B its viscosity decreases
Answer:
490 in^3 = 8.03 L
Explanation:
Given:
The engine displacement = 490 in^3
= 490 in³
To determine the engine piston displacement in liters L;
(NOTE: Both in^3 (in³) and L are units of volume). Hence, to find the engine piston displacement in liters (L), we will convert in^3 to liters (L)
First, we will convert in³ to cm³
Since 1 in = 2.54 cm
∴ 1 in³ = 16.387 cm³
If 1 in³ = 16.387 cm³
Then 490 in³ = (490 in³ × 16.387 cm³) / 1 in³ = 8029.63 cm³
∴ 490 in³ = 8029.63 cm³
Now will convert cm³ to dm³
(NOTE: 1 L = 1 dm³)
1 cm = 1 × 10⁻² m = 1 × 10⁻¹ dm
∴ 1 cm³ = 1 × 10⁻⁶ m³ = 1 × 10⁻³ dm³
If 1 cm³ = 1 × 10⁻³ dm³
Then, 8029.63 cm³ = (8029.63 cm³ × 1 × 10⁻³ dm³) / 1 cm³ = 8.02963 dm³
≅ 8.03 dm³
∴ 8029.63 cm³ = 8.03 dm³
Hence, 490 in³ = 8029.63 cm³ = 8.03 dm³
Since 1L = 1 dm³
∴ 8.03 dm³ = 8.03 L
Hence, 490 in³ = 8.03 L
Answer:
Gravity is so strong that even light cannot escape a black hole.
Explanation:
A is obviously out because it leads to a volume of 125.0 milliliters of the new solution and gives you a lower concentration than you were aiming for.
D is out because you are adding 75 milliliters of the stock solution, so your concentration would be too high. You only need 25.0 milometers of stock solution per 100 milliliters of the new solution.
C is also out because it leads to 50.0 milliliters stock solution per 100 milliliters of the new solution and hence the wrong concentration.
B is by default the correct answer. It also details the correct technique. First you add the stock solution (This you know from your calculations to be 25 milliliters.) then you add the water up to the volume you needed. (Because the calculations only tell you the total volume of water not what you need to add) You also add the water last so you can rinse the neck of the flask to make sure you also get all the stock solution residue into the stock solution.
I would add the final step of stirring, but B is the only answer that can be correct.
What were Lamarck's ideas about evolution and why were those ideas incorrect
