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
Hey there!
The correct answer to your question is a compound.
A substance of two or more elements chemically combined in a set ratio, or proportion, is called a compound.
This is because compounds are made of two or more elements. Table salt is an example of a compound because it is made up of the elements chlorine and sodium.
Hope this helps you.
Have a great day!
Explanation:
Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.
Equilibrium reaction of strontium sulfate and sodium sulfate follows the equation:
According to Le-Chateliers principle: If there is any change in the variables of the reaction, the equilibrium will shift in the direction in order to minimize the effect.
In the equilibrium reactions, hypochlorite ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of hydrogen hypochlorite.
Thus, the addition hypochlorite ions will shift the equilibrium in the left direction.
The dissociation of hydrogen hypochlorite is suppressed due to the common ion effect.
