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3 years ago

Girls are usually completely physically mature by _____ years old.

Chemistry

2 answers:

fomenos3 years ago

6 0

Answer:

Explanation:

edge 2020 i just took the topic test

krok68 [10]3 years ago

5 0

Answer:

a 17 bc it is how it works

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If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?

Semenov [28]

Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

Consider the following reaction: 

SO2Cl2 -----> SO2(g) + Cl2(g) 

Kc= 2.99 x 10^-7 at 227 degrees celcius 

If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

SOCl2 ---------> SO2 + Cl2 Kc = 2.99x10⁻⁷

i) 0.168 0 0

c) -x +x +x

e) 0.168-x x x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

This means then, that the concentration of Cl2 in equilibrium would be:

[Cl₂] = 2.24x10⁻⁴ M

5 0

3 years ago

Negatively charged particles within the atom are called what?

Ivahew [28]

Negatively charged particles in atom - Electrons

5 0

3 years ago

As part of a soil analysis on a plot of land, a scientist wants to determine the ammonium content using gravimetric analysis wit

jeka94

Answer:

Mass percentage of NH₄Cl = 3.54%

Mass percentage of K₂CO₃ = 1.01%

Explanation:

If a 200.0 mL aliquot produced 0.105 g of KB(C₆H₅)₄, then a 100.0 mL aliquot would produce 1/2 * 0.105 g = 0.0525 g of KB(C₆H₅)₄.

Therefore, mass of NH₄B(C₆H₅)₄ in the 100.0 ml aliquot = (0.277 - 0.0525)g = 0.2245 g

Number of moles of NH₄B(C₆H₅)₄ in 0.2245 g = 0.2245 g/ 337.27 g/mol = 0.0006656 moles

In 500 ml solution, number of moles present = 0.0006656 * 500/100 = 0.003328 moles.

From equation of the reaction; mole ratio of NH₄⁺ and NH₄B(C₆H₅)₄ = 1:1

Similarly, mole ratio of NH₄⁺ and NH₄Cl = 1:1

Therefore, moles of NH₄Cl in 500 ml sample = 0.003328 moles

Mass of NH₄Cl = 0.003328 mol * 53.492 g/mol = 0.178 g

Mass percentage of NH₄Cl = (0.178/5.025) * 100% = 3.54%

Number of moles of KB(C₆H₅)₄ in 0.105 g (precipitated from 200.0 ml aliquot) = 0.105 g/ 358.33 g/mol = 0.000293 moles

In 500 ml solution, number of moles present = 0.000293 * 500/200 = 0.0007326 moles.

From equation of the reaction; mole ratio of K⁺ and KB(C₆H₅)₄ = 1:1

Similarly, mole ratio of K⁺ and K₂CO₃ = 2:1

Therefore, moles of K₂CO₃ in 500 ml sample = 0.0007326/2 moles = 0.0003663 moles

Mass of K₂CO₃ = 0.0003663 mol * 138.21 g/mol = 0.05063 g

Mass percentage of K₂CO₃ = (0.05063/5.025) * 100% = 1.01%

7 0

4 years ago

Chemical reactions forming ions _____.

Sergeu [11.5K]

The answer should be balance electrically

Chemical reactions that form ions should have a balanced charge. The example of the reaction is HCl. When forming ions, the equation should be:
HCl => $H^{+}$ + $Cl^{-}$
In this case, the hydrogen has one plus charge and chlorine has one negative charge. The resultant should be zero, so it's balanced.

4 0

3 years ago

Which college classes would a student most likely take to become an oceanographer? Select 4 choices.

Setler [38]

Answer:

marine biology

coastal ecology

astronomy

meteorology

Explanation:

As a college student, to study oceanography one will have to take classes in the field of marine biology, coastal ecology, astronomy and meteorology.

An oceanographer is someone or a professional that studies the ocean in order have more scientific knowledge about them.

Oceanography is a merger of geology, biology, chemistry, physics as it pertains to the ocean.

There is no need to study human anatomy to study oceanography.
Marine biology and coastal ecology deals with study of life forms in their marine environment.
Astronomy and meteorology helps to gain insight about the formation of the ocean and how weather relates to the ocean.

5 0

3 years ago