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tresset_1 [31]
3 years ago
14

Are hydrocarbons approved for retrofit applications

Chemistry
1 answer:
Alborosie3 years ago
3 0
Yes they are what are your options
You might be interested in
when atoms react, they often lose, gain, or share electrons to form a more stable version of themselves. for example, alkali met
AfilCa [17]

The electron configuration of alkali metals would then resemble those of group 17 of the periodic table in the compounds they form.

<h3></h3><h3>What is periodic table?</h3>

Periodic table is defined as a tabular approach of showing the items so that they appear in the same vertical column or group when their attributes are similar. Phosphorus is the oldest chemical element, and hassium is the newest. Please take note that, unlike in the Periodic system, the elements do not exhibit their natural relationships with one another.

The elements that make up group 17 of the periodic table are the halogens. They are nonmetals that are reactive, such as iodine, bromine, chlorine, and fluorine. Halogens are non-metals that are very reactive. These substances share a lot of characteristics with one another.

Thus, the electron configuration of alkali metals would then resemble those of group 17 of the periodic table in the compounds they form.

<h3></h3>

To learn more about periodic table, refer to the link below:

brainly.com/question/11155928

#SPJ2

8 0
1 year ago
Read 2 more answers
This exercise uses the radioactive decay model.
Anna11 [10]
For radioactive decay, we can relate current amount, initial amount, decay constant and time using:
N = No x exp(-λt)

Half-life = ln(2)/λ
λ = ln(2) / 5730
N/No = 80% = 0.8
0.8 = exp( -ln(2)/5730 x t)
t = 1844 years
3 0
3 years ago
A generic weak acid with formula HA has a Ka = 2.76 x 10-8. Calculate the Kb for the conjugate base of the acid.
Reika [66]

Answer:

3.62x10⁻⁷ = Kb

Explanation:

The acid equilibrium of a weak acid, HX, is:

HX + H₂O ⇄ X⁻ + H₃O⁺

Where Ka = [X⁻] [H₃O⁺] / [HX]

And basic equilibrium of the conjugate base, is:

X⁻ + H₂O ⇄ OH⁻ + HX

Where Kb = [OH⁻] [HX] / [X⁻]

To convert Ka to Kb we must use water equilibrium:

2H₂O ⇄ H₃O⁺ + OH⁻

Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]

Thus, we can obtain:

Kw = Ka*Kb

Solving for Kb:

Kw / Ka = Kb

1x10⁻¹⁴ /  2.76x10⁻⁸ =

3.62x10⁻⁷ = Kb

4 0
3 years ago
A pycnometer is a precisely weighted vessel that is used for highly accurate density determinations. Suppose that a pycnometer h
dexar [7]

Answer:

5.758  is the density of the metal ingot in grams per cubic centimeter.

Explanation:

1) Mass of pycnometer = M = 27.60 g

Mass of pycnometer with water ,m= 45.65 g

Density of water at 20 °C = d =998.2 kg/m^3

1 kg = 1000 g

1 m^3=10^6 cm^3

998.2 kg/m^3=\frac{998.2 \times 1000 g}{10^6 cm^3}=0.9982 g/cm^3

Mass of water ,m'= m - M = 45.65 g -  27.60 g =18.05 g

Volume of pycnometer = Volume of water present in it = V

Density=\frac{Mass}{Volume}

V=\frac{m'}{d}=\frac{18.05 g}{0.9982 g/cm^3}=18.08 cm^3

2) Mass of metal , water and pycnometer = 56.83 g

Mass of metal,M' =  9.5 g

Mass of water when metal and water are together ,m''= 56.83 g - M'- M

56.83 g - 9.5 g - 27.60 g = 19.7 g

Volume of water when metal and water are together = v

v=\frac{m''}{d}=\frac{19.7 g}{0.9982 g/cm^3}=19.73 cm^3

Density of metal = d'

Volume of metal = v' =\frac{M'}{d'}

Difference in volume will give volume of metal ingot.

v' = v - V

v'=19.73 cm^3-18.08 cm^3=

v'=1.65 cm^3

Since volume cannot be in negative .

Density of the metal =d'

=d'=\frac{M'}{v'}=\frac{9.5 g}{1.65 cm^3}=5.758 g/cm^3

5 0
3 years ago
In h2o, the type of bond that holds one of the hydrogen atoms to the oxygen atom is a
Murrr4er [49]
The hydrogen and oxygen<span> atoms from H</span>₂O are <span>bonded together through covalent </span>bonding. 
3 0
3 years ago
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