The correct answer to your question is noble gases are stable <span>due to having the maximum number of valence electrons their outer shell can hold. Meaning their outer shells are stable.
Hope this helps let me know!</span>
Here's an example of a bar graph.
Answer: It's equal to 10^(-2.3), or 0.00501 M, or 5.01 * 10^-3 moles/Liter
Explanation:
Well, pH = - log[H+]
Or, in words, pH is equal to -1 multiplied by the logarithm (base 10) of the hydrogen ion concentration.
So you have 2.3 = -log[H+]. We want to isolate the H+, so let's start simplifying the right hand side of the equation. First, we multiply both sides by -1.
-2.3=log[H+]
Now, the definition of a logarithm says that if the log (base 10) of [H+] is -2.3, then 10 raised to the -2.3 power is [H+]
So on each side of the equation, we raise 10 to the power of that side of the equation.
10^(-2.3) = 10^(log[H+])
and because 10^log cancels out...
10^(-2.3) = [H+]
Now we've solved for [H+], the hydrogen ion concentration!
Answer:
P=12.16 atm
Explanation:
Using the formula of ideal gas law:
PV = nRT
P= nRT/V
n= number of moles
R= Avogadro constant = 0.0821
T= Temperature in K => ºC + 273.15 K
P= (1.50 moles)(0.0821)( 296.15 K)/ 3.00L
P= 12.15
Answer:
The concentration the student should write down in her lab is 2.2 mol/L
Explanation:
Atomic mass of the elements are:
Na: 22.989 u
S: 32.065 u
O: 15.999 u
Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.
Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.
For mole(s) of Na2S2O3 = (mass taken)/(molar mass)
= (17.240 g)/(158.105 g/mol) = 0.1090 mole.
Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)
= 0.05029 L.
To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:
= (moles of sodium thiosulfate)/(volume of solution in L)
= (0.1090 mole)/(0.05029 L)
= 2.1674 mol/L
