For which of the following elements (in their normal, stable, forms) would it be correct to describe the bonding as involving "a sea of electrons"?

a. hydrogen
b. nellum
c. sulfur
d. Iodine
e. Ethium

Answer:

1.49e24

Explanation:

Standars temperature and pressure are 273.15K and 1atm, respectively.

Using ideal gas law, we can find moles of an ideal gas if we know its pressure, temperature and volume as follows:

PV = nRT

PV / RT = n

Where P is pressure (1atm), V is volume (55.3L), R is gas constant (0.082atmL/molK), T is temperature (273.15K) and n moles of the ideal gas.

Replacing:

PV / RT = n

1atm*55.3L / 0.082atmL/molK*273.15K = n

2.47 moles = n

Now, the question is about the number of molecules in the jar. By definition, 1 mole = 6.022x10²³ molecules.

As we have 2.47 moles:

2.47 mol × (6.022x10²³ molecules / 1 mole) =

1.49x10²⁴ molecules that are in the jar

In scientific notation:

Answer:

0.500 moles of oxygen

Explanation:

Full question says: "Calculate the moles of oxygen needed to produce 0.500 moles of acetic acid.

The reaction of ethanol (C₂H₅OH) with oxygen (O₂) is:

C₂H₅OH + O₂ → CH₃COOH + H₂O

Where 1 mole of ethanol reacts per mole of oxygen to produce 1 mole of acetic acid (CH₃COOH) and 1 mole of water

Based on the chemical equation (1 mole of oxygen produce 1 mole of acetic acid; Ratio 1:1). Thus, if you want to produce 0.500 moles of acetic acid you will need:

Answer:

0.0583g

Explanation:

The equation of the reaction is;

2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)

From the question, number of moles of HNO3 reacted= concentration × volume

Concentration of HNO3= 0.100 M

Volume of HNO3 = 20.00mL

Number of moles of HNO3= 0.100 × 20/1000

Number of moles of HNO3 = 2×10^-3 moles

From the reaction equation;

2 moles of HNO3 reacts with 1 mole of Mg(OH)2

2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2

But

n= m/M

Where;

n= number of moles of Mg(OH)2

m= mass of Mg(OH)2

M= molar mass of Mg(OH)2

m= n×M

m= 1×10^-3 moles × 58.3 gmol-1

m = 0.0583g

Answer:

H+/PO-4^-2

Explanation:

hydrogen has dissolved completely

In the given reaction conjugate base is H₂PO₄⁻, it also behave as a weak acid.

An acid and conjugate base pairs are those pairs in which they are differentiated by the one atom of hydrogen atom.

Given chemical reaction is:

H₃PO₄ + H₂O → H₂PO₄⁻ + H₃O⁺

In the above reaction H₃PO₄ is an acid as it gives H⁺ ion to the solution and formed H₂PO₄⁻, which is a conjugate base of H₃PO₄ acid. H₂PO₄⁻ will also behave as an acid because it have H⁺ ion to gives in the solution but nature of this acid is weak as they not readily dissociates.

Hence, H₂PO₄⁻ is a conjugate base.

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Answer:

315 g

Explanation:

Step 1: Write the thermochemical equation

2 H₂O(l) → 2 H₂(g) + O₂(g)      ΔH = +572 kJ

Step 2: Calculate the molar of water decomposed by 5.00 × 10³ kJ of energy

According to the thermochemical equation, 572 kJ are required to decompose 2 moles of water.

5.00 × 10³ kJ × (2 mol/572 kJ) = 17.5 mol

Step 3: Calculate the mass corresponding to 17.5 moles of water

The molar mass of water is 18.02 g/mol.

17.5 mol × 18.02 g/mol = 315 g

Answer:

a. Methanol remains the same

b. Methanol decreases

c. Methanol increases

d. Methanol remains the same

e. Methanol increases

Explanation:

Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.

a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.

b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.

c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.

d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.

e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.

Answer:

Here are four possible voltaic cells.  

Explanation:

1. Standard reduction potentials

                                         E°/V

I₂(s) + 2e⁻ ⟶ 2I⁻(aq);        0.54

Cu²⁺(aq) + 2e⁻ ⟶ Cu(s);   0.34

Fe²⁺(aq) + 2e⁻ ⟶ Fe(s);   -0.41

Zn²⁺(aq) + 2e⁻ ⟶ Zn(s);   -0.76

2. Possible Voltaic cells

(a) Zn/I₂

                                                                       E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                 0.76

Cathode:  I₂(s) + 2e⁻ ⟶ 2I⁻(aq);                     0.54

Cell:         Zn(s) +  I₂(s) ⟶  Zn²⁺(aq) + 2I⁻(aq); 1.30

Zn(s)|Zn²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)

Zn is the anode; graphite is the cathode.

(b) Zn/Cu²⁺

                                                                          E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                    0.76

Cathode:  Cu²⁺(aq) + 2e⁻ ⟶ Cu(s);                  0.34

Cell:          Zn(s) +  Cu²⁺(s) ⟶  Zn²⁺(aq) + Cu(s); 1.10

Zn(s)|Zn²⁺(aq)∥Cu²⁺(aq)|Cu(s)

Zn is the anode; Cu is the cathode.

(c) Zn/Fe²⁺

                                                                            E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                     0.76

Cathode:  Fe²⁺(aq) + 2e⁻ ⟶ Fe(s);                    -0.41

Cell:          Zn(s) +  Fe²⁺(s) ⟶  Zn²⁺(aq) + Fe(s);  0.35

Zn(s)|Zn²⁺(aq)∥Fe²⁺(aq)|Fe(s)

Zn is the anode; Fe is the cathode.

(d) Fe/I₂

                                                                         E°/V

Anode:     Fe(s) ⟶ Fe²⁺(aq) + 2e⁻;                   0.41

Cathode: I₂(s) + 2e⁻ ⟶ 2I⁻(aq);                     0.54

Cell:         Zn(s) +  I₂(s) ⟶  Zn²⁺(aq) + 2I⁻(aq); 0.95

Fe(s)|Fe²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)

Fe is the anode; graphite is the cathode.

Answer:

0.17 lb

Explanation:

78 g * (1 lb/454 g)=0.17 lb

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Answer:

D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice

Explanation:

Copper is a metal with an atomic number of 29. This metal is soft and reddish in color which explains why it is very malleable(beaten to form various shapes without breaking).

All metals are good conductors of electricity including copper which is also a metal. Metals generally are insoluble in water. Copper also has a high melting point which is a characteristic of metals due to their giant structure and metallic lattice which makes it difficult to be broken down.

Answer:

Half-life at 629K = 252.4min

Explanation:

Using Arrhenius equation:

[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]

And as Half-life in a first order reaction is:

[tex]t_{1/2}=\frac{ln2}{K}[/tex]

We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:

[tex]58.0min=\frac{ln2}{K}[/tex]

K = 0.01195min⁻¹ = K₁

[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]

[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]

[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]

K₂ = 2.75x10⁻³ min⁻¹

And, replacing again in Half-life expression:

[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]

The half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.   

The activation energy of a reaction is related to its rate constant as follows:  

[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex]   (1)

Where:

We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:

[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex]   (2)

Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min

Hence, the rate constant at 652 K is:                            

[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]

Now, from equation (1) we can find the pre-exponential factor (A):

[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]  

With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):

[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]

Finally, the half-life at 629 K is (eq 2):

[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]

Therefore, the half-life at 629 K is 251.1 min.

Find more about activation energy here:

I hope it helps you!

Answer:

[tex][F^-]_{max}=4x10{-3}\frac{molF^-}{L}[/tex]

Explanation:

Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

[tex]Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)[/tex]

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:

[tex][F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}} \\[/tex]

[tex][F^-]_{max}=4x10{-3}\frac{molF^-}{L}[/tex]

Best regards.

Answer:

CO2

Explanation:

Mass of carbon = 27.29

Mass of oxygen = 72.21

Step 1:

We have to first convert these masses to moles

Carbon = 2.29/2.01 = 2.274

Oxygen = 72.71/16 = 4.544

Step 2:

We have to divide the mols by the smallest mol to get simplest ratio of whole number.

The smallest mol is 2.274. we have to divide the mols by this.

2.274/2.274 = 1

4.544/2.274 = 2

Our empirical formula is therefore CO2

Answer:

1. metals

2. atom

3. homogeneous

4. compounds

5. lustrous

6. saturated

7. colloidal

8. homogeneous

Explanation:

Answer:

[tex]\mathbf{S^0_{rxn} = -140.41 \ J/mol.K}[/tex]

Based upon the stoichiometry of the reaction the sign of Sºrxn should be negative

Explanation:

Consider the reaction:

H2CO(g) + O2(g) --------> CO2(g) + H2O(l)

Using standard thermodynamic data;

Based upon the stoichiometry of the reaction the sign of Sºrxn should be _________ . calculate Sºrxn at 25°C. Sºrxn = J/K•mol

At standard thermodynamic data

[tex]\mathtt{S^0_{rxn} = \sum S^0 _{product} - \sum S^0 _{reactant}}[/tex]

[tex]S^0(CO_2)[/tex] = 213.79 J/mol.K

[tex]S^0(H_2O)=[/tex]  69.95  J/mol.K

[tex]S^0 ({H_2CO}) =[/tex] 218.95 J/mol.K

[tex]S^0 (O_2)[/tex]  = 205.2 J/mol.K

[tex]\mathtt{S^0_{rxn} = (213.79 + 69.95) J/mol.K - (218.95+ 205.2) J/mol.K}[/tex]

[tex]\mathtt{S^0_{rxn} = (283.74) J/mol.K - (424.15) J/mol.K}[/tex]

[tex]\mathbf{S^0_{rxn} = -140.41 \ J/mol.K}[/tex]

Based upon the stoichiometry of the reaction the sign of Sºrxn should be negative

Answer:

A (Atomic mass number or Nucleon number)

Explanation:

The mass number is the total number of protons and nucleons in an atomic nucleus.

Hope this helps.

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Answer:

I. dipole-dipole

III. dispersion

IV. hydrogen bonding

Explanation:

Intermolecular forces are weak attraction force joining nonpolar and polar molecules together.

London Dispersion Forces are weak attraction force joining non-polar and polar molecules together. e.g O₂, H₂,N₂,Cl₂ and noble gases. The attractions here can be attributed to the fact that a non -polar molecule sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant.

Dispersion forces are the weakest of all electrical forces that act between atoms and molecules. The force is responsible for liquefaction or solidification of non-polar substances such as noble gas an halogen at low temperatures.

Dipole-Dipole Attractions are forces of attraction existing between polar molecules ( unsymmetrical molecules) i.e molecules that have permanent dipoles such as HCl, CH3NH2 . Such molecules line up such that the positive pole of one molecule attracts the negative pole of another.

Dipole - Dipole attractions are more stronger than the London dispersion forces but weaker than the attraction between full charges carried by ions in ionic crystal lattice.

Hydrogen Bonding is a dipole-dipole intermolecular attraction which occurs when hydrogen is covalently bonded to highly electronegative elements such as nitrogen, oxygen or fluorine. The highly electronegative elements have very strong affinity for electrons. Hence, they attracts the shared pair of electrons in the covalent bonds towards themselves, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom ( nitrogen in the case of CH3NH2 ) . This attractive force is know as hydrogen bonding.

Answer:

The intermolecular forces present in CH_3NH_2 includes

Explanation:

The intermolecular forces present in CH_3NH_2 includes II. (ion-dipole) and IV. (hydrogen bonding)

It is a polar molecule due to NH polar bond and it can form Hydrogen bond also due to NH bond.

Interaction will be dipole- dipole and Hydrogen dispersion forces can always be taken into account.

For more information on intermolecular forces, visit

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Answer:

C

Explanation:

According to the mole ratio, using 2NH3 will give you 3H2. Which means in order to find the moles of H2 you would only need to divide 2 and multiply 3 to get the amount of moles of H2 produced.

Answer:

I think it's C

Explanation:

Please, tell me if I'm incorrect.

Answer:

ΔHrxn = -635.14kJ/mol

Explanation:

We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:

(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol

(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole

(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole

Reaction (1) - (3) produce:

Ca(s) + H2O(l) → H2(g) + CaO(s)

ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol

Now this reaction + 1/2(2):

Ca(s) + ½ O2(g) → CaO(s)

ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)

Answer:

[tex]\Delta G=-97.14kJ[/tex]

Explanation:

Hello,

In this case, the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

[tex]\Delta G=-RTln(K)[/tex]

Hence, we compute it as required:

[tex]\Delta G=-8.314\frac{J}{mol\times K}*298K*ln(1.5297x10^7)\\\\\Delta G=-40.99kJ/mol[/tex]

And for 2.37 moles of hydrogen bromide, we obtain:

[tex]\Delta G=-40.99kJ/mol*2.37mol\\\\\Delta G=-97.14kJ[/tex]

Best regards.

Explanation:

this is the ans

hope this helps

Answer:

There are fifteen molecular orbitals in benzene filled with electrons.

Explanation:

Benzene is an aromatic compound. Let us consider the number of bonding molecular orbitals that should be present in the molecule;

There are 6 C-C σ bonds, these will occupy six bonding molecular orbitals filled with electrons.

There are 6 C-H σ bonds, these will occupy another six molecular orbitals filled with electrons

The are 3 C=C π bonds., these will occupy three bonding molecular pi orbitals.

All these bring the total number of bonding molecular orbitals filled with electrons to fifteen bonding molecular orbitals.

Answer:

[tex]SF_4[/tex]

Explanation:

The first thing would be to calculate the number of moles of each element in the compound.

No of moles of sulfur (S) = mass/molar mass

                                        = 29.68/32.065 = 0.9256

No of moles of fluorine (F) = mass/molar mass

                                       = 70.32/18.998 = 3.7014

Then, let us find the atomic ratio of each of the element in the compound by dividing by the no of moles by the smallest no of mole:

            S            :             F

      [tex]\frac{0.9256}{0.9256}[/tex] = 1       :       [tex]\frac{3.7014}{0.9256}[/tex] = 4

Therefore, the empirical formula for the compound is [tex]SF_4[/tex]

Answer:

there is one atom of oxygen and two atoms of hydrogen

Explanation:

Answer:

The pH of the solution is 9.06.

Explanation:

The reaction of the dissociation of NH₃ in water is:

NH₃(aq) + H₂O(l)  ⇄  NH₄⁺(aq) + OH⁻(aq)     (1)

[NH₃] - x                     [NH₄⁺] + x     x  

The concentration of NH₃ and NH₄⁺ is:

[tex] [NH_{3}] = \frac{n_{NH_{3}}}{V_{T}} = \frac{C_{i}_{(NH_{3})}*Vi_{NH_{3}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.12 M*0.2 L}{0.2 L + 0.25 L} = 0.053 M [/tex]

[tex] [NH_{4}^{+}] = \frac{C_{i}_{(NH_{4}^{+})*V_{NH_{4}^{+}}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.15 M*0.25 L}{0.2 L + 0.25 L} = 0.083 M [/tex]

From equation (1) we have:

[tex]Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}[/tex]

[tex] 1.8 \cdot 10^{-5} = \frac{(0.083 + x)*x}{0.053 - x} [/tex]

[tex] 1.8 \cdot 10^{-5}(0.053 - x) - (0.083 + x)*x = 0 [/tex]

By solving the above equation for x we have:

x =  1.15x10⁻⁵ = [OH⁻]

The pH of the solution is:

[tex] pOH = -log([OH^{-}]) = -log(1.15 \cdot 10^{-5}) = 4.94 [/tex]

[tex] pH = 14 - pOH = 14 - 4.94 = 9.06 [/tex]

Therefore, the pH of the solution is 9.06.

I hope it helps you!

Answer:

Atomic mass is calculated by adding protons and neutrons.

Explanation:

Atomic mass is the sum of protons and neutrons in an atomic nucleus. For example, the element Oxygen has 8 protons (derived from the atomic number) and 8 neutrons (derived from subtracting the amount of protons from the atomic mass).

We can craft an equation to show the relationship between these variables.

M - N = P, where M = Mass, N = Neutrons, and P = Protons

This equation can be rearranged to show the relationship between the neutrons and protons leading to the atomic mass. Simply add N to both sides of the equation.

M = N + P

This shows that atomic mass is equivalent to the sum of protons and neutrons in an atom's nucleus.

Answer:

The correct option is C.

Note the full question and structure of the moleculesis found in the attachment below.

Explanation:

Triglycerides or triacylglycerols are non-polar, hydrophobic lipid molecules composed of three fatty acids linked by ester bonds to a molecule of glycerol.

The fatty acids linked to the glycerol molecule are denoted by R and may be of the same kind or different. when the R group is the same, the R is attached in all the three positions for ester bonding in the glycerol molecule but when they are different are denoted by R, R' and R'' respectively.

During the hydrolysis of triglycerides, the three fatty acids molecules are obtained as well as a glycerol molecule.

From the question, when 1 mole of the triglyceride is hydrolysed, 2 moles of RCOOH, 1 mole of R'COOH and 1 mole of glycerol is obtained. The triglyceride must then be composed of two fatty acids which are the same denoted by R, and a different fatty acid molecule denoted by R'.

The correct option therefore, is C

Answer:

d. HF(g)

Explanation:

Hello,

In this case, the standard entropy S° could be predicted by looking at the amount of bonds the compound has, thus, the fewer the number bonds, the lower the standard entropy, it means that d. HF(g) has lowest value as it has one bond only whereas methane has four bonds, water two bonds and ammonia three bonds.

Best regards.

Answer:

Sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon are the elements of third period.

Explanation:

There are three energy levels in sodium atom. It has 11 electrons revolving around the nucleus. the atomic radius of sodium atom is 227 ppm. Magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon has also three energy levels like sodium because all these elements belongs to third period. There are 12 electrons present in magnesium, 13 in aluminium, 14 in silicon, 15 in phosphorus, 16 in sulfur, 17 in chlorine, and 18 electrons in argon. The atomic radius of magnesium atom is 173 ppm.  The atomic radius of aluminium atom is 143 ppm.  The atomic radius of silicon atom is 111 ppm.  The atomic radius of phosphorus atom is 98 ppm.  The atomic radius of sulfur atom is 87 ppm. The atomic radius of chlorine atom is 79 ppm and the atomic radius of argon atom is 71 ppm.

Answer:

NSAIDs are steroidal anti-inflammatories, their action is on the phospholipase A2 enzyme, this enzyme is responsible for breaking down the phospholipids of the membrane to trigger an inflammatory response. This is how steroidal anti-inflammatory drugs inhibit ALL inflammatory pathways (not like NSAIDs that they only inhibit the COX pathway).

These corticosteroid drugs cannot exceed the systemic mineralocorticoid value 1 in the body, since this corticosteroid hormone is also released by the adrenal cortex.

The NSAIDs generate: sporadic peaks in blood glucose, hypertension, fluid retention, increase in body fat mass, possible suppression of the adrenal cortex over time, inhibiting endogenous synthesis of corticosteroids.

On the other hand, naproxen and ibuprofen are NSAIDs, that is, non-steroidal anti-inflammatory drugs that do not influence both routes of inflammation, but only COX, this enzyme is abbreviated as COX but is called cyclooxygenase, and is responsible for a single route of inflammation.

NSAIDs such as naproxen and ibuprofen can cause gastric disorders such as ulcers or gastritis if they are consumed in a very repetitive manner.

In addition, both drugs are anti-inflammatory, analgesic and antipyretic. Although its two main functions are the first two, it was shown to have an effect in lowering body temperature.

That they are anti-inflammatory means that they inhibit the path of inflammation and analgesics the path of pain.

Explanation:

Both types of drugs generate the same effect but by different mechanisms.

Some are steroids and others are not, although steroids are considered to have a greater risk of benefit that is why they are administered against more systematically compromised instances such as anaphylactic shock.

NSAIDs such as naproxen and ibuprofen are the most prescribed today, since they have few risks and very good benefits, meaning that their adverse effects are not lethal or highly relevant and have a good effect on symptoms.

Both must be administered with care when treating a diabetic patient since corticosteroids generate glycemic peaks or increase in blood glucose, and NSAIDs compete for plasma protein with oral hypoglycemic agents, thus generating that these are in higher free concentrations. high diffusing better through the tissues and increases the potency of the adverse effects of these.