Answer:
The correct answer is pOH= 11
Explanation:
From the aqueous acid-base equilibrium we know that
pH + pOH = 14
If we know pH, we can calculate pOH as follows:
pOH = 14 - pH
In this problem, the solution has a pH of 3, so:
pOH = 14 - 3 = 11
Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
1) since we are given percentages, we can assume we have 100 grams of the molecule.
55.6 % Cu ----> 55.6 grams Cu
16.4 % Fe------> 16.4 grams Fe
28.0% S--------> 28.0 grams S
2) convert each gram to moles using the molar masses given
3) we divide the smallest value of moles (0.293) to each one.
Cu --> 0.876 / 0.293= 3
Fe---> 0.293 / 0.293= 1
S-----> 0.875 / 0.293= 3
4) let's write the empirical formula
Cu₃FeS₃
Answer:
Because u would have to find the undercorse of 010-1 witch makes the out of part by 6
Explanation:
Given :
Juan rolled a six-sided number cube 18 times.
The number two occurred four times.
To Find: Juan claimed the experimental probability of rolling a two was approximately 1/9. Why is Juan’s experimental probability incorrect?
Solution:
Total events = number of times cube rolled = 18
Favorable events = The number two occurred four times. = 4
So, Experimental probability of rolling a two was approximately 1/9
