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grandymaker [24]
3 years ago
8

What property do the following elements have in common? Li, C, and F A) They are poor conductors of electricity. B) Each element

participates in covalent bonding. C) They all exist in the same phase at room temperature. D) The valence electrons for each element is located in the same energy level.
Physics
2 answers:
Aloiza [94]3 years ago
7 0

Answer: Option (D) is the correct answer.

Explanation:

The given elements Li, C and F are all second period elements. So, when we move from left to right across a period then there occurs increase in number of valence electrons as there occurs increase in total number of electrons.

So, it means more electrons are added to the same energy level.

Thus, we can conclude that a property of valence electrons for each element is located in the same energy level is common in the given elements.

erik [133]3 years ago
4 0

Answer:

the answer is d

Explanation:

i did the usa test prep

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What is the volume V of a sample of 4.00 mol of copper? The atomic mass of copper (Cu) is 63.5 g/mol, and the density of copper
rusak2 [61]

Answer : The volume of a sample of 4.00 mol of copper is 28.5cm^3

Explanation :

First we have to calculate the mass of copper.

\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}

\text{ Mass of copper}=(4.00moles)\times (63.5g/mole)=254g

Now we have to calculate the volume of copper.

Formula used :

Density=\frac{Mass}{Volume}

Now put all the given values in this formula, we get:

8.92\times 10^3kg/m^3=\frac{254g}{Volume}

Volume=\frac{254g}{8.92\times 10^3kg/m^3}=2.85\times 10^{-2}L=2.85\times 10^{-2}\times 10^3cm^3=28.5cm^3

Conversion used :

1kg/m^3=1g/L\\\\1L=10^3cm^3

Therefore, the volume of a sample of 4.00 mol of copper is 28.5cm^3

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3 years ago
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2 years ago
A car runs at a constant speed of 15ms-1 for 30secs, and then accelerates uniformly to a speed of 25ms-1 over a time of 20secs.
Otrada [13]

Answer:

a. Please find the attached velocity time graph of the car's motion created with Microsoft Excel

b. The total distance traveled by the car is 8,725 meters

c. The average speed of the car is 22.9605263 m/s

Explanation:

The given parameters of the motion are;

The initial speed of the car, v₁ = 15 m/s

The time during which the car runs at the initial speed, t₁ = 30 seconds

The new speed the car then accelerates at 'a₁' to, v₂ = 25 m/s

The duration it takes for the car to accelerate to the new speed = 20 seconds

The time during which the car runs at the initial speed = 300 seconds

The time it takes the car to be brought to rest with a deceleration, 'a₂' from the new speed (20 m/s) = 30 seconds

The final speed of the car at rest, v₃ = 0 m/s

The acceleration, a₁ = (v₂ - v₁)/t₁ = (25 - 15)/20 = 1/2 m/s²

The deceleration , a₂ = (v₃ - v₂)/t₁ = (0 - 25)/30 = -5/6 m/s²

a. Please find attached the drawing of the velocity time graph of the motion created with Microsoft Excel

b. The total distance traveled by the car, 'Δx', is given b the area under the velocity time graph as follows;

Area of trapezoid, A₁ = (320 + 300)/2 × 10 = 3,100

Area of rectangle, A₂ = 15 × 350 = 5,250

Area of triangle, A₃ = 1/2×30×25 = 375

The total area under the velocity time graph = A₁ + A₂ + A₃ = 3,100 + 5,250 + 375 = 8,725

The total area under the velocity time graph = The total distance traveled by the car, Δx = 8,725 meters

c. The average speed of the car is given as follows;

The \ average \  speed  \ of  \ the \  car, \overline v =\dfrac{\Delta x}{\Delta t}  = \dfrac{The \  total \  distance  \ traveled by \  the \  car}{The  \ total  \  time \  the \  car \ travels}

Where;

Δt = The total time during which the car travels

∴ The average speed of the car = 8,725 m/(380 s) = 22.9605263 m/s

4 0
3 years ago
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