Who found chemistry who is the father of chemistry

Give your familiarity for following terms

KiRa [710]

Answer:

The roasting process is a delicate combination of art and science . Roasters are familiar with how the beans look and the smells Well, familiarity with the machine makes things much easier to predict, but the best way is to do many different tests .Well, familiarity with the machine makes things much easier to predict, but the best way is to do many different tests.
Smelting is a process of applying heat to ore in order to extract a base metal. It is a form of extractive metallurgy. It is used to extract many metals from their ores, including silver, iron, copper, and other base metals.In zone refining, solutes are segregated at one end of the ingot in order to purify the remainder, or to concentrate the impurities. For example, in the preparation of a transistor or diode semiconductor, an ingot of germanium is first purified by zone refining.
 In zone refining, solutes are segregated at one end of the ingot in order to purify the remainder, or to concentrate the impurities. ... For example, in the preparation of a transistor or diode semiconductor, an ingot of germanium is first purified by zone refining.
Polling is the process where the computer or controlling device waits for an external device to check for its readiness or state, often with low-level hardware. For example, when a printer is connected via a parallel port, the computer waits until the printer has received the next character.

Explanation:

hope it heloed

3 0

3 years ago

The photodissociation of ozone by ultraviolet light in the upper atmosphere is a first-order reaction with a rate constant of 1.

atroni [7]

Answer:

[O₃]= 8.84x10⁻⁷M

Explanation:

The photodissociation of ozone by UV light is given by:

O₃ + hν → O₂ + O (1)

The first-order reaction of the equation (1) is:

$rate = k [O_{3}] = - k \frac{\Delta [O_{3}]}{\Delta t}$ (2)

where k: is the rate constant and Δ[O₃]/Δt: is the variation in the ozone concentration with time, and the negative sign is by the decrease in the reactant concentration

We can get the following expression of the first-order integrated law of the reaction (1), by resolving the equation (2):

$[O_{3}]_{t} = [O_{3}]_{0} \cdot e^{-kt}$ (3)

where [O₃](t): is the ozone concentration in the elapsed time and [O₃]₀: is the initial ozone concentration

We can calculate the initial ozone concentration using equation (3):

$[O_{3}]_{t} = 5.0 \cdot 10^{-3}M \cdot e^{-(1.0\cdot 10^{-5}s^{-1}) (\frac{10d \cdot 24h \cdot 3600 s}{1d \cdot 1h})} = 8.84 \cdot 10^{-7}M$

So, the ozone concentration after 10 days is 8.84x10⁻⁷M.

I hope it helps you!

3 0

4 years ago

What is the electron configuration of bromine in this oxidation state? Express your answer in condensed form, in order of increa

Ymorist [56]

Answer:

[Ar]3d^{10}4s^{2}4p^{6}

Explanation:

Electronic Configuration of Bromine : [Ar]3d^{10}4s^{2}4p^{5}

At oxidation state -1, electronic Configuration of Bromine : [Ar]3d^{10}4s^{2}4p^{6}

8 0

3 years ago

2. (8 pts) Boric acid (H3BO3) has three hydrogens in a molecule, but effectively acts as a monoprotic acid (Ka = 5.8∙10-10), sin

Andrew [12]

Answer:

7.85 g H₃BO₃

2.92 g NaOH

Explanation:

The strategy for solving this question is to first utilize the Henderson- Hasselbach equation to calculate the ratio of conjugate base concentration to weak acid:

pH = pKa + log [A⁻]/ [HA]

In this case:

pH = pKa + log [H₂BO₃⁻]/[H₃BO₃]

We know pH and indirectly pKa ( = - log Ka ).

9.00 = -log(5.8 x 10⁻¹⁰) + log [H₂BO₃⁻]/[H₃BO₃]

9.00 = 9.24 + log [H₂BO₃⁻]/[H₃BO₃]

log [H₂BO₃⁻]/[H₃BO₃] = - 0.24

taking inverse log function to both sides of the equation:

[H₂BO₃⁻]/[H₃BO₃] = 10^-0.24 = 0.58

We are also told we want to have a total concentration of boron of 0.200 mol/L, and if we call x the concentration of H₂BO₃⁻ and y the concentration of H₃BO₃, it follows that:

x + y = 0.200 ( since we have 1 Boron atom per formula of each compound)

and from the Henderson Hasselbach calculation, we have that

x / y = 0.58

So we have a system of 2 equations with two unknowns, which when solved give us that

x = 0.073 and y = 0.127

Because we are told the volume is one liter it follows that the number of moles of boric acid and the salt are the same numbers 0.073 and 0.127

gram boric acid = 0.127 mol x molar mass HBO₃ = 0127 mol x 61.83 g/mol

= 7.85 g boric acid

grams NaOH = 0.073 mol x molar mass NaOH = 0.073 x 40 g/mol

= 2.92 g NaOH

4 0

4 years ago

What is the number of electrons shared between the atoms in a I2 molecule

vova2212 [387]

The number of electrons shared between the atoms in a I₂ molecule is two (2).

Explanation:

An iodine atom has 53 electrons and its electron configuration is;

2. 8. 8. 18. 17

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁵

Note that the outermost p shell has 5 electrons while it can accommodate a maximum of 6 for stable electron configuration. Therefore an Iodine atom needs one (1) electron to acquire nobility. Two Iodine atoms will, therefore, share two electrons (each from one of their p sub-shell ) by covalent bonds to form I₂

Learn More:

For more on electron configuration and nobility check out;

brainly.com/question/5246773

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brainly.com/question/10642978

#LearnWithBrainly

5 0

4 years ago