Based on their locations on the periodic table, which two elements would youexpect to form positive ions with a +3 charge?
O A. Nitrogen, N
O B. Aluminum, Al
O C. Phosphorus, P
D. Boron, B

Answers

Answer 1

Answer:

Final answer:

The elements that would form positive ions with a +3 charge are aluminum (Al) and boron (B).

Explanation:

Based on their locations on the periodic table, the two elements that would form positive ions with a +3 charge are aluminum (Al) and boron (B).

When an element gains or loses electrons to form an ion, it takes on a charge that corresponds to the number of electrons gained or lost. Aluminum, located in Group 13 of the periodic table, tends to lose its three valence electrons to form an ion with a +3 charge. Likewise, boron, located in Group 13 as well, will also lose its three valence electrons to achieve a +3 charge.

Learn more about positive ions with a +3 charge here:

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Answer 2

Answer:

Answer:

i think it would be B. Aluminum, Al and D. Boron, B

Explanation:

since they're both in group 13 and they forms a 3+ ion

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A motorist traveling at 17 m/s encounters a deer in the road 39 m ahead. If the maximum acceleration the vehicle’s brakes are capable of is −7 m/s 2 , what is the maximum reaction time of the motorist that will allow her or him to avoid hitting the deer? Answer in units of s. 019 (part 2 of 2) 10.0 points If his or her reaction time is 1.21983 s, how fast will (s) he be traveling when (s)he reaches the deer? Answer in units of m/s.

Answers

Answer:

1. t_reaction = 1.08 s

2. v₀₁ = 16.365 m/s

Explanation:

1. This is a kinematics exercise, let's analyze the situation a bit, we can calculate the braking distance and the rest of the distance we can use to calculate the reaction time.

Braking distance

v² = v₀² + 2 a x

when he finishes braking the speed is v = 0

0 = v₀² + 2 a x

x = -v₀² / 2a

x = - 17²/2 (-7)

x = 20.64 m

the distance for the reaction is

d = x_reaction + x

x_reaction = d - x

x_reaction = 39 - 20.64

x_reaction = 18.36 m

as long as it has not reacted the vehicle speed is constant

v = x_reaction / t_reaction

t_reaction = x_reaction / v

t_reaction = 18.36 / 17

t_reaction = 1.08 s

2. Let's find the distance traveled in the reaction time of t1 = 1.21983 s

as the speed is constant

v = x / t

x₁ = v t₁

the distance traveled during braking is

v² = v₀² + 2a x₂

0 = v₀² + 2 a x₂

x₂ = -v₀² / 2a

v = v₀

the total distance is

x_total = x₁ + x₂

x_total = v₀ t₁ + v₀² / 2a

39 = v₀ 1.21983 + v₀²/14

v₀² + 17.08 vo - 546 =0

we solve the second degree equation

v₀ = [ -17.08 ±√(17.08² + 4 546) ]/2

v₀ = [-17.08 ± 49.81 ]/2

v₀₁ = 16.365 m/s

v₀₂ = - 33.445 m/s

as the acceleration is negative the correct result is v₀₁ = 16.365 m/s

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Answers

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