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
I’m not sure but I think it’s Earthquake Detector
Answer:
<em>The empirical formula is Ag2O.</em>
<em>The empirical formula is Ag2O.Explanation:</em>
<em>The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.</em>
<em>The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of Ag to </em><em>2</em><em>O.</em>
<em>do</em><em> </em><em>the</em><em> </em><em>steps</em><em> </em><em>.</em><em>.</em><em>.</em>
To get this into an integer ratio, we divide both numbers by the smaller value.
From this point on, I like to summarize the calculations in a table.
ElementAgMass/gXMolesXllRatiomllIntegers
mAgXXXm7.96Xm0.07377Xll2.00mmm2
mlOXXXXl0.59mm0.0369Xml1mmmml1
There are 2 mol of Ag for 1 mol of O.
Answer:
Boron would have a
+
3
charge.
Explanation:
Boron has an atomic number of
5
. For an atom to be an ion, it wants to have full electron shells with the "minimum number" (this is hard to explain). So that means that boron would either want
2
electrons or
10
electrons. Of course, it would be easier to have
2
electrons from
5
electrons, so there it will lose
3
electrons, and would have a final charge of
+
3
.
