Answer:Chemical reaction
Explanation:
The main concept that must be applied to determine the coefficients (amount of each item) is that there must be equal amounts of each element on each side of the equation. We are not destroying or creating new atoms. In this case, the unbalanced reaction formula is:
B
r
2
+
K
I
=
K
B
r
+
I
2
There are a two problems we need to solve before it will be balanced:
There are two moles of Iodine atoms (
I
) on the right side of the equation, while there is only one mole on the right side.
There are two moles of bromine (
B
r
) atoms on the left side, while there is only one on the right.
Since there are two moles of bromine atoms on the left side, we need two moles on the right as well. We can do this by adding a coefficient of two to the
'
K
B
r
'
term in the equation. Our now modified equation looks like this:
B
r
2
+
K
I
=
2
K
B
r
+
I
2
There is one mole of Iodine atoms on the left, and two on the right. To fix this, we add a coefficient of two to the
'
K
I
'
term. The resulting equation is below.
B
r
2
+
2
K
I
=
2
K
B
r
+
I
2
Bonus step: We can also put ones in front of the coefficient-less species. This is like changing a phrase from "an apple" to "1 apple". It is the exact same thing, but makes it a little more clear sometimes. This would like like this:
1
B
r
2
+
2
K
I
=
2
K
B
r
+
1
I
2
Can you see that there is now an equal amount of each element on each side of the equation? That means that it is balanced.
Answer:
Too high a value
Explanation:
HA + NaOH ⟶ NaA +H₂O
If the student has gone slightly past the equivalence point, they have added too much base.
The moles of HA are directly proportional to the moles of NaOH, so the moles of acid that the student calculates will be too high.
The calculated concentration of acid will also be too high.
Boron: isotope data. Both isotopes ofBoron, B-10 and B-11, are used extensively in the nuclear industry. B-10 is used in the form of boric acid as a chemical shim in pressurized water reactors while in the form of sodium pentaborate it is used for standby liquid control systems in boiling water reactors
Answer:
The answer to your question is: 30 g
Explanation:
Data
mass of water = 25 g
mass of salt = 5 g
Process
total mass = mass of water + mass of salt
= 25 + 5
= 30 g
