Need science help (screenshot included)

Chemistry

1 answer:

Dima020 [189]3 years ago

6 0

A. False. If it is high tide in one place on Earth, the place exactly opposite to it will also have a high tide.

The gravitational attraction of the Moon and the inertia of the oceans cause two tidal bulges on opposite sides of the Earth.

B. True. Cassini used flybys of Venus, Earth and Jupiter as slingshots to reach Saturn.

C. True. The whole solar system moves around the galaxy.

D. True. If a planet’s gravity is not strong enough, the molecules in its atmosphere will have enough kinetic energy to escape into space.

E. False. The mass of an object is constant, but its weight changes according to the gravity of the planet.

F. False. To find the mass of an object, divide its weight by gravity.

$F = mg$ or weight = mass × gravity

∴ Mass = weight/gravity

You might be interested in

Convert 1.5 km to millimeters, and express the result in scientific notation.

Grace [21]

The answer is 1.5e+6. Hope this helped!

7 0

3 years ago

The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be

Mrac [35]

Answer: The freezing point of solution is 5.35°C

Explanation:

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 4.90°C/m

$m_{solute}$ = Given mass of solute (naphthalene) = 2.60 g

$M_{solute}$ = Molar mass of solute (naphthalene) = 128.2 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC$