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2 years ago

HELPPPPP

Chemistry

1 answer:

AVprozaik [17]2 years ago

4 0

A. electron, B. Nucleus

Proton is positive charge, electron is negative charge

proton weight, 1 AMU

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Find the energy in kJ for an x-ray photon with a frequency of 2.4 × 10^18 1/s.

Ilya [14]

Answer:

15912 × 10∧-19 KJ

Explanation:

Given data;

frequency of photon = 2.4 × 10^18 1/s.

Planck's constant = 6.63 × 10∧-34 j.s

Energy = ?

Formula:

E = h × ν

E = 6.63 × 10∧-34 j.s × 2.4 × 10^18 1/s

E= 15.912 × 10∧-16 j

now we will convert the joule into kilo joule,

E = 15.912 × 10∧-16 j /1000 = 15.912 × 10∧-19 KJ

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3 years ago

375 mL of a 0.88 M potassium hydroxide solution is added to 496 mL of a 0.76 M cesium hydroxide solution. Calculate the pOH of t

AysviL [449]

Answer:

pOH of resulting solution is 0.086

Explanation:

KOH and CsOH are monoacidic strong base

Number of moles of $OH^{-}$ in 375 mL of 0.88 M of KOH = $\frac{0.88\times 375}{1000}moles$ = 0.33 moles

Number of moles of $OH^{-}$ in 496 mL of 0.76 M of CsOH = $\frac{0.76\times 496}{1000}moles$ = 0.38 moles

Total volume of mixture = (375 + 496) mL = 871 mL

Total number of moles of $OH^{-}$ in mixture = (0.33 + 0.38) moles = 0.71 moles

So, concentration of $OH^{-}$ in mixture, $[OH^{-}]$ = $\frac{0.71}{871}\times 1000M=0.82M$

Hence, $pOH=-log[OH^{-}]=-log(0.82)=0.086$

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3 years ago

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Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C

ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)

The enthalpy of reaction (1) is given by:

$\Delta H = \Delta H_{r} - \Delta H_{p}$ (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond

7O₂: 7 moles of 1 O=O bond

4CO₂: 4 moles of 2 C=O bonds

6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

$\Delta H = \Delta H_{r} - \Delta H_{p}$

$\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})$

$\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})$

$\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol$

$\Delta H = -2860 kJ/mol$

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.