This group of elements contains brittle solids at room temperature. They

Y_Kistochka [10]

D. Radio waves

In order it would be Radio wave, Microwave, Infrared, visible, Ultraviolet, x-ray, Gamma ray

5 0

3 years ago

The molecular structure of water contains two atoms of hydrogen and one atom of oxygen. When water reaches its boiling point and

Alexxandr [17]

It remains the same . The molecules are further apart thats all.

7 0

3 years ago

"An increase in the price of gasoline will increase the demand for hybrid vehicles." This statement is an example of a positive

-Dominant- [34]

Answer:

True

Explanation:

This is because the statement above fits the definition of a positive economic statement. A Positive economic statement refers to objective statements that can be verified, tested , amended or discarded by consulting available evidence .The statement above can be tested objectively to establish its veracity.

3 0

3 years ago

Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→f

Zanzabum

K = 5.0 × 10^25

<h2>Part (a). Calculate E° for the reaction </h2>

Step 1. Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

Step 2. Identify the cathode and the anode

The half-cell with the more negative E° (Zn) is the anode.

Step 3. Calculate E°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

E° = +0.76 V

<h2>Part (b). Calculate K for the reaction </h2>

The relation between E° and K is

E° = (RT)/(nF)lnK

where

R = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

T = the Kelvin temperature

n = the moles of electrons transferred

F = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]lnK

0.76 = 0.012 85 lnK

lnK = 0.76/0.012 85 = 59.16

K =e^59.16 = 5.0 × 10^25

4 0

3 years ago

If 36.2g of Acetic Acid (HC2H302) was dissolved in 300. mL of water, what is the

uysha [10]

Answer:

2.01 M

Explanation:

Step 1: Calculate the moles of acetic acid (HC₂H₃O₂)

The molar mass of acetic acid is 60.05 g/mol. We will use this data to calculate the moles corresponding to 36.2 g of acetic acid.

$36.2g \times \frac{1mol}{60.05g} = 0.603mol$

Step 2: Convert the volume of solution to liters

We will use the relation 1000 mL = 1 L. We assume that the volume of solution is that of water (300 mL)

$300mL \times \frac{1L}{1000mL} = 0.300L$

Step 3: Calculate the molarity of the solution

The molarity is equal to the moles of solute (acetic acid) divided by the liters of solution

$M = \frac{0.603mol}{0.300L} = 2.01 M$

7 0

3 years ago