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hodyreva [135]
3 years ago
9

Describe the electron transfers that occur in the formation of calcium fluoride from elemental calcium and elemental fluorine.

Physics
1 answer:
boyakko [2]3 years ago
3 0

Answer:

Check explanation

Explanation:

In the formation of calcium fluoride we take calcium and fluorine.

in elemental form calcium exist in solid form and fluorine in gaseous form.

formation of compound takes place to complete their octet, in case of calcium  need to remove two electron and need to add one elecron in fluorine to complete their octet so two electron will ransferred from calcium to two fluorine atom.  

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A 1000kg ar accelerates from rest to 25.0m/s in 4.20
bezimeni [28]

Answer:

74.4 kilowatts or 99.8 horsepower

Explanation:

The explanation is in the attachment.

7 0
3 years ago
Which of the following has the lowest U value?
jarptica [38.1K]

Answer:

c. expanded polyurethane

Explanation:

Thermal performance of a building fabric is measured in terms of heat loss and is expressed as U-value or R-value. U-value is the rate of heat transferred through a structure divided by the difference in temperature across the structure with a unit of measurement of W/m²K.You can calculate the U-value of a by getting the reciprocal of the sum of thermal resistances , R, making the building material.

If you have the value of R, then U=1/R

Material                         size            R                      U

plywood                          1"              1.25                0.8

Poured concrete            2"              0.99               1.010

Expanded polyurethane  1"            6.5                   0.1538

Asbestos shingles            1"             0.03                33.33

The material with lowest U-value is expanded polyurethane

4 0
3 years ago
A train started from rest and moved with constant acceleration. At one time it was traveling 27 m/s, and 150 m farther on it was
AlekseyPX

Explanation:

(a) Given:

Δx = 150 m

v₀ = 27 m/s

v = 54 m/s

Find: a

v² = v₀² + 2aΔx

(54 m/s)² = (27 m/s)² + 2a (150 m)

a = 7.29 m/s²

(b) Given:

Δx = 150 m

v₀ = 0 m/s

a = 7.29 m/s²

Find: t

Δx = v₀ t + ½ at²

150 m = (0 m/s) t + ½ (7.29 m/s²) t²

t = 6.42 s

(c) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: t

v = at + v₀

27 m/s = (7.29 m/s²) t + 0 m/s

t = 3.70 s

(d) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: Δx

v² = v₀² + 2aΔx

(27 m/s)² = (0 m/s)² + 2 (7.29 m/s²) Δx

Δx = 50 m

7 0
3 years ago
Help please, i don't know what the answer is but i accidently clicked D .....
GaryK [48]

Answer:

You are right correct option is D.

4 0
3 years ago
If a farsighted person has a near point that is 0.600 m from the eye, what is the focal length f2 of the contact lenses that the
Doss [256]

Answer:

0.84 cm

Explanation:

u = Object distance =  0.35 cm

v = Image distance = -0.6 cm (near point is considered as image distance and negative due to sign convention)

f = Focal length

From lens equation

\frac{1}{f_2}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f_2}=\frac{1}{0.35}+\frac{1}{-0.6}\\\Rightarrow \frac{1}{f_2}=\frac{25}{21}\\\Rightarrow f_2=\frac{21}{25}=0.84\ cm

Focal length of the lens is 0.84 cm

3 0
3 years ago
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