<h2>Answer:</h2>
5.65moles
<h2>Explanations:</h2>
The formula for calculating the number of moles the compound contain is given as:
Given the following parameters
Mass of Ag = 700grams
Determine the molar mass of AgO
Molar mass = 107.87 + 16
Molar mass = 123.87g/mol
Determine the moles of AgO
Hence the moles of AgO present is 5.65moles
Answer:
Mass = 135.66 ×10⁻²¹ g
Explanation:
Given data:
Number of molecules of CuSO₄= 5.119×10²
Mass of CuSO₄= ?
Solution:
The given problem will solve by using Avogadro number.
1 mole contain 6.022×10²³ molecules
5.119×10² molecules ×1 mol / 6.022×10²³ molecules
0.85×10⁻²¹ mol
Mass in grams:
Mass = number of moles × molar mass
Mass = 0.85×10⁻²¹ mol × 159.6 g/mol
Mass = 135.66 ×10⁻²¹ g
Answer:
I believe the answer is "b". "During the experiment, the scientist has only one element, or variable, that is changed to test the hypothesis."
Explanation:
I remember from last year but I'm not totally sure. Good luck!
Answer:
Sorry pal! Didn't understand your language.
:(
Explanation:
Answer:
pKa = 3.675
Explanation:
∴ <em>C</em> X-281 = 0.079 M
∴ pH = 2.40
let X-281 a weak acid ( HA ):
∴ HA ↔ H+ + A-
⇒ Ka = [H+] * [A-] / [HA]
mass balance:
⇒<em> C</em> HA = 0.079 M = [HA] + [A-]
⇒ [HA] = 0.079 - [A-]
charge balance:
⇒ [H+] = [A-] + [OH-]... [OH-] is negligible; it comes from to water
⇒ [H+] = [A-]
∴ pH = - log [H+] = 2.40
⇒ [H+] = 3.981 E-3 M
replacing in Ka:
⇒ Ka = [H+]² / ( 0.079 - [H+] )
⇒ Ka = ( 3.981 E-3 )² / ( 0.079 - 3.981 E-3 )
⇒ Ka = 2.113 E-4
⇒ pKa = - Log ( 2.113 E-4 )
⇒ pKa = 3.675
