Science Reviewer Answer Key Flashcards

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A collection of vocabulary terms, answer key entries, and step-by-step problem calculations from the Science Reviewer.

Last updated 1:12 PM on 10/3/26
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32 Terms

1
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Absolute humidity

Correct answer for Question 1 (Option C) in the Science Reviewer Answer Key.

2
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Bose-Einstein condensate

Correct answer for Question 21 (Option A) in the Science Reviewer Answer Key.

3
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Law of conservation of energy

Correct answer for Question 77 (Option D) in the Science Reviewer Answer Key.

4
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Law of conservation of mass

Correct answer for Question 78 (Option A) in the Science Reviewer Answer Key.

5
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Law of definite proportion

Correct answer for Question 79 and Question 80 (Option C) in the Science Reviewer Answer Key.

6
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Law of multiple proportion

Correct answer for Question 81 (Option D) in the Science Reviewer Answer Key.

7
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Density Calculation (Question 106)

Calculated as Volume=3×2×2=12 cm3\text{Volume} = 3 \times 2 \times 2 = 12\,\text{cm}^3 and Density=150÷12=12.5 g/cm3\text{Density} = 150 \div 12 = 12.5\,\text{g/cm}^3. The closest listed choice is Option A (13 g/cm313\,\text{g/cm}^3).

8
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Mass Calculation (Question 107)

Calculated as Mass=density×volume=1.03×1.020=1.0506 g≈1.05 g\text{Mass} = \text{density} \times \text{volume} = 1.03 \times 1.020 = 1.0506\,\text{g} \approx 1.05\,\text{g} (Option D).

9
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Volume Calculation (Question 108)

Calculated as Volume=mass÷density=220÷0.659=333.85 cm3\text{Volume} = \text{mass} \div \text{density} = 220 \div 0.659 = 333.85\,\text{cm}^3 (Option D).

10
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Mass Calculation (Question 109)

Calculated as Mass=density×volume=1.6×33=52.8 g\text{Mass} = \text{density} \times \text{volume} = 1.6 \times 33 = 52.8\,\text{g} (Option A).

11
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Fahrenheit Temperature Conversion (Question 110)

Calculated as ∘F=(35×95)+32=95∘F^\circ\text{F} = (35 \times \frac{9}{5}) + 32 = 95^\circ\text{F} (Option B).

12
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Celsius Temperature Conversion (Question 111)

Calculated as ∘C=(95−32)×59=35∘C^\circ\text{C} = (95 - 32) \times \frac{5}{9} = 35^\circ\text{C} (Option C).

13
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Kelvin Temperature Conversion (Question 112)

Calculated as K=80+273=353 K\text{K} = 80 + 273 = 353\,\text{K} (Option B).

14
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Work Calculation 1 (Question 113)

Calculated as Work=F×d=25×2=50 J\text{Work} = F \times d = 25 \times 2 = 50\,\text{J} (Option C).

15
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Work Calculation 2 (Question 114)

Calculated as Work=F×d=25×0.25=6.25 J\text{Work} = F \times d = 25 \times 0.25 = 6.25\,\text{J} (Option B).

16
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Work Calculation 3 (Question 115)

Calculated as Work=F×d=10×5=50 J\text{Work} = F \times d = 10 \times 5 = 50\,\text{J} (Option B).

17
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Work Calculation 4 (Question 120)

Calculated as Work=F×d=100×5=500 J\text{Work} = F \times d = 100 \times 5 = 500\,\text{J} (Option C).

18
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Power Calculation (Question 121)

Calculated as Power=Work÷time=500÷5=100 W\text{Power} = \text{Work} \div \text{time} = 500 \div 5 = 100\,\text{W} (Option C).

19
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Work Application Condition (Question 125)

Work is done when you applied a force and a displacement on an object (Option A).

20
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Speed Calculation (Question 131)

Calculated as Speed=distance÷time=15÷10=1.5 m/s\text{Speed} = \text{distance} \div \text{time} = 15 \div 10 = 1.5\,\text{m/s} (Option B).

21
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Distance Calculation (Question 132)

Calculated as Distance=speed×time=15×2=30 m\text{Distance} = \text{speed} \times \text{time} = 15 \times 2 = 30\,\text{m} (Note: No correct option is listed in the reviewer).

22
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Velocity Calculation (Question 134)

Calculated as v=u+at=5+(2×10)=25 m/sv = u + at = 5 + (2 \times 10) = 25\,\text{m/s} (Option C).

23
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Distance Traveled Calculation (Question 137)

With 1 min=60 s1\,\text{min} = 60\,\text{s}, distance is calculated as Distance=3×60=180 m\text{Distance} = 3 \times 60 = 180\,\text{m} (Option D).

24
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Total Distance Calculation (Question 138)

Calculated as Total distance=4+2=6 m\text{Total distance} = 4 + 2 = 6\,\text{m} (Option D).

25
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Displacement Calculation (Question 139)

Calculated as Displacement=4−2=2 m\text{Displacement} = 4 - 2 = 2\,\text{m} (Option B).

26
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Potential Energy Calculation 1 (Question 141)

Calculated as PE=mgh=10×9.8×0.50=49 J\text{PE} = mgh = 10 \times 9.8 \times 0.50 = 49\,\text{J} (Option C).

27
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Potential Energy Calculation 2 (Question 142)

Calculated as PE=mgh=5×9.8×0.35=17.15 J≈17.2 J\text{PE} = mgh = 5 \times 9.8 \times 0.35 = 17.15\,\text{J} \approx 17.2\,\text{J} (Option D).

28
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Kinetic Energy Calculation (Question 143)

Calculated as KE=12mv2=12(1500)(252)=468,750 J≈469,000 J\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2}(1500)(25^2) = 468,750\,\text{J} \approx 469,000\,\text{J} (Option B).

29
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Velocity from Kinetic Energy (Question 144)

Calculated as v=2KEm=34,00045≈27.5 m/sv = \sqrt{\frac{2\text{KE}}{m}} = \sqrt{\frac{34,000}{45}} \approx 27.5\,\text{m/s}, with closest listed choice Option A (27 m/s27\,\text{m/s}).

30
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Asthenosphere

Correct answer for Question 146 (Option A) in the Science Reviewer Answer Key.

31
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Covalent bonding

Correct answer for Question 149 (Option C) in the Science Reviewer Answer Key.

32
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Double replacement reaction

Correct answer for Question 150 (Option B) in the Science Reviewer Answer Key.