Answer:
330 mL of (NH₄)₂SO₄ are needed
Explanation:
First of all, we determine the reaction:
(NH₄)₂SO₄ + 2NaOH → 2NH₃ + 2H₂O + Na₂SO₄
We determine the moles of base:
(First, we convert the volume from mL to L) → 62.6 mL . 1L/1000 mL = 0.0626L
Molarity . volume (L) = 2.31 mol/L . 0.0626 L = 0.144 moles
Ratio is 2:1. Therefore we make a rule of three:
2 moles of hydroxide react with 1 mol of sulfate
Then, 0.144 moles of NaOH must react with (0.144 .1) /2 = 0.072 moles
If we want to determine the volume → Moles / Molarity
0.072 mol / 0.218 mol/L = 0.330 L
We convert from L to mL → 0.330L . 1000 mL/1L = 330 mL
Answer: Salt and Water
Explanation:
An Arrhenius acid (HCl) can best be defined as any substance that when added to water increases the concentration of H+ ions.
While an Arrhenius base (KOH) is any substance that when added to water increases the concentration of OH- ions.
When an Arrhenius acid such as HCl reacts with an Arrhenius base such as KOH, the end products will be salt and water, in a process called Neutralization Reaction.
HCl (aq) + KOH (aq) -------> KCl (aq) + H2O (l)
Answer:
The metalloids; boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), polonium (Po) and astatine (At) are the elements found along the step like line between metals and non-metals of the periodic table.
Elements: Germanium; Boron; Arsenic
Explanation:
Answer:
Explanation:
For a first order reaction the rate law is:
![v=\frac{-d[A]}{[A]}=k[A]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B-d%5BA%5D%7D%7B%5BA%5D%7D%3Dk%5BA%5D)
Integranting both sides of the equation we get:
![\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%7D%7D%20%5C%2C%20dx%20%3D-k%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt)
where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.
From that integral we get the integrated rate law:
![ln\frac{[A]}{[A]_{0} } =-kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA%5D%7D%7B%5BA%5D_%7B0%7D%20%7D%20%3D-kt)
![[A]=[A]_{0}e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_%7B0%7De%5E%7B-kt%7D)
![ln[A]=ln[A]_{0} -kt](https://tex.z-dn.net/?f=ln%5BA%5D%3Dln%5BA%5D_%7B0%7D%20-kt)
![k=\frac{ln[A]_{0}-ln[A]}{t}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bln%5BA%5D_%7B0%7D-ln%5BA%5D%7D%7Bt%7D)
therefore k is
To calculate the number of atoms of Cr, we first find the number of moles per unit of cubic centimeter of Cr. Then, use avogadros number for the number of atoms. Calculations are as follows:
1 cm^3 (7.15 g/cm^3) (1 mol / 51.996 g Cr) = 0.14 mol Cr
0.14 mol Cr ( 6.022 x 10^23 atoms Cr / 1 mol Cr ) = 8.28 x 10^22 atoms Cr
