Answer:
Mine is language arts brainliest?
Explanation:
Answer:
To prepare a 15L 0.920 % NaCl solution we need 138 grams of NaCL
Explanation:
Step 1: Data given
Normal saline is 0.920% (w/v)
⇒ 0.92 g NaCl / 100 mL H2O
Step 2: Calculate grams of NaCl needed to prepare 15 L
We need 0.92 grams of NaCl for 100 mL
We need 9.2 grams of NaCL for 1L
For 15L we need 9.2 grams * 15 = 138 grams of NaCl
To prepare a 15L 0.920 % NaCl solution we need 138 grams of NaCL
Answer : The equilibrium concentration of is, 0.16 M
Explanation :
First we have to calculate the concentration of
and,
The given chemical reaction is:
Initial conc. 0.4 2.0 0
At eqm. (0.4-x) (2.0+x) x
The expression for equilibrium constant is:
Now put all the given values in this expression, we get:
x = -5.57 and x = 0.24
We are neglecting the value of x = -5.57 because equilibrium concentration can not be more than initial concentration.
Thus, the value of x = 0.24
The equilibrium concentration of = (0.4-x) = (0.4-0.24) = 0.16 M
Therefore, the equilibrium concentration of is, 0.16 M
Explanation:
hope this answer was helpful
The relation between moles and mass is given by the formula:
# of moles = (mass in grams) / (molar mass)
So, given the molar mass of 18.02 and 0.025 mol, you can solve the formula for the mass of water in grams:
mass of water in grams = # of moles of water * molar mass of water
mass of water in grams = 0.025 mol * 18.02 g / mol = 0.4505 g of water.
Then, she is wrong.