If every oxygen ion is combined with an aluminium ion has a charge of -2,the charge of each aluminum ion would be -3.
Uncharged Aluminum atom must need to lose it's electrons,in order to form the bond with oxygen which has vacant orbitals
ion
atom that has a positive or negative charge because it lost or gained one or more electrons
chemical bond
the attractive force that holds atoms or ions together
ionic bond
a chemical bond in which one atom loses an electron and the other atom gains electrons to form ions
chemical formula
a combination of chemical symbols and numbers to represent a substance
covalent bond
bond formed by the sharing of electrons between atoms
Answer:
4,1,5,3,2 (from left to right)
Answer:
A
Explanation:
CH4+O2-CO2+ H20
that mean methane has burn in oxygen to produce CO2
Answer:
[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.
Explanation:
Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.
Analysis:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷
C(i) 0.115M 0 0
ΔC -x +x +x
C(eq) 0.115M - x x x
≅ 0.115M
Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M
= 4.3 x 10⁻⁷ => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.
In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion concentration, the hydroxide ion concentration is then calculated from
[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.
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NOTE: The 2.32 x 10⁻⁴M value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.
Answer:
The Flow rate = 0.0208 mL/min
Explanation:
Data provided:
Rate of dose = 39 mg every 30 min = (39/30) mg/min = 1.3 mg/min
also,
125mg of methylprednisolone is present in every 2 mL
thus,
concentration = (125/2) mg/mL = 62.5 mg/mL
Now,
The flow rate is given as:
Flow rate = Rate / Concentration
on substituting the respective values, we get
Flow rate = (1.3 mg/min) / (62.5mg/mL)
or
The Flow rate = 0.0208 mL/min
