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2 years ago

Elemental sulfur is one of the products of the gas-phase reaction of nitric acid and hydrogen sulfide. The other

Chemistry

1 answer:

Goshia [24]2 years ago

8 0

Answer:

The manufacturing processes for liquefied petroleum gas are designed so that the majority, if not all, of the sulfur compounds are removed. The total sulfur level is therefore considerably lower than for other crude oil-based fuels and a maximum limit for sulfur content helps to define the product more completely. The sulfur compounds that are mainly responsible for corrosion are hydrogen sulfide, carbonyl sulfide and, sometimes, elemental sulfur. Hydrogen sulfide and mercaptans have distinctive unpleasant odors. A control of the total sulfur content, hydrogen sulfide and mercaptans ensures that the product is not corrosive or nauseating. Stipulating a satisfactory copper strip test further ensures the control of the corrosion.

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Calculate Î”HÂ° = for the chemical reaction

babymother [125]

Answer: $\Delta H^0$ for the reaction $Cl_2(g)+F_2(g)\rightarrow 2ClF(g)$ is -108kJ

Explanation:

The balanced chemical reaction is,

The expression for enthalpy change is,

$\Delta H=[n\times B.E_{reactants}]-[n\times B.E{products}]$

where, n = number of moles

Now put all the given values in this expression, we get

$\Delta H=[1\times B.E_{Cl_2}+1\times B.E_{F_2}]-[2\times B.E_{ClF}]$

$\Delta H=[(1\times 243)+(1\times 159)]-[2\times 255]$

$\Delta H=-108kJ$

Thus $\Delta H^0$ for the reaction $Cl_2(g)+F_2(g)\rightarrow 2ClF(g)$ is -108kJ

4 0

3 years ago

What is neutralization reaction?

Verizon [17]

It's basically when a base and an acid of some kind, come together to make H2O or Water.

7 0

3 years ago

Read 2 more answers

The formula for bismuth(III) arsenide is..?

vlabodo [156]

Answer:

BiAs

Explanation:

6 0

3 years ago

Find the de Broglie wavelength lambda for an electron moving at a speed of 1.00 \times 10^6 \; {\rm m/s}. (Note that this speed

masya89 [10]

(A) $7.28\cdot 10^{-10} m$

The De Broglie wavelength of an electron is given by

$\lambda=\frac{h}{p}$ (1)

where

h is the Planck constant

p is the momentum of the electron

The electron in this problem has a speed of

$v=1.00\cdot 10^6 m/s$

and its mass is

$m=9.11\cdot 10^{-31} kg$

So, its momentum is

$p=mv=(9.11\cdot 10^{-31} kg)(1.00\cdot 10^6 m/s)=9.11\cdot 10^{-25}kg m/s$

And substituting into (1), we find its De Broglie wavelength

$\lambda=\frac{6.63\cdot 10^{-34}Js}{9.11\cdot 10^{-25} kg m/s}=7.28\cdot 10^{-10} m$

(B) $1.16\cdot 10^{-34}m$

In this case we have:

m = 0.143 kg is the mass of the ball

v = 40.0 m/s is the speed of the ball

So, the momentum of the ball is

$p=mv=(0.143 kg)(40.0 m/s)=5.72 kg m/s$

And so, the De Broglie wavelength of the ball is given by

$\lambda=\frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{5.72 kg m/s}=1.16\cdot 10^{-34}m$

(C) $9.02\cdot 10^{-9}m$

The location of the first intensity minima is given by

$y=\frac{L\lambda}{a}$

where in this case we have

$y=0.492 cm = 4.92\cdot 10^{-3} m$

L = 1.091 is the distance between the detector and the slit

$a=2.00\mu m=2.00\cdot 10^{-6}m$ is the width of the slit

Solving the formula for $\lambda$ , we find the wavelength of the electrons in the beam:

$\lambda=\frac{ya}{L}=\frac{(4.92\cdot 10^{-3}m)(2.00\cdot 10^{-6} m)}{1.091 m}=9.02\cdot 10^{-9}m$

(D) $7.35\cdot 10^{-26}kg m/s$

The momentum of one of these electrons can be found by re-arranging the formula of the De Broglie wavelength:

$p=\frac{h}{\lambda}$

where here we have

$\lambda=9.02\cdot 10^{-9}m$ is the wavelength

Substituting into the formula, we find

$p=\frac{6.63\cdot 10^{-34}Js}{9.02\cdot 10^{-9}m}=7.35\cdot 10^{-26}kg m/s$

7 0

3 years ago

The surface of the moon is

babymother [125]

Answer:

...gray

Explanation:

5 0

2 years ago

Read 2 more answers