why graphite is a nonmetal, why it is not included in metalloid.

Answer: D

Explanation:

A reducing agent is a species that reduces other compounds, and is thereby oxidized. The whole compound becomes the reducing agent. In other words, of a compound is oxidized, then they are the reducing agent. On the other hand, if the compound is reduced, it is an ozidizing agent.

Since we have established that a reducing agent is the compound being oxidized, we know that A is not our answer. An oxidized compound is losing electrons. Choice A states exactly this.

For B, this is true as we have established this already.

C is also correct. Since a reducing agent loses electrons, it becomes more positive. This makes the oxidation number increase.

D would be our correct answer. It is actually a good oxidizing agent is a metal in a high oxidation state, such as Mn⁷⁺.

Answer:

During complete oxidation of the fatty acid CH3(CH2)14COOH, eight molecules of acetyl-CoA are produced, and the fatty acid goes through the β-oxidation cycle seven times.

Explanation:

In the β-oxidation of fatty acids, an acetyl-CoA molecule is removed from the fatty acid chain after every β-oxidation cycle that the fatty acid undergo, leaving behind a fatty acyl-CoA molecule shortened by two cabon atoms..

The removal of the acetyl-CoA molecule starts from the carboxyl end and shortens the fatty acid molecule by two carbon units. Successive β-oxidation cycles results in the complete oxidaton of the fatty acid molecle to acetyl -CoA molecules.

The compound CH3(CH2)14COOH, is a 16-carbon saturated fatty acid molecule known as palmitic acid.. It undergoes seven passes through the β-oxidation cycle to yield eight molecules of acetyl-CoA with each cycle yielding an acetyl-CoA molecule and a fatty acyl-CoA shortened by two carbon atoms. Finally, the seventh step yields two acetyl-CoA molecules.

During complete oxidation of CH3(CH2)14COOH, eight (8) molecules of acetyl-CoA are produced, and the fatty acid goes through the β-oxidation cycle seven (7) times.

In conclusion, during complete oxidation of CH3(CH2)14COOH, eight (8) molecules of acetyl-CoA are produced, and the fatty acid goes through the β-oxidation cycle seven (7) times.

Learn more in:

https://brainly.com/question/13254687

Answer:

Explanation:

Be careful. The tricky part of the problem is that there are 4 places of sig digs.

m = 6.557 grams

V = 11.16 - 10.50 = 0.66

density = mass/ volume

density = 6.557/0.66 = 9.935 g/mL

Answer:

It gives the same result when an experiment is repeated.

Explanation:

Below are the possible answers to the question:

It is biased.

It cannot be verified.

It cannot add new evidence to existing evidence.

It gives the same result when an experiment is repeated.

The correct answer would be that it gives the same result when an experiment is repeated.

If a reaction is conducted in chemistry and certain results are obtained, once a detailed procedure of the experiment is known along with all the chemicals involved, such reaction/experiment can be repeated anywhere in the world and the same result would be obtained.

The repeatability of experiments always makes the experiments to be reliable. Hence, chemistry is reliable because it gives the same result without any variation when experiments are repeated under similar conditions.

Answer:

6.20 mol

Explanation:

Step 1: Write the balanced reaction

FeO + CO → Fe + CO₂

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of carbon monoxide (CO) to iron (Fe) is 1:1.

Step 3: Calculate the moles of Fe produced from 6.20 moles of CO

6.20 mol CO × (1 mol Fe/1 mol CO) = 6.20 mol Fe

Answer:

Good precision because he is hitting the same spot 3 times but we don't know if he is accurate. SO the answer is 2

Answer:

e is the correct one , hope I helped u

Answer:

6 H⁺ + BrO₃⁻ + 3 Sb³⁺ ⟶ Br⁻ + 3 H₂O + 3 Sb⁵⁺

Explanation:

Step 1: Write the unbalanced reaction

BrO₃⁻ + Sb³⁺ ⟶ Br⁻ + Sb⁵⁺

Step 2: Identify both half-reactions

Reduction: BrO₃⁻ ⟶ Br⁻

Oxidation: Sb³⁺ ⟶ Sb⁵⁺

Step 3: Perform the mass balance, adding H⁺ and H₂O where appropriate

6 H⁺ + BrO₃⁻ ⟶ Br⁻ + 3 H₂O

Sb³⁺ ⟶ Sb⁵⁺

Step 4: Perform the charge balance, adding electrons where appropriate

6 H⁺ + BrO₃⁻ + 6 e⁻ ⟶ Br⁻ + 3 H₂O

Sb³⁺ ⟶ Sb⁵⁺ + 2 e⁻

Step 5: Multiply both half-reactions by numbers that assure that the number of electrons gained and lost is the same

1 × (6 H⁺ + BrO₃⁻ + 6 e⁻ ⟶ Br⁻ + 3 H₂O)

3 × (Sb³⁺ ⟶ Sb⁵⁺ + 2 e⁻)

Step 6: Add both half-reactions and cancel what is repeated in both sides

6 H⁺ + BrO₃⁻ + 3 Sb³⁺ ⟶ Br⁻ + 3 H₂O + 3 Sb⁵⁺

Answer:

-133.2 kJ

Explanation:

Let's consider the following balanced equation.

4 KClO₃(s) → 3 KClO₄(s) + KCl(s)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression.

ΔG°rxn = 3 mol × ΔG°f(KClO₄(s)) + 1 mol × ΔG°f(KCl(s)) - 4 mol × ΔG°f(KClO₃(s))

ΔG°rxn = 3 mol × (-303.1 kJ/mol) + 1 mol × (-409.1 kJ/mol) - 4 mol × (-296.3 kJ/mol)

ΔG°rxn = -133.2 kJ

Answer:

Number of Significant Figures

0.0000620  = 6

0.05600  = 3

70820.0  = 5

309100 = 6

Standard Scientific Notation

0.0000620  = 6.2 * [tex]10^{-5}[/tex]

0.05600  = 5.6 * [tex]10^{-2}[/tex]

70820.0  = 7.082 * [tex]10^{4}[/tex]

309100 = 3.091 * [tex]10^{5}[/tex]

Answer:

unchanged

Explanation:

The density of an object or of a substance is independent of the amount of substance. From the definition of density as mass per unit volume, the ratio of the mass of an object to the volume of the same object is always constant. The more the mass of the object, the more the volume, and vice-versa.

Hence, if a brownie that measures 15 cm by 7 cm by 7cm and has a density of 2.3 g/cm^3 is cut into half, both the mass and the volume would change in equal proportion and the density would remain the same.

Answer:

3.3 L

Explanation:

Step 1: Given data

Step 2: Calculate the final volume of the balloon

Inside the balloon we have gas. If we consider it behaves as an ideal gas, we can calculate the final volume using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁ / P₂

V₂ = 2.7 atm × 1.6 L / 1.3 atm

V₂ = 3.3 L

The given question is incomplete. The complete question is :

For the reaction [tex]PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)[/tex] at equilibrium, which statement correctly describes the effects of increasing pressure and adding [tex]PCl_5[/tex], respectively

a) Increasing pressure causes shift to reactants, adding [tex]PCl_5[/tex] causes shift to products.

b) Increasing pressure causes shift to products ,adding [tex]PCl_5[/tex] causes shift to reactants.

c) Increasing pressure causes shift to products, adding [tex]PCl_5[/tex] causes shift to products.

d) Increasing pressure causes shift to reactants,adding [tex]PCl_5[/tex] causes shift to reactants

Answer: Increasing pressure causes shift to reactants, adding [tex]PCl_5[/tex] causes shift to products.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

[tex]PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)[/tex]

a)  If the pressure is increased, the volume will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. As the number of moles of gas molecules is lesser at the reactant side. So, the equilibrium will shift in the left direction. i.e. towards reactants.

b) If [tex]PCl_5[/tex] is added, the equilibrium will shift in the direction where [tex]PCl_5[/tex] is decreasing. So, the equilibrium will shift in the right direction. i.e. towards products.

Answer:

31.2 g of Ag₂SO₄

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2AgNO₃(aq) + H₂SO₄ (aq) → Ag₂SO₄ (s) + 2HNO₃ (aq)

From the balanced equation above,

2 moles of AgNO₃ reacted with 1 mole of H₂SO₄ to produce 1 mole of Ag₂SO₄ and 2 moles of HNO₃.

Next, we shall determine the limiting reactant.

This can obtained as follow:

From the balanced equation above,

2 moles of AgNO₃ reacted with 1 mole of H₂SO₄.

Therefore, 0.2 moles of AgNO₃ will react with = (0.2 x 1)/2 = 0.1 mole of H₂SO₄.

From the calculations made above, only 0.1 mole out of 0.155 mole of H₂SO₄ given is needed to react completely with 0.2 mole of AgNO₃. Therefore, AgNO₃ is the limiting reactant.

Next,, we shall determine the number of mole of Ag₂SO₄ produced from the reaction.

In this case we shall use the limiting reactant because it will give the maximum yield of Ag₂SO₄ as all of it is consumed in the reaction.

The limiting reactant is AgNO₃ and the number of mole of Ag₂SO₄ produced can be obtained as follow:

From the balanced equation above,

2 moles of AgNO₃ reacted to produce 1 mole of Ag₂SO₄.

Therefore, 0.2 moles of AgNO₃ will react to produce = (0.2 x 1)/2 = 0.1 mole of Ag₂SO₄.

Therefore, 0.1 mole of Ag₂SO₄ is produced from the reaction.

Finally, we shall convert 0.1 mole of Ag₂SO₄ to grams.

This can be obtained as follow:

Molar mass of Ag₂SO₄ = (2x108) + 32 + (16x4) = 312 g/mol

Mole of Ag₂SO₄ = 0.1

Mass of Ag₂SO₄ =?

Mole = mass /Molar mass

0.1 = Mass of Ag₂SO₄ /312

Cross multiply

Mass of Ag₂SO₄ = 0.1 x 312

Mass of Ag₂SO₄ = 31.2 g

Therefore, 31.2 g of Ag₂SO₄ were obtained from the reaction.

Taking into account the definition of reaction stoichiometry and limiting reagent, the mass of Ag₂SO₄ that could be formed is 31.18 grams.

First of all, the balanced reaction is:

2 AgNO₃ +  H₂SO₄  → Ag₂SO₄  + 2 HNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 1 mole of H₂SO₄ reacts with 2 moles of AgNO₃, 0.155 moles of H₂SO₄ react with how many moles of AgNO₃?

[tex]amount of moles of AgNO_{3} =\frac{0.155 moles of H_{2}SO_{4} x 2 moles of AgNO_{3} }{1 mole of H_{2}SO_{4} }[/tex]

moles of AgNO₃= 0.31 moles

But 0.31 moles of AgNO₃ are not available, 0.200 moles are available. Since you have less moles than you need to react with 0.155 moles of H₂SO₄,  AgNO₃ will be the limiting reagent.

Then, it is possible to determine the amount of moles of Ag₂SO₄ produced by another rule of three, using the limiting reagent: if by stoichiometry 2 moles of AgNO₃ produce 1 mole of Ag₂SO₄, 0.200 moles of AgNO₃ how many moles of Ag₂SO₄ will be formed?

[tex]amount of moles of Ag_{2} SO_{4} =\frac{1 mole of Ag_{2} SO_{4} x 0.200 moles of AgNO_{3} }{2 moles of AgNO_{3} }[/tex]

amount of moles of Ag₂SO₄ =0.100 moles

Finally, with 311.8 g/mole being the molar mass of Ag₂SO₄, then the mass produced of the compound can be calculated as:

[tex]0.100 molesx311.8 \frac{g}{mole} = 31.18 grams[/tex]

In summary, the mass of Ag₂SO₄ that could be formed is 31.18 grams.

Learn more with this examples:

Answer:

1.63×10¹⁰ L.

Explanation:

We'll begin by calculating the number of mole in 125 g of silver (ii) oxide, AgO. This can be obtained as follow:

Mass of AgO = 125 g

Molar mass of AgO = 108 + 16 = 124 g/mol.

Mole of AgO =?

Mole = mass /Molar mass

Mole of AgO = 125 /124

Mole of AgO = 1.01 moles.

Next, we shall convert 6.19×10¯⁵ μmol/L to mol/L. This can be obtained as follow:

1 μmol/L = 1×10¯⁶ mol/L

Therefore,

6.19×10¯⁵ μmol/L = 6.19×10¯⁵ × 1×10¯⁶ = 6.19×10¯¹¹ mol/L

Finally, we shall determine the volume as follow:

Molarity = 6.19×10¯¹¹ mol/L

Mole of AgO = 1.01 moles

Volume =?

Molarity = mole /Volume

6.19×10¯¹¹ = 1.01/Volume

Cross multiply

6.19×10¯¹¹ × volume = 1.01

Divide both side by 6.19×10¯¹¹

Volume = 1.01/6.19×10¯¹¹

Volume = 1.63×10¹⁰ L

Therefore, the volume of the solution is 1.63×10¹⁰ L

Answer:

Amphetamine | C9H13N | CID 3007 - structure, chemical names, physical and chemical properties , ... Defined Bond Stereocenter Count, 0, Computed by PubChem ... Collision Energy, 10 eV .

Taking into account the definition of units/mL, the correct option is option C. The nurse should administer 2.15 mL.

Heparin is an anticoagulant drug used in the treatment and prevention of blood clots caused by certain medical conditions or medical procedures. Heparin is also used before surgery to reduce the risk of blood clots. That is, this drug works by decreasing the blood's ability to clot.

The results of some medical tests are reported in units per milliliter (u/mL). This unit of concentration expresses unit of activity per unit of volume.

In this case, Heparin is available in 2,000 units/mL ands this means that for every 1 mL, there are Heparin 2,000 units.

Then you can apply the following rule of three: if 2,000 units are present for every 1 mL, 4,300 units are present in how many mL?

volume= (4,300 units× 1 mL) ÷ 2,000 unita

Solving:

volume= 2.15 mL

In summary, the correct option is option C. The nurse should administer 2.15 mL.

Learn more with this examples:

Answer:

a melting of lead Is the answer

Answer:

[tex]P_2=0.393atm[/tex]

Explanation:

Hello,

In this case, we can use the combined ideal gas law in order to analyze its behavior as a function of changing temperature, volume and pressure:

[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]

Thus, since we know the volume, temperature and pressure at the initial condition, we can compute the final pressure as shown below:

[tex]P_2=\frac{P_1V_1T_2}{T_1V_2} =\frac{1.00atm*225mL*(25+273)K}{(0+273)K*625mL}\\ \\P_2=0.393atm[/tex]

Regards.

Answer:

A. True

Explanation:

Carboxypeptidases are enzymes which function in the digestion of short peptides known as oligopeptides in the small intestine. Oligopeptides contain between 10 to 50 amino acid residues.

The two carboxypeptidases A and B involved in the digestion of proteins in the small intestine are secreted by the exocrine glands of the pancreas.They are both zinc-containing enzymes which remove successive carboxyl-terminal  (C-terminal) residues from oligopeptides until free amino acids are obtained.

Since they function in this way, they can be used to determine the entire sequence of short peptides or oligopeptides.

Answer:

See answer below

Explanation:

First, you are not putting the whole data, however, I found an exercise very similar to this, so I'm gonna use this data, as an example, and you can use it as guide to solve yours. The exercise is the following:

A chemist must prepare 275. mL of 1967 uM aqueous copper(II) fluoride (Cur) working solution. He'll do this by pouring out some 2.63 mmol/L aqueous copper(II) fluoride stock solution into a graduated cylinder and diluting it with distilled water. Calculate the volume in mL of the copper(II) fluoride stock solution that the chemist should pour out.

According to this, we want a more dilluted solution fo copper Fluoride solution. To do this, we need to know the moles that are required in the desired solution.

1967 uM in just mol/L is:

1967 uM * 1 M / 1x10⁶ uM = 1.967x10⁻³ M

Now that we have the concentration, we can calculate the moles required to prepare this solution:

moles = 1.967x10⁻³ mol/L * 0.275 L = 5.41x10⁻⁴ moles

These are the moles that we need to have to prepare this solution. Now, with the concentration of the stock solution, we just solve for the volume required:

V = moles / M

V = 5.41x10⁻⁴ / 2.63x10⁻³

V = 0.2057 L ----> 205.7 mL or 206 mL

Answer:

Octane - Formula: C8H^18 = Carbon^8 + Hydrogen^18

Explanation: Octane is a compound because there are 8 atoms of carbon and 18 atoms of hydrogen in one molecule of C8H18. There are also 8 moles of carbon and 18 moles of hydrogen.

Answer:

see below

Explanation:

Aqueous hydroiodic acid reacts with liquid water in a reversible reaction.

2 HI  +  2 H₂O  ⇔  2 H₃O  +  I₂

This is an acid-base reaction.  HI is the acid, and H₂O is the base.  The conjugate acid is H₃O, and the conjugate base is I₂.  You can figure out the acid and base by a using a table of strong/weak acids and bases.  The conjugate will have one less or one more H⁺ depending on if it's an acid or a base.

The expression for Kₐ of the reaction is below.  You have to put the acid and its conjugate base on the top and the base and its conjugate on the bottom.  I didn't put in H₂O since, although it is the base in this equation, it isn't really a base.

[tex]K_{a}=\frac{[HI][H_{3}O] }{[I_{2}] }[/tex]

Solid zinc reacts with solid manganese (IV) oxide producing solid zinc oxide and solid manganese (III) oxide.

Zn  +  2 MnO₂  ⇒  ZnO  +  Mn₂O₃

This is a redox reaction.  The Zn is being oxidized, and the Mn is being reduced.

Oxidation:    Zn  ⇒  Zn⁺²  +  2 e⁻

Reduction:   Mn⁺⁴  +  e⁻  ⇒  Mn⁺³

Aqueous silver nitrate reacts with aqueous calcium chloride.

2 AgNO₃  +  CaCl₂  ⇒  2 AgCl  +  Ca(NO₃)₂

This is a precipitation reaction.  You have two soluble salts combing to form a solution.  You now need to figure out which of the compounds produced are not soluble.  AgCl is the compound that is not soluble in water, making it the precipitate.

I think I covered everything.  If I missed something let me know.

Answer:

1.87x10⁻³ M SO₄²⁻

Explanation:

The reaction of SO₄²⁻ with Ba²⁺ (From Ba(NO₃)₂) is:

SO₄²⁻(aq) + Ba²⁺(aq) → BaSO₄(s)

Where 1 mole of  SO₄²⁻ reacts per mole of Ba²⁺

To reach the end point in this titration, we need to add the same moles of Ba²⁺ that the moles that are of SO₄²⁻.

Thus, to find molarity of SO₄²⁻ we need to find first the moles of Ba²⁺ added (That will be the same of SO₄²⁻). And as the volume of the initial sample was 100mL we can find molarity (As ratio of moles of SO₄²⁻ per liter of solution).

Moles Ba²⁺:

7.48mL = 7.48x10⁻³L ₓ (0.0250moles / L) = 1.87x10⁻⁴ moles of Ba²⁺ = Moles of SO₄²⁻

Molarity SO₄²⁻:

As there are 1.87x10⁻⁴ moles of SO₄²⁻ in 100mL = 0.1L, molarity is:

1.87x10⁻⁴ moles of SO₄²⁻ / 0.1L =

Answer:

The product formed is orthonitrophenol

Explanation:

The compound orthonitrophenol is formed when ortho-bromonitrobenzene is treated with NaOH at elevated temperature.

The reaction leads to the formation of an intermediate stabilized by resonance. This intermediate can not be formed by the meta isomer because a resonance stabilized intermediate can not be formed and the product is not stabilized by intramolecular hydrogen bonding hence this reaction will not occur.

Answer:

1. Ice at 0 degrees C.

2. N₂ at STP.

3. N₂ at STP.

4. Water vapor at 150 degrees C and 1 atm.

Explanation:

First, we need to remember that entropy (S) is a measure of how spread out or dispersed the energy of a system is among the different possible ways that system can contain energy. The greater the dispersal, the greater is the entropy.

When the temperature is increased, the energies associated with all types of molecular motion increase. Consequently, the entropy of a system always increases with increasing temperature.

With this in mind, we consider the pairs:

1. Since the ice at 0ºC has a greater temperature than the ice at -40 ºC, the first has the higher entropy.

2. The N₂ at STP (that is, 1 atm and 25 ºC) has higher entropy than N₂ at 0ºC and 10 atm because it has a higher temperature and less pressure, which allows a greater dispersal of energy by the molecules of the gas.

3. The N₂ at STP has a higher entropy since it has a higher temperature than N₂ at 0ºC, even though it the first has a lower volume (24,4 L vs. 50 L).

4. The water vapor at 150 ºC and 1 atm have a higher temperature and a lower pressure. This means that its molecules will have an increased molecular motion than the molecules of water vapor at a lower temperature and higher pressure. Therefore, the first has the highest entropy.

Answer:

See figure 1

Explanation:

For this question, we can start with the structure of pyruvic acid. In this molecule, we have two functional groups, the ketone group, and a carboxylic acid group.

In the acid group, we have an acidic hydrogen. That is, this hydrogen can leave the molecule to produce a hydronium ion ([tex]H^+[/tex]).

Now, to know what is the pH at which this hydrogen leaves the molecule we must look for the pKa value. Which for the case of this molecule is 2.45

If there is a pH value greater than 2.45, the molecule will lose hydrogen. In this case, we will have a value above 2.45 (7.4), which is why the conjugated acid of pyruvic acid will be produced, which is pyruvate.

Therefore, in the structure, we will have a negative charge on the acid group.

See figure 1

I hope it helps!

Answer:

The lower the temperature, the lower the pressure inside of the sphere. If the exact values were plotted out, a linear relationship would be apparent. Extrapolating this line to the point where there would be no pressure yields absolute zero, which is about -273.15 degrees Celsius.