = 1000 g)

\text{ Moles of ibuprofen}=\frac{\text{ Mass of ibuprofen}}{\text{ Molar mass of ibuprofen}}=\frac{0.2g}{206.29g/mole}=0.000969moles

The moles of ibuprofen = 0.000969 moles

3) Now we have to calculate the number of moles of ibuprofen for four doses.

Number of tablets in one dose = 2

Total number of tablets in 4 doses = 4 × 2 = 8

Number of moles of ibuprofen in 8 tablets =

\text{ Number of moles of ibuprofen in 1 tablet}\times \text{ Total number of tablets}=0.000969\times 8=0.007752moles

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2 years ago
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Irina18 [472]

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Relative motion of source and observer.

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8 0
3 years ago
If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×
Elodia [21]

Answer:

2.30 × 10⁻⁶ M

Explanation:

Step 1: Given data

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Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³

Step 2: Write the reaction for the solution of Mg(OH)₂

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Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂

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