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1 year ago

In this experiment, simple chemical reactions are used to generate small quantities of hydrogen and

1 answer:

6 0

One of the industrial method for the production of hydrogen is the steam reforming process while oxygen is air separation.

<h3>Industrial method of hydrogen preparation </h3>

One of the industrial method for the production of hydrogen is the steam reforming process also known catalytic steam–hydrocarbon process.

In catalytic steam–hydrocarbon process, gaseous or vaporized hydrocarbons are treated with steam at high pressure over a nickel catalyst at 650°–950° C to produce carbon oxides and hydrogen.

<h3>Industrial method of oxygen preparation </h3>

One of the industrial method of oxygen production is the separation of air using either a cryogenic distillation process or a vacuum swing adsorption process.

Learn more about hydrogen production here: brainly.com/question/25597694

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Is BH2F a gas or a liquid and what is the boiling point

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3 years ago

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3 years ago

Consider the solubilities of a particular solute at two different temperatures.

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2 years ago

At its critical point, ammonia has a density of 0.235 g cm23. You have a special thickwalled glass tube that has a 10.0mm outs

salantis [7]

Answer:

$\large \boxed{\text{69.3 mg}}$

Explanation:

1. Volume of sealed tube

Assume the sealed tube is a right circular cylinder in which the cap and the base are also 4.20 mm thick.

Its outside dimensions are 155 mm long × 10.0 mm diameter.

Its inside dimensions are

h = 155 mm - 2 × 4.20 mm = 146.6 mm

r = 5.0 mm - 4.20 mm = 0.8 mm

V = πr²h = π(0.8)²× 146.6 mm³ = 294.8 mm³ = 0.2948 cm³

2. Calculate the mass of NH₃

$\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\0.235 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{0.2948 cm}^{3}}\\\\m & = & \text{0.0693 g}\\& = & \textbf{69.3 mg}\\\end{array}\\\text{You must seal $\large \boxed{\textbf{69.3 mg}}$ of ammonia in the tube.}$

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3 years ago

Calculate the energy of a photon of 765 nm wavelength light in J.

weqwewe [10]

Answer:

2.6×10^-19 J/photon

Explanation:

E of photon = h × ν

where h= 6.63 × 10^-34 j.s

v= C ÷ λ

E = ( h × c) ÷ λ

E = (6.63 × 10^-34 × 3.00 ×10^8 ) ÷ ( 765 × 10^-9)

E = 2.6×10^-19 J/photon

3 0

2 years ago