The bicarbonate ion [tex](HCO_3-)[/tex]is an important buffer system in the blood that helps to maintain the pH of the blood within a narrow range.
The concentration of [tex]HCO_3[/tex]- can be calculated from the pH and [tex]pCO_2[/tex] levels through the Henderson-Hasselbalch equation. This equation relates the pH, [tex]pCO_2[/tex], and [tex]HCO_3[/tex]- concentration in the blood as follows: pH = pKa + log ([[tex]HCO_3[/tex]-]/[CO2]), where pKa is the dissociation constant for [tex]H_2CO_3[/tex] (carbonic acid), [[tex]HCO_3[/tex]-] is the bicarbonate concentration, and [CO2] is the dissolved carbon dioxide concentration. By rearranging this equation, we can solve for the bicarbonate concentration [[tex]HCO_3[/tex]-] = (0.03 × [tex]pCO_2[/tex])/(10^(pH - pKa)). Therefore, if we know the pH and [tex]pCO_2[/tex] levels, we can calculate the [tex]HCO_3[/tex]- concentration.
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be sure to answer all parts. write the equation for the acid-base reaction that takes place when nitric acid (hno3) dissolves in h2o. (include the phase of each substance.)
The equation for the acid-base reaction that takes place when nitric acid (HNO3) dissolves in water (H2O) is:
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
What makes an acid strong?When a strong acid like HNO3 is dissolved in water, it completely dissociates into its ions. The HNO3 molecule transfers a proton (H+) to a water molecule, which acts as a base by accepting the proton. The result is the formation of hydronium ion (H3O+) and nitrate ion (NO3-):
HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq)
The hydronium ion (H3O+) is an important species in acidic solutions and is responsible for the acidic properties of the solution. It can donate a proton to a base, thus catalyzing acid-catalyzed reactions. The nitrate ion (NO3-) is the conjugate base of nitric acid and is itself a weak base.
The state symbols "(aq)" and "(l)" indicate that the reactants and products are in solution and liquid phases, respectively. In other words, HNO3 is in the form of an aqueous solution, meaning it is dissolved in water. The water molecule is also in the liquid phase, as it is not a gas or a solid in this reaction.
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what is the equilibrium concentration of the product c given the initial concentration of a was 0.70 m. 2a(g) <----> 1b(g) 2c(g) k = 2.2 x 10-8
The equilibrium concentration of C is 1.04 x 10⁻⁴ M.
The equilibrium constant expression for the given reaction is Kc = [c]²/[a]², where [a] is the initial concentration of A, [b] is the equilibrium concentration of B, and [c] is the equilibrium concentration of C.
Using the equilibrium constant (Kc) value given (2.2 x 10⁻⁸) and the initial concentration of A (0.70 M), we can solve for the equilibrium concentration of C:
Kc = [c]²/[a]²
2.2 x 10⁻⁸ = (2[c]²)/(0.7)²
2.2 x 10⁻⁸ x 0.7² = 2[c]²
[c]² = 1.078 x 10⁻⁸
[c] = 1.04 x 10⁻⁴ M
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Calculate:
a) ΔG 0
.
b) the equilibrium constant for the formation of NO 2
from NO and O 2
at 298 K.
NO(g)+ 2
1
O 2
(g)⇌NO 2
(g) where Δ f
G 0
(NO 2
)=52.0 kJ/mol Δ f
G 0
(NO)=87.0 kJ/mol Δ f
G 0
(O 2
)=0 kJ/mol.
a. The standard Gibbs free energy change for the reaction can be calculated as follows: [tex]K = e^{(-2.56 kJ/mol).[/tex]
b. The equilibrium constant can be calculated using the expression: K = [tex]e^{(-2.56 kJ/mol) .[/tex]
a) ΔG 0 = -ΔG_rxn + ΔG_m
ΔG_rxn = -ΔH_rxn + RT ln Q
The standard enthalpy change for the reaction can be calculated as:
ΔH_rxn = H_f - H_c
The standard enthalpy of formation of [tex]NO_2[/tex] can be found to be 52.0 kJ/mol.
The reaction quotient Q = [Products] / [Reactants] is a dimensionless quantity that measures the relative concentrations of the products and reactants.
Using the reaction quotient, we can determine the equilibrium constant K:
K = [Products] / [Reactants]
At equilibrium, the concentrations of the products and reactants will be equal, and the reaction quotient will be close to 1.
The equilibrium constant can be calculated using the expression:
[tex]K = e^{(-ΔG rxn / RT)}\\\\K= e^{(-(52.0 kJ/mol - (87.0 kJ/mol + 0 kJ/mol)) / (298 J/mol * 3.14))}\\\\K = e^{(-2.56 kJ/mol)[/tex]
b) The standard Gibbs free energy change for the formation of NO2 can be calculated as follows: ΔG 0 = -ΔG_rxn + ΔG_m
ΔG_rxn = -ΔH_rxn + RT ln Q
The standard enthalpy change for the reaction can be calculated as this:
The standard enthalpy of formation of [tex]NO_2[/tex] can be found to be 52.0 kJ/mol.
The reaction quotient Q = [Products] / [Reactants] is a dimensionless quantity that measures the relative concentrations of the products and reactants.
Using the reaction quotient, we can determine the equilibrium constant K:
K = [Products] / [Reactants]
At equilibrium, the concentrations of the products and reactants will be equal, and the reaction quotient will be close to 1.
The equilibrium constant can be calculated using the expression:
[tex]K = e^{(-ΔG rxn / RT)}\\\\K = e^{(-(52.0 kJ/mol - (87.0 kJ/mol + 0 kJ/mol)) / (298 J/mol * 3.14))}\\\K = e^{(-2.56 kJ/mol)[/tex]
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what is the formal charge on the oxygen atom in n2o (the atomic order is n–n–o)? group of answer choicesA. -1B. 2C. 0 D.4E. 1
The oxygen atom in N₂O has a formal charge of -1, as indicated by option A.
To determine the formal charge on the oxygen atom in N₂O, we need to compare the number of valence electrons the atom has to the number of lone pair electrons and shared electrons it possesses.
In N₂O, each nitrogen atom contributes 5 valence electrons, and oxygen contributes 6 valence electrons.
To calculate the formal charge on the oxygen atom, we use the formula:
Formal charge = Valence electrons - Lone pair electrons - Shared electrons/2
For oxygen in N₂O, the formal charge can be calculated as:
Formal charge = 6 - 4 - (6/2) = 6 - 4 - 3 = -1
Therefore, the formal charge on the oxygen atom in N₂O is -1.
The correct answer is option A. -1.
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if 200 g of carbon reacts with excess sulfur dioxide to produce 225 g of carbon disulfide (cs2 ; molar mass 76.139 g/mol), what is the percent yield for the reaction?
The percent yield for the reaction is approximately 17.74%.
To calculate the percent yield for the reaction, we need to first determine the theoretical yield and then compare it with the actual yield.
1. Balanced equation for the reaction:
C + 2SO2 → CS2 + 2SO
2. Calculate moles of carbon:
200 g C × (1 mol C / 12.01 g C) = 16.66 mol C
3. Calculate theoretical yield:
16.66 mol C × (1 mol CS2 / 1 mol C) × (76.139 g CS2 / 1 mol CS2) = 1268.5 g CS2
4. Calculate percent yield:
Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (225 g CS2 / 1268.5 g CS2) × 100
≈ 17.74%
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Use the data in Appendix L to determine the equilibrium constant for the following reactions. Assume 298. 15 K if no temperature is given. (a) AgCl(s) ⇌ Ag+(aq) + Cl−(aq) (b) CdS(s) ⇌ Cd2+(aq) + S2−(aq) at 377 K (c) Hg2+(aq) + 4Br−(aq) ⇌ [HgBr4 ] 2−(aq) (d) H2 O(l) ⇌ H+(aq) + OH−(aq) at 25 °C
Kc = [ [tex]Cl^-[/tex]][[tex]Ag^+[/tex]]/[AgCl] is the equilibrium constant for the given balanced equation AgCl(s) ⇌ [tex]Ag^+[/tex](aq) + [tex]Cl^-[/tex](aq).
When a chemical process reaches equilibrium, the equilibrium constant (typically indicated by the symbol K) offers information on the relationship among the products and reactants. The equilibrium constant of concentration (denoted by Kc) between a chemical reaction at equilibrium, for example, can be defined as the ratio of product concentration to reactant concentration, each increased to their respective stoichiometric coefficients.
Kc = [product]/[reactant]
Kc = [ [tex]Cl^-[/tex]][[tex]Ag^+[/tex]]/[AgCl]
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A gas occupies 505 cm' at STP. A weather change reduces volume to 710 cm?, calculate the temperature.
Answer:
Temperature = 382.913 Kelvin
Explanation:
To calculate the temperature, use the ideal gas law:-
PV = nRT
Here
- P stands for Pressure, measured in atm.
- V stands for Volume, measured in liters.
- n represents the Number of moles.
- R is the Ideal gas constant, with a value of 0.0821 L·atm/(mol·K).
- T stands for Temperature, measured in Kelvin.
The conditions defined as STP (Standard Temperature and Pressure) consist of a temperature of 273.15 Kelvin (equivalent to 0 degrees Celsius) and a pressure of 1 atmosphere.
We Have
Initial volume (V1) = 505 cm³
Final volume (V2) = 710 cm³
Initial temperature (T1) = 273.15 K (STP)
First, convert the volumes to liters:
V1 = 505 cm³ = 505/1000 L = 0.505 L
V2 = 710 cm³ = 710/1000 L = 0.710 L
Pressure and the number of moles are constant, we can write the equation as:
V1/T1 = V2/T2
Solving for T2:
T2 = (V2 * T1) / V1
T2 = (0.710 L * 273.15 K) / 0.505 L
On Solving the equation :-
T2 = 383.913 K (Approx)
Based on the weather change, the gas temperature is estimated to be around 383.913 Kelvin.
