Well....a beaker fits your description. It can store substances, you can heat substances on it and you can transport substances with it.
Rate = k * [A]^2 * [B]^1
<span>Use the data from any trial to calculate k. </span>
<span>k = (rate)/([A]^2 * [B]^1) </span>
<span>E.g., for Trial 1, we have </span>
<span>rate = 3.0×10−3 M/s </span>
<span>[A] = 0.50 M </span>
<span>[B] = 0.010 M </span>
<span>Plug those numbers in and crank out the answer. </span>
<span>Now with the calculated value of k, calculate the initial rate for [A] = 0.50 M and [B] = 0.075 M </span>
<span>rate = k * [A]^2 * [B]^1 </span>
<span>k = calculated value </span>
<span>[A] = 0.50 M </span>
<span>[B] = 0.075 M</span>
Potassium Hydroxide (KOH) when added to the solution would give the highest pH
Explanation:
Bases or Alkali are associated with high pH while Acidic substances are represented by lower pH value.
In the given option
HCl- is a strong acid hence would have pH less than 7
H2SO4- also an acid with a pH less than 7
KOH- base with a pH higher than 7
H20-neutral compound with pH as 7
KOH is a very strong base and dissociates in aqueous solution to give it's corresponding metallic ion and hydroxyl ions (OH-) which are characteristic property of any base.
Solution of 0.25 M is prepared in two steps,
1) Calculate Amount of Solute:
Molar Mass of Solute: 342.3 g/mol
As we know,
Molarity = Moles / 1 dm³
or,
Moles = Molarity × 1 dm³
Putting Values,
Moles = 0.25 mol.dm⁻³ × 1 dm³
Moles = 0.25 moles
Now, find out mass of sucrose,
As,
Moles = Mass / M.mass
or,
Mass = Moles × M.mass
Putting Values,
Mass = 0.25 mol × 342.3 g.mol⁻¹
Mass = 85.57 g
2) Prepare Solution:
Take Volumetric flask and add 85.57 g of sucrose in it. Then add distilled water up to the mark of 1 dm³. Shake well! The solution prepared is 0.25 M in 1 Liter.
The solubility of nitrogen in water at 25 °C= 4.88 x 10⁻⁴ mol/L
<h3>Further explanation</h3>
Given
78% Nitrogen by volume
Required
The solubility of nitrogen in water
Solution
Henry's Law states that the solubility of a gas is proportional to its partial pressure
Can be formulated
S = kH. P.
S = gas solubility, mol / L
kH = Henry constant, mol / L.atm
P = partial gas pressure
In the standard 25 C state, the air pressure is considered to be 1 atm, so the partial pressure of N₂ -nitrogen becomes:
Vn / Vtot = Pn / Ptot
78/100 = Pn / 1
Pn = 0.78 atm
Henry constant for N₂ at 25 °c = 1600 atm/mol.L=6.25.10⁻⁴ mol/L.atm
The solubility :
