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2 years ago

Which best describes the effect of jj Thomson discovery

1 answer:

7 0

JJ Thomson discovered electrons using the cathode ray. Electrons are negatively charged particles that orbit around the nucleus of an atom. Finding out that there was a negative charge to balance out the positive charge. This later helped develop Hund's Rule and the Pauli Exclusion Principle.

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Conociendo el volumen de la solución y la masa del soluto y su masa molar, ¿qué concentración es posible determinar

NISA [10]

Answer:

Conociendo el volumen de solución, masa de soluto y su masa molar, es posible determinar: B) Concentración molar

La molaridad es la relación entre el número de moles de soluto y los litros de solución. Más:

M = No moles de solución de soluto / volumen (L)

Y a su vez los moles de soluto se encuentran por:

No moles de soluto = masa soluto / masa molar soluto

8 0

2 years ago

Which of kbr or ch3br is likely to have the higher normal boiling point?

allsm [11]

Answer:

KBr

Explanation

KBr is an ionic compound which have very high BP since they exist as 3D crystals

Methyl Bromide, CH3Br. is a covalent compound which have low BP since they exist as individual molecules

8 0

2 years ago

Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m

erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

$n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3} =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2} =0.06882molBr^-$

Now, we compute the total moles of bromide:

$n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol$

Then, the total volume in liters:

$150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L$

Therefore, the concentration of total bromide is:

$M=\frac{0.12338mol}{0.325L}\\\\M=0.380M$

Best regards!

8 0

2 years ago

Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is lib

KengaRu [80]

Answer:

For methanol: Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol: Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol: Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

Explanation:

Given:

Mass of Methanol = 1.0 g

Mass of ethanol = 1.00 g

Mass of n-propanol = 1.00 g

For methanol:

2 CH₃OH + 3 O₂ ----> 2 CO₂ + 4 H₂O, ∆H₀ = -22.6 kJ/g (negative sign signifies release of heat)

1 g of methanol on combustion gives 22.6 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 32.04 g/mol

Thus moles of methanol = 1 g/ (32.04 g/mol) = 0.0312 moles

Hence energy in kJ/mol:

Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol:

C₂H₅OH + 3 O₂ ----> 2 CO₂ + 3 H₂O, ∆H₀ = -29.7 kJ/g (negative sign signifies release of heat)

1 g of ethanol on combustion gives 29.7 kJ of energy

Calculation of moles of ethanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of ethanol = 46.07 g/mol

Thus moles of ethanol = 1 g/ (46.07 g/mol) = 0.0217 moles

Hence energy in kJ/mol:

Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol:

2 C₃H₇OH + 9 O₂ ----> 6 CO₂ + 8 H₂O, ∆H₀ = -33.4 kJ/g , (negative sign signifies release of heat)

1 g of methanol on combustion gives 33.4 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 60.09 g/mol

Thus moles of methanol = 1 g/ (60.09 g/mol) = 0.0166 moles

Hence energy in kJ/mol:

Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

5 0

2 years ago

What is the hybridization and shape for an XeF6 2+ molecule?

Anna11 [10]

Answer:

Xe:[Kr]4d¹⁰5(sp³d³)₆⁺² => Octahedral Geometry (AX₆)⁺²

Explanation:

Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆²

Ca. #Valence e⁻ = Xe + 6F - 2e⁻ = 1(8) + 6(7) - 2 = 48

Ca. #Substrate e⁻ = 6F = 6(8) = 48

#Nonbonded free pairs e⁻ = (V - S)/2 = (48 - 48)/2 = 0 free pairs

#Bonded pairs e⁻ = 6F substrates = 6 bonded pairs

BPr + NBPr = 6 + 0 = 6 e⁻ pairs => Geometry => [AX₆]⁺² => Octahedron

Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆⁺²

XeF₆⁺² => 6(sp³d³) hybrid orbitals => Octahedral Geometry (AX₆)

8 0

2 years ago