Answer:
120 g of NaCl in 300 g H20 at 90 C
Explanation:
At x = 90 go vertical to the line for NaCl...then go left to the y-axis to find the solubility in 100 g H20 = 40
we want 300 g H20 so multiply this by 3 to get 120 gm of NaCl in 300 g
Answer: place a 2 in front of NaNo3 on left side of equation while leaving the other blanks empty or you can place a 1 in those blanks
Explanation:
Step 1 count and write down the amount of each given element for both sides
Step 2 begin placing numbers (coefficients) to each side to balance
<span>PV=nRT= a universal constant
For any condition
P1V1/n1T1=R
and
P2V2/n2T2=R
i.e
P1V1/n1T1=P2V2/n2T2
Becomes
V1/n1=V2/n2
rearranging and solving
V2=V1X(n2/n1)= 750 mLx((0.65+0.35)/(0.65))=1200ml=1.2L...2 sig figs</span>