Which of the following is the most likeley example of an favorable mutation

Why can scientist assume elastic collisions as long as the temperature remains constant?

Tema [17]

Answer:

Because in elastic collisions there is no heat emission or absorption.

Explanation:

A collision is considered elastic when the total kinetic energy of the study system is conserved during the collision. Since the total kinetic energy is conserved, heat is not emitted or absorbed during the collision. Since the emission or absorption of heat is what produces changes in temperature, If the system remains at a constant temperature, there were only elastic collisions.

8 0

3 years ago

A 24.00 mL sample of a solution of Pb(ClO3)2 was diluted with water to 52.00 mL. A 17.00 mL sample of the dilute solution was fo

klio [65]

Answer:

0.238 M

Explanation:

A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.

C₁.V₁ = C₂.V₂

C₁ × 24.00 mL = 0.220 M × 52.00 mL

C₁ = 0.477 M

The concentration of Pb(ClO₃)₂ is:

$\frac{0.477molClO_{3}^{-} }{L} \times \frac{1molPb(ClO_{3})_{2}}{2molClO_{3}^{-}} =0.238M$

4 0

3 years ago

A mechanical pencil has a mass of 47.4 grams. The volume of the pencil is 15.8 cubic centimeters. What is the density of the pen

BlackZzzverrR [31]

Density = mass over volume
so
47.4g/15.8cm^3
=3g/cm^3

4 0

3 years ago

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Which transition metals usually form only one monatomic cation? select all that apply.

lyudmila [28]

Zn, Cd, and Ag are transition metals that usually form only one monoatomic cation.

A monatomic cation is a cation made of only one atom.

Cations are positively charged ions, in this example Ag⁺, Cd²⁺ and Zn²⁺.

These cations form only one type of ion, while iron and copper form more than one type of cations.

Iron and copper form cations with different charges (Fe²⁺, Fe³⁺, Cu⁺, Cu²⁺).

It depends on electron configuration which type would be formed.

Electron configuration of zinc atom: ₃₀Zn 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²

Transition metals are elements in the d-block of the Periodic table.

More about transition metals: brainly.com/question/12843347

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5 0

1 year ago

7. The equilibrium constant Kc for the reaction H2(g) + I2(g) ⇌ 2 HI(g) is 54.3 at 430°C. At the start of the reaction there are

Juli2301 [7.4K]

Answer:

[H2] = 0.0692 M

[I2] = 0.182 M

[HI] = 0.826 M

Explanation:

Step 1: Data given

Kc = 54.3 at 430 °C

Number of moles hydrogen = 0.714 moles

Number of moles iodine = 0.984 moles

Number of moles HI = 0.886 moles

Volume = 2.40 L

Step 2: The balanced equation

H2 + I2 → 2HI

Step 3: Calculate Q

If we know Q, we know in what direction the reaction will go

Q = [HI]² / [I2][H2]

Q= [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Q =(n(HI)²) /(nH2 *nI2)

Q = 0.886²/(0.714*0.984)

Q =1.117

Q<Kc This means the reaction goes to the right (side of products)

Step 2: Calculate moles at equilibrium

For 1 mol H2 we need 1 mol I2 to produce 2 moles of HI

Moles H2 = 0.714 - X

Moles I2 = 0.984 -X

Moles HI = 0.886 + 2X

Step 3: Define Kc

Kc = [HI]² / [I2][H2]

Kc = [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Kc =(n(HI)²) /(nH2 *nI2)

KC = 54.3 = (0.886+2X)² /((0.714 - X)*(0.984 -X))

X = 0.548

Step 4: Calculate concentrations at the equilibrium

[H2] = (0.714-0.548) / 2.40 = 0.0692 M

[I2] = (0.984 - 0.548) / 2.40 = 0.182 M

[HI] = (0.886+2*0.548) /2.40 = 0.826 M

6 0

3 years ago

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