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3 years ago

Calculate the wavelength of light with an energy of 3.60x10^-19 J *

Chemistry

1 answer:

Travka [436]3 years ago

3 0

Explanation:

from E = mc^2 and de Broglie wavelength=h/mc

combine the eqns with respect to m(considering the body is moving with the speed of light(c))

then

E= hc/de Broglie wavelength

since, c= 3× 10^ 8

h= 6.626 × 10^ -34 and E is given, substitute the values.

sorry I don't have my calculator.

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A student performed a chemical reaction in which 35 grams of hydrogen and 65 grams of oxygen reacted to form water. What is the

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Answer:

The mass of the product is 73,06 g

Explanation:

The reaction to form water is:

2H2 + O2 ---> 2H2O

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Molar mass O2 = 32g/m ---> I have 2,03 moles (65g / 32g/m)

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17,5 moles of H2 __ reacts with ___ (17,5 m . 1 m) / 2 m = 8,75 m

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2,03 moles of O2 __ reacts with (2,03 m . 2 m) / 1 m = 4,06 m

I have 17,5 moles of H2 and I need 4,06 m. H2 is my excess reagent

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1 mol O2 are required____ to form 2 H2O

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3 years ago

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2 years ago

A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base

tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ = 0.00375 mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

$HN_3 + OH- ---> N^-_{3} + H_2O$

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L

Concentration of $HN_3$ = $\dfrac{0.001755}{0.0383}$ = 0.0458 M

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GIven that :

Ka = $1.9 x 10^{-5}$

Thus; it's pKa = 4.72

$pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})$

$pH =4.72 + log(1.1376)$

$pH =4.72 + 0.05598$

$pH =4.77598$

pH ≅ 4.80

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