The molar mass of K is 39, Cr is 52, and O is 16. The rough calculation of these numbers are 294.
Answer:
the mass required to inflate a 72 L bag is 191.491 g
Explanation:
reaction:
conditions:
- V = 72.0 L
- STP: P = 1 atm ∧ T = 298 K
gas law:
- PV = RTn
- R = 0.082 atm * L / K * mol
⇒ n = PV / RT
⇒ n = ((1 atm) * ( 72.0 L)) / (0.082 atm*L / K*mol) * (298 K)
⇒ n = 2.946 mol
⇒ m = n * Mw = ( 2.946 mol ) * ( 64.99 g/mol)
⇒ m = 191.491 g
Answer:
True
Explanation:
This is because the statement above fits the definition of a positive economic statement. A Positive economic statement refers to objective statements that can be verified, tested , amended or discarded by consulting available evidence .The statement above can be tested objectively to establish its veracity.
Answer:
27.4 gram is the solution it's simple dude...
Explanation:
don't be afraid of huge question they confuse you you need not to be confused
now see simple solution
molality is denoted by m
so
m= moles of solute / mass of solvent in kg.
i hope your know the meaning of solute and solvent....
so moles are given 0.467
and molar mass is given 58.44
so just take out the gram means
by applying formula
58.44×0.467
it will give 27.4 grams simple.....
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).

Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).

To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
You can learn more about solutions here: brainly.com/question/2412491