Mass = molarity x molar mass( NaNO₃) x volume
mass = 1.50 x 85.00 x 4.50
mass = 573.75 g of NaNO₃
hope this helps!
<span>3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)
The unbalanced equation is:
ZnBr2 (aq) + Al (s) ==> AlBr3 (aq) +Zn (s)
First, count the atoms of each element on each side of the equation:
Zn 1,1
Br 2,3
Al 1,1
The zinc and aluminum, but the bromine doesn't match with 2 and 3. So look for the least common multiple of 2 and 3 which is 6 and adjust the quantities on both sides to have 6 bromine atoms on both sides. Do this by having 3 zinc bromide on the left and 2 aluminum bromide on the right, getting:
3 ZnBr2 (aq) + Al (s) ==> 2 AlBr3 (aq) +Zn (s)
Now check the atom counts again for both sides:
Zn 3,1
Br 6,6
Al 1,2
Now bromine matches, but zinc and aluminum doesn't. But it's easy enough to add an extra aluminum to the left and 2 more zinc to the right. Giving:
3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)
Now check the atom counts again:
Zn 3,3
Br 6,6
Al 2,2
And they match. So the balanced equation is:
3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)</span>
Answer:
Explanation:
Hello,
Among the options given on the attached document, since phenolic functional group is characterized by a benzene ring bonded with a hydroxyl group (C₆H₅OH) we can see that the first option correctly points out such description. Thus, answer is on the second attached picture. Other options are related with other sections found in eugenol that are not phenolic.
Best regards.
<u>The given reaction is:</u>
F2 + ClO2 → 2FClO2
Rate = k[F2][ClO2]
<u>Explanation:</u>
The possible mechanism for this reaction can be broken down into two steps with the slow step being the rate determining step
Step 1: F2 + ClO2 → FClO2 + F ----------- Slow
Step 2: F + ClO2 → FClO2 ----------- Fast
-----------------------------------------------------------
Overall: F2 + 2ClO2 → 2FClO2
Rate = k[F2][ClO2]
Answer:
Explanation:
Given:
For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.
To find: fraction of the whole athletic field reserved for each fifth class
Solution:
Fraction of the whole athletic field reserved for four fifth classes = 
So, fraction of the whole athletic field reserved for each fifth class = 
