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3 years ago

Which of the following questions can be answered by science?

Chemistry

2 answers:

Irina18 [472]3 years ago

7 0

Answer:

A. What is the composition of air?, B. Should the driving age be reduced?, D.

Should students be allowed to bring toys to school? Answer: A,B,D.

Hunter-Best [27]3 years ago

6 0

Answer:

<h2>(a), (b) and (c) I guess</h2>

You might be interested in

Write the significance of Faraday's law?

Gala2k [10]

It helps shape and create the study of electromagnetism

3 0

3 years ago

At 35°C, K = 1.6 × 10^-5 for the reaction

TEA [102]

Answer:

a) [NOCl] = 0.968 M

[NO] = 0.032M

[Cl²] = 0.016M

b) [NOCl] = 1.992M

[NO] = 0.008 M

[Cl2] = 1.004 M

Explanation:

Step 1: Data given

Temperature = 35°C = 308K

K = 1.6 × 10^-5

Step 2: The reaction

2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)

For 2 moles NOCl we'll have 2 moles NO and 1 mol Cl2

Step 3

a. 2.0 mol pure NOCl in a 2.0 L flask

Concentration at the start:

Concentration = mol / volume

[NOCl] = mol / volume

[NOCl] = 2.0 / 2.0 L

[NOCl] = 1.0 M

[NO] = 0 M

[Cl] = 0M

Concentration at the equillibrium

[NOCl] = 1.0M - 2x

[NO] = 2x

[Cl2]= x

K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5

1.6*10^-5 = ((2x)² * x) / (1.0-2x)²

x = 0.016

[NOCl] = 1.0 - 2*0.016 = 0.968 M

[NO] = 2*0.016 = 0.032M

[Cl²] = 0.016M

b. 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0 L flask

Concentration at the equillibrium

[NOCl] = 2.0 mol / 1.0 L = 2.0 M

[NO] = 0 M

[Cl2]= 1.0 mol / 1.0 L = 1.0 M

Concentration at the equillibrium

[NOCl] = 2.0M - 2x

[NO] = 2x

[Cl2]= 1.0 + x

K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5

1.6 *10^-5 = (2x)²*(1.0+x) / ((2.0-2x)²)

1.6 *10^-5= (2x)² * 1 )/2.0²

1.6 *10^-5= 4x² / 4 = x²

x = $\sqrt{1.6 *10^-5}$ = 4.0*10^-3

[NOCl] = 2.0 - 2*0.004 = 1.992M

[NO] = 2*0.004 = 0.008 M

[Cl2] = 1+ 0.004M = 1.004 M

5 0

3 years ago

one parsec equals 3.26 light years, where a light year is the distance light travels in one year. if the speed of light is 18600

Tpy6a [65]

Parsec is a unit of distance (as stated 1 Parsec = 3.26 light years, wihch is 3.26 times the distance run by light in one year).

That distance is:

1 parsec = 3.26 mi× 186,000 mi/s × 3600 s/h × 24 h/day × 365 day/ year × 1 year = 19,122,168,960,000.mi.

So, the question of how many parsecs it takes for light to reach Mars from Earth does not make sense because parsecs is not a unit of time.


You can calculate how many parsecs is equivalent to the distance between Mars and Earth, 60,000,000 km. For this you can first calculate the equivalence of which you do in this way:


60,000,000 km × 0,621 mi/km × 1 parsec / ( 19,122,168,960,000 mi) = 1.94E-6 parsecs.

7 0

3 years ago

For the reaction Cu2S(s)⇌2Cu+(aq)+S2−(aq)Cu2S(s)⇌2Cu+(aq)+S2−(aq), the equilibrium concentrations are as follows: [Cu+]=1.0×10−5

klio [65]

Answer:

1x10⁻¹²

Explanation:

Cu₂S(s) ⇌ 2Cu⁺(aq) + S²⁻(aq)

At equilibrium:

[Cu⁺] = 1.0x10⁻⁵ M

[S²⁻] = 1.0x10⁻² M

The equilibrium constant for the the reaction can be written as:

Keq = [Cu⁺]² * [S²⁻]

[Cu⁺] is squared because it has a stoichiometric coefficient of 2 in the reaction. Cu₂S has no effect on the constant because it is a solid.

Now we can calculate the equilibrium constant:

Keq = (1.0x10⁻⁵)² * 1.0x10⁻² = 1x10⁻¹²

8 0

3 years ago

How much time would it take for 9000 atoms of nobellium-253 (half-life is 97 s) to decay to 2250 atoms?

Svetradugi [14.3K]

Answer:it’s 828292atoms

Explanation

7 0

3 years ago