Molarity can be used to calculate the volume of solvent or the amount of solute. The relationship between two solutions with the same amount of moles of solute can be represented by the formula c1V1 = c2V2, where c is concentration and V is volume.
Answer:
-169°C to -104°C
Explanation:
Ethene, also known as ethylene exists in solid, liquid and gaseous states. Ethene is an aliens with condensed structural formula C2H4. Athens is a colourless gas. It is flammable and is also a sweet smelling gas in its pure form. It is the monomer in the production of polyethylene which is of great importance in the plastic industry. In agriculture, it is used to induce the ripening of fruits. It can be hydrated in order to produce ethanol.
The liquid range of ethene refers to the temperatures at which ethene is found in the liquid state of matter. It is actually the difference between the melting point and the boiling points of ethene. Hence the liquid range of ethene is -169°C to -104°C
Answer:
[Br₂] = 1.25M
Explanation:
2NO (g) + Br₂ (g) ⇄ 2NOBr (g)
Eq 0.80M ? 0.80M
That's the situation told, in the statement.
Let's make the expression for Kc
Kc = [NOBr]² / [Br₂] . [NO]²
Kc = 0.80² / [Br₂] . [0.80]²
0.80 = 1 / [Br₂]
[Br₂] = 1 / 0.80 → 1.25
Answer:
It would take 3.11 J to warm 3.11 grams of gold
Explanation:
Step 1: Data given
Mass of gold = 3.11 grams
Temperature rise = 7.7 °C
Specific heat capacity of gold = 0.130 J/g°C
Step 2: Calculate the amount of energy
Q = m*c*ΔT
⇒ Q = the energy required (in Joules) = TO BE DETERMINED
⇒ m = the mass of gold = 3.11 grams
⇒ c = the specific heat of gold = 0.130 J/g°C
⇒ ΔT = The temperature rise = 7.7 °C
Q = 3.11 g * 0.130 J/g°C * 7.7 °C
Q = 3.11 J
It would take 3.11 J to warm 3.11 grams of gold
1.5 × 10^-6 i would think
