Answer:Reaccionan 54gr de Nitrato de plata al 39 % de pureza, con 72gr de ácido clorhídrico al 83% de pureza, en un proceso donde se obtienen 93 gr de cloruro de - 154… ... 93 gr de cloruro de plata. El otro producto es el ácido nítrico: Calcular el porcentaje de rendimiento de la reacción y balancearlo.
Answer:
A
Explanation:
Definitely not in the cell theory
Answer:
0.645 L
Explanation:
To find the volume, you need to (1) convert grams to moles (using the molar mass) and then (2) calculate the volume (using the molarity ratio). The final answer should have 3 sig figs to match the sig figs of the given values.
(Step 1)
Molar Mass (KOH): 39.098 g/mol + 15.998 g/mol + 1.008 g/mol
Molar Mass (KOH): 56.104 g/mol
19.9 grams KOH 1 mole
-------------------------- x ----------------------- = 0.355 moles KOH
56.014 grams
(Step 2)
Molarity = moles / volume <----- Molarity ratio
0.550 M = 0.355 moles / volume <----- Insert values
(0.550 M) x volume = 0.355 moles <----- Multiply both sides by volume
volume = 0.645 L <----- Divide both sides by 0.550
I will say this is True….?
The relative molecular mass of acid A : 50 g/mol
<h3>Further explanation</h3>
Given
40.0 cm³(40 ml) of 0.2M sodium hydroxide
0.2g of a dibasic acid
Required
the relative molecular mass of acid A
Solution
Titration formula
M₁V₁n₁=M₂V₂n₂
n=acid/base valence(number of H⁺/OH⁻)
NaOH ⇒ n = 1
Dibasic acid = diprotic acid (such as H₂SO₄)⇒ n = 2
mol = M x V
Input the value in the formula :(1 = NaOH, 2=dibasic acid)
0.2 x 40 x 1 = M₂ x V₂ x 2
M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A
The relative molecular mass of acid A (M) :
