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lapo4ka [179]
3 years ago
12

Which of the following causes high and low tides on Earth? A. The moon's gravitational pull on Earth causes water to bulge on tw

o sides of the Earth. B. Water swells when tsunamis occur in two areas of Earth at the same time. C. The Earth's rotation is so fast, water will swell and shrink twice a day. D. The Earth is tilted on an axis, causing water to spread unevenly throughout the planet.
Chemistry
2 answers:
arlik [135]3 years ago
8 0
A is the answer
hope it helps!
Finger [1]3 years ago
5 0
A. the moon's gravitational pull on earth causes water to bulge on two sides of the earth
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Calculate the vapor pressure in torr of a solution containng 24.5 g of glycerin (C3H8O3) in 135 mL water at 30.0* C; the vapor p
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2 years ago
Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.
Marianna [84]

The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
  • Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
  • Initial temperature (T₁) = –25 °C
  • Final temperature (T₂) = 0 °
  • Change in temperature (ΔT) = 0 – (–38) = 38 °C
  • Specific heat capacity (C) = 2050 J/(kg·°C)
  • Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
  • Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
  • Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 100 °C
  • Final temperature (T₂) = 160 °C
  • Change in temperature (ΔT) = 160 – 100 = 60 °C
  • Specific heat capacity (C) = 1996 J/(kg·°C)
  • Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
  • Heat for –38 °C to 0°C (Q₁) = 31160 J
  • Heat for melting (Q₂) = 133600 J
  • Heat for 0 °C to 100 °C (Q₃) = 167200 J
  • Heat for vaporization (Q₄) = 904000 J
  • Heat for 100 °C to 160 °C (Q₅) = 47904 J
  • Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

brainly.com/question/10286596

#SPJ1

7 0
2 years ago
Please help me on 9 and 10
Fantom [35]
C and A because the product of a matter will always be the same and just google A.
4 0
3 years ago
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