Answer is: <span>the mass of the glucose is 81,07 grams.
</span>c(C₆H₁₂O₆) = 0,3 M = 0,3 mol/L.
V(C₆H₁₂O₆) = 1,500 L.
n(C₆H₁₂O₆) = c(C₆H₁₂O₆) · V(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 0,3 mol/L · 1,5 L.
n(C₆H₁₂O₆) = 0,45 mol.
m(C₆H₁₂O₆) = n(C₆H₁₂O₆) · M(C₆H₁₂O₆).
m(C₆H₁₂O₆) = 0,45 mol · 180,156 g/mol.
m(C₆H₁₂O₆) = 81,07 g.
Answer: Balanced molecular equation :
Total ionic equation:
The net ionic equation:
Explanation:
Complete ionic equation : In complete ionic equation, all the substances that are strong electrolyte are present in an aqueous state as ions.
Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
When sodium phosphate and zinc acetate then it gives zinc phosphate and sodium acetate as product.
The balanced molecular equation will be,
The total ionic equation in separated aqueous solution will be,
In this equation, and are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
16) Na (s) + H2O(L) ---> H2 (g) + NaOH (aq)
17) O2 (g) + NH3 (g) --->H2O (L) + HNO3 (aq)
18) K (s) + Cl2 (g) ---> KCl (s)
19) Al (s) + HCl (aq) ---> H2 (g) + AlCl (aq)
20) Na3PO4 (aq) + CaCl2 (aq) ---> NaCl (s) + Ca3(PO4)2 (s)
Answer:
Ionization is defined as the process by which the atoms accept or donate electrons and form positively or negatively charged species, called ions.
The ionization of water is known as the self-ionization or autoionization of water. This process involves the deprotonation of a water molecule and protonation of another water molecule to give hydroxide ion and hydronium ion, which exist in the state of chemical equilibrium.
H₂O + H₂O ⇌ H₃O⁺ + OH⁻
The pH of pure water is 7, neutral.
Answer:
43.75 ml
Explanation:
Given that the equation of the reaction is;
2HNO3(aq) + Ca(OH)2(aq) ---> Ca(NO3)2(aq) + 2 H20(l)
Concentration of acid CA= 0.05 M
Concentration of base CB = 0.02 M
Volume of acid VA = 35.00ml
Volume of base VB= ???
Number of moles of acid NA= 2
Number of moles of base NB=1
From
CAVA/CBVB= NA/NB
Making VB the subject of the formula;
VB= CAVANB/CBNA
VB= 0.05 × 35 × 1/ 0.02 × 2
VB=1.75 /0.04
VB= 43.75 ml
