A compound that changes color when it is in contact with an acid or base is called

Which statement correctly describes an atom of the element helium? A. An atom of helium has a full outer energy level and is the

Citrus2011 [14]

The correct description for an atom of helium would be option C. An atom of helium has its valence electrons in its first energy level, it wouldn't and can't satisfy the Octet rule as it only has 2 electrons, but with 2, it has a full shell, as the first energy level can hold only 2 electrons.

4 0

3 years ago

Bubba falls out of a plane flying over his peanut field. What will bubba speed be after falling for 8 seconds?

chubhunter [2.5K]

Variables we know:

t = 8 seconds

Vi = 0 m/s

g = -9.81

Δy = ?

Vf = ?

Equation we will be using to solve for Vf: Vf = Vi + gt

Steps to solve:

Vf = (0) + (-9.81)(8)

Vf = -78.48 m/s

Hope this helps!! :)

5 0

3 years ago

Read 2 more answers

Using the Brønsted-Lowry concept of acids and bases, identify the Brønsted-Lowry acid and base in each of the following reaction

Luden [163]

Answer:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

There are no Brønsted-Lowry acids and bases in this reaction.

Explanation:

According to the Brønsted-Lowry concept, when an acid (HA) and a base (B) undergoes a chemical reaction, the acid (HA) loses a proton and forms its conjugate base (A⁻), whereas the base gains (B) the proton to form its conjugate acid (HB⁺).

The chemical equation for this reaction is:

HA + B ⇌ A⁻ + HB⁺

Given reactions:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻

Reason: In this reaction, the acid H₂O loses a proton and forms its conjugate base, OH⁻. Whereas, the base HSO³⁻ gains a proton to form its conjugate acid, H₂SO₃.

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

There are no Brønsted-Lowry acids and bases in this reaction.

Reason: In this reaction, there is no exchange of proton between the acid and the base.

4 0

3 years ago

The line graph shows the heights of plants grown in fertilized and unfertilized soil. Based on this information, what will most

AfilCa [17]

Answer:

g) Neither plant should increase by 1 cm in height.

Explanation:

See the graph for this question on the figure attached.

The growing of the plant A is represented by the line that goes above the other. At start, that line has a slope that rises about 0.75 cm ( height increase) in 1 day. From the day 2 and forward the slope of the line decreases. The line reaches its highest point about at day 4 and seems to start decreasing. Thus, you should predict that on the day six it most likely does not increase in height.

The growing of the plant B is represented by the line drawn below the other. As for the plant B, the growing decreases with the number of days. Between the days 4 and 5 the line is almost flat, which means that most likely this plant will not grow on the day six or grow less than 0.5 cm.

Thus, for both plants you can say that on day six, most likley, neither should increase by 1 cm in height (option g).

4 0

3 years ago

Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the r

Kazeer [188]

Answer:

∆H° rxn = - 93 kJ

Explanation:

Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.

We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.

N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)

1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)

∆H° rxn = ∑ H bonds broken - ∑ H bonds formed

∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]

∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ

be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .

8 0

3 years ago