A(n) 7.26−mol sample of carbon monoxide was stored in a 30.0−L container at 46.0°C. What is the pressure of the carbon monoxide in the container?

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Answer:

AZ₂

Explanation:

The formed compound will not contain Q because mass percentages, %A and %B, form 100% of the compound.

Steps

Solving

Taking into account the reaction stoichiometry, 8.8488 grams of (NH₄)₃PO₄ is produced by the reaction of 5.82g of phosphoric acid.

In first place, the balanced reaction is:

H₃PO₄ + 3 NH₃  → (NH₄)₃PO₄

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 molar mass of the compounds is:

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

The following rule of three can be applied: if by reaction stoichiometry 98 grams of H₃PO₄ form 149 grams of (NH₄)₃PO₄, 5.82 grams of H₃PO₄ form how much mass of (NH₄)₃PO₄?

mass of (NH₄)₃PO₄=\(\frac{5.82 grams of H_{3} PO_{4}x149 grams of (NH_{4} )_{3}PO_{4} }{98 grams of H_{3} PO_{4}}\)

mass of (NH₄)₃PO₄= 8.8488 grams

Finally, 8.8488 grams of (NH₄)₃PO₄ is produced by the reaction of 5.82g of phosphoric acid.

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In the given reaction, hydrogen undergoes both oxidation and reduction. Hydrogen first oxidises from 0 to +1 oxidation state and then reduced  to -1 in LiH.

Oxidation is the process by which element combines with oxygen or loses electron forming its higher oxidation state. Whereas reduction is the process of combining with hydrogen or gaining electron to form the lower oxidation state of the element.

If an element undergoes both oxidation and reduction in the same reaction, it is called as disproportionation reaction. In the given reaction hydrogen undergoes disproportionation reaction.

In the molecular state H₂ is in 0 oxidation state and when it turns to LiNH₂, it oxidises to + 1 oxidation state and then reduces to -1 in LiH. Hence, option C is correct.

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Your question is incomplete. But your complete question probably was:

Because pure H2 is a hazardous substance, safer and more cost effective techniques to store it as a solid for shipping purposes have been developed. One such method is the reaction represented above, which occurs at 200 degrees Celsius. Which of the following happens to H atoms in the forward reaction?

Li₃N(s) + 2 H₂(g) ←→ LiNH₂(s) + 2 LiH(s)

(A) H atoms are oxidized only.

(B) H atoms are reduced only.

(C) H atoms are both oxidized and reduced.

(D) H atoms are neither oxidized nor reduced.

Answer:

41826.05 Calories

Explanation:

1 J = 0.239006 Calories

175 KJ

= 175 x 1000 J

= 175000 J

175000 J to Calories

= 175000 x 0.239006

= 41826.05 Calories

Answer: 41.8

Explanation:

Acellus verified ✅

Answer:

The dolphin's fast cruising speed (a travelling speed they can maintain for quite a while) is about 3-3.5 m/s (6-7 knots, 11-12.5 km/hr)

Explanation:

so 37 kilometers

A. The patch's area in square kilometers (km²) is 1.61×10⁻⁹ km²

B. The cost of the patch to the nearest cent is 734 cents

We can convert 16.1 cm² to km² as illustrated below:

Conversion scale

1 cm² = 1×10⁻¹⁰ km²

Therefore,

16.1 cm² = 16.1 × 1×10⁻¹⁰

16.1 cm² = 1.61×10⁻⁹ km²

Thus, 16.1 cm² is equivalent to 1.61×10⁻⁹ km²

We'll begin by converting 16.1 cm² to in². This can be obtained as illustrated below:

1 cm² = 0.155 in²

Therefore,

16.1 cm² = 16.1 × 0.155

16.1 cm² = 2.4955 in²

Finally, we shall the determine the cost in centas fo r llow:

1 in² = 294 cent

Therefore,

2.4955 in² = 2.4955 × 294

2.4955 in² = 734 cents

Thus, the cost of the patch is 734 cents

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Compose balanced molecule, gross ionic, as well as net ionic formulae for the interaction between sodium hydroxide plus nitric acid in aqueous solutions. The net ionic equation's coefficient sum is three.

Ammonium nitrate, a crucial component of fertilizers, is created using nitric acid. Additionally, it is used to oxidize materials and to make explosives like trinitrotoluene (TNT) as well as nitroglycerin.

The liquid leads to severe burns when it comes in contact with your eyes, which could cause lifelong damage and vision loss. The liquefied or concentrated vapor causes immediate, severe, and profound burns on the skin, while acidic solution result in deep ulcers and leave a brilliant yellow stain.

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The moles of glucose that produces that amount of energy is 0.825 moles of glucose.

The process of metabolism is the process by which food is broken down in the body to generate the energy that we need to carry out our daily tasks. We know that the energy that is burnt by the individual that is weighing 150 pounds jogging at 5 mph for 30 minutes is 231 kcal but the energy that is produced by the metabolism of one mole of the glucose molecule is 280 kilocalories of energy.

Now we are required to obtain the moles of glucose would need to be metabolized to generate this required energy. We can write that;

231 = 280n

n = 231/280

n = 0.825 moles of glucose.

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Answer: D

Explanation:  Irregular galaxies

Answer:

Your answer is D

Explanation:

Irregular galaxies do not have a regular structure.

Answer:

Explanation:

The Law of Conservation of Mass is defined and explained using examples of reacting mass calculations using the law are fully explained with worked out examples using the balanced symbol equation. The method involves reacting masses deduced from the balanced symbol equation.

Answer:

53.7 L

Explanation:

The volume occupied by 35.2 g of methane gas at 25 °C and 1.0 atm. is 55.9 L.

Given to us is temperature, pressure, universal gas constant, and mass of methane gas, we need to find the volume occupied by the gas.

To calculate the volume occupied by the methane gas, we can use the ideal gas law equation:

PV = nRT

Where:

P = pressure (in atm)

V = volume (in liters)

n = number of moles

R = ideal gas constant (0.08206 Latm/(Kmol))

T = temperature (in Kelvin)

First, we need to convert the given temperature from Celsius to Kelvin:

T(K) = T(°C) + 273.15

T(K) = 25 °C + 273.15

T(K) = 298.15 K

Next, we calculate the number of moles of methane gas using its molar mass:

molar mass of CH₄ = 12.01 g/mol + 4(1.008 g/mol) = 16.04 g/mol

n = mass/molar mass

n = 35.2 g / 16.04 g/mol

n = 2.19 mol

Now we can substitute the values into the ideal gas law equation:

PV = nRT

V = (nRT) / P

V = (2.19 mol × 0.08206 Latm/(Kmol) ×298.15 K) / 1.0 atm

V = 55.9 L

Therefore, the volume occupied by 35.2 g of methane gas at 25 °C and 1.0 atm is 55.9 liters.

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Answer:  **

H-N-H

   |

  H

Explanation:

Look at a periodic table to determine how many electrons you need to account for. Hydrogen (H) only has 1 electron, while Nitrogen (N) has 5. We have three Hydrogen atoms and one Nitrogen atom, so the total number of electrons will be 3 * 1 + 5 = 8 e-.

Now, place the center atom, which will be Nitrogen and place the three Hydrogens on three sides of it as above in the answer. You should use single bonds for this. Each single bond is a pair of electrons, so since we have three single bonds so far, we have accounted for 2 * 3 = 6 electrons. However, we need 2 more electrons for the total of 8. We put these electrons in as a lone pair above Nitrogen.

We check to see if everything follows the octet rule: Nitrogen has three single bonds, so that's 6 e-, as well as one lone pair, so that's another 2 e- for a total of 8 e-. Check. Now look at Hydrogen: H is the only element whose full orbital is 2 e-. Each H has a single bond with Nitrogen, so each does have 2 e-.

Thus, we know this is the correct diagram, and we are done.

Explanation:

A bond line structure of a compound has H N H in linear plane and a hydrogen is branching upward, and the compound is H N (H) H. The nitrogen has two dots at its bottom represents a lone pair of electrons. So ,the correct answer is option C.

The correct Lewis structure for ammonia (\(NH_3\)) is option C. It shows a bond line structure with three hydrogen atoms (H) bonded to a central nitrogen atom (N) in a linear plane.

One hydrogen atom branches upward from the plane. Additionally, the nitrogen atom in this structure has two dots at its bottom, indicating a lone pair of electrons. This arrangement follows the octet rule, as nitrogen has formed three covalent bonds with hydrogen, completing its valence shell. The lone pair on nitrogen gives ammonia its characteristic properties.

Thus, option C accurately represents the Lewis structure of ammonia, showing the bonding and lone pair arrangement of its atoms.

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Answer:

Kinematics is a branch of physics concerned with the geometrically possible motion of a body or system of bodies without considering the  forces which are involved

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Answer:  Two cold air masses surround a warm air mass during a warm front, but a cold air mass moves against a warm air mass during an occluded front.

Answer:

A

Explanation:

The weak base ionization constant (Kb) for \(C_{4} H_{4} N_{2}\) is equal to Option A.

A weak foundation is a base that makes ionize only slightly in an aqueous solution. Remember that a foundation can be defined as an object, which receives a hydrogen ion from another object.

To obtain an aqueous solution of a weak acid, the dissociation constant is called the acid ionization constant (Ka). Similarly, the constant equilibrium of a weak base reaction with water is the base ionization constant (Kb). In any conjugate acid-base pair, KaKb = Kw.

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Answer:

The observed temperature when the flame momentarily ignites the vapor/air mixture is the flashpoint. The observed temperature when the burning becomes continuous is the fire point.

Answer:

Water has both the higher surface tension and viscosity

Explanation:

Water has stronger intermolecular forces

Water has a higher viscosity and surface tension than methanol.

Water has a high melting and boiling point irrespective of its low relative molecular mass because of strong intermolecular hydrogen bonding in the liquid.

This strong intermolecular hydrogen bonding also accounts for its high viscosity and surface tension. Surface tension is a force acting along the surface of a liquid that makes it behave like a stretched elastic skin.

Methanol has a lesser magnitude of intermolecular hydrogen bonding compared to water hence it has less surface tension and viscosity compared to water.

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The growth of the new plant depicts the asexual reproduction characteristic. The characteristic that describes the growth of the new plant in Stephan's mother cutting a twig from a rose bush and planting it in the soil is asexual reproduction.

Asexual reproduction is the mode of reproduction by which organisms generate offspring that are identical to the parent's without the fusion of gametes. Asexual reproduction is a type of reproduction in which the offspring is produced from a single parent.

The offspring created are clones of the parent plant, meaning they are identical to the parent.The new plant in Stephan’s mother cutting a twig from a rose bush and planting it in the soil depicts the process of asexual reproduction, which is the ability of a plant to reproduce without seeds. In asexual reproduction, plants can reproduce vegetatively by cloning themselves using their roots, bulbs, or stems.

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Answer:

Third option, because copper (Cu) is a stronger oxidizing agent than zinc (Zn).

Explanation:

The table shows the Oxidation Half-Reaction, so the elements oxidizes.

When an element is oxidized, it causes the other element in the reaction to be reduced, therefore the element being oxidized is a reducing agent.

In this case, the correct reaction is:

because, copper (Cu) is a stronger oxidizing agent than zinc (Zn).

Answer:

Remain the same

Explanation:

If water dose no effect why would the moist air make a change

If you increase the pressure of the reaction 2A(g) + B(g)C(g) + 2D(g)the reaction will shift forward to the side with little moles of gas.

Le Chatelier's principle states that if pressure is raised, the reaction will favor the direction with a lower number of gas moles.

On this  case, the cleared out side includes a add up to of three gas moles (two moles of A and one mole of B), whereas the correct side covers four gas moles (one mole of C and two moles of D). In this way, the balance will move towards the left-hand side in arrange to diminish the collective gas particles and relieve the rise in weight. The levels of A and B will rise, as the levels of C and D will decrease.

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Answer:

G3P, ADP, and NADP+

Explanation:

Therefore in photosynthesis, NADPH is used for storing energy. The light-dependent reaction of photosynthesis utilizes water (H2O) and produces oxygen (O2).

...

Products of light-independent reactions.

Purpose Utilize stored chemical energy for carbon dioxide fixation and form glucose as a product

Input ATP, NADPH, CO2

Chemical elements are classified according to their atomic radius, which is typically the average or typical distance between the centre of the nucleus and the outermost isolated electron.

The total distance from an atom's nucleus to its outermost electron orbital is typically defined as the atomic radius. It can be described more simply as something akin to the radius of a circle, with the nucleus acting as the circle's centre and the outermost orbital of the electron as the circle's perimeter. Trends explain how atomic radii change start to appear as you start moving across or down the periodic table. Some of the properties are:

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Answer:

B. PROTONS EXHIBIT STRONGER PULL ON OUTER f ORBITALS

Explanation:

Lanthanide contraction is the greater than normal decrease in the ionic radius of the lanthanide series from atomic number 57 to atomic number 71. This decrease is rather not expected of the ionic radii of these elements and they result in the greater decrease in the subsequent series of the lanthanides from the atomic number 72. The cause of which is as a result of the poor shielding effects of the nuclear charge around the electrons of the f orbitals. So therefore, protons are strongly pulled out of the 4f orbital and as a result of the poor shielding effect which causes the electrons of the 6s orbitals to be drawn more closer to the nucleus and hence resulting in a smaller atomic radii. It is worthy to note that the shielding effects of the inner electrons decreasing from s orbital to the f orbital; that is s > p > d > f. So from the decrease in the shielding effects from s to the f orbitals, lanthanide contraction results from the inability of the orbitals far away from s like the 4f orbiatls to shield the outermost shells of the lanthanide elements. So the cause of lanthanide contraction is the action of the protons which strongly pull the electrons of the f orbitals because of the poor shielding effects due to the distance of this orbital from the nucleus.

Answer:

B) Protons exhibit a stronger pull on outer f orbitals than on d orbitals.

Explanation:

Conservation of charge is a principle of chemistry that states that the total electric charge of the system in an isolated system remains constant. It is conserved in the universe and does not varies.

For that reason, the equation that shows conservation of charge is:

Which is option 2.

Answer:

B

Explanation:

There is no reason to run these solutes at the 20 MWCO because they are both much smaller than the MWCO of the membrane.

The MWCO (molecular weight cut off) is the molecular weight of a solute at which it will be retained by a membrane during a process such as ultrafiltration or dialysis. If a solute has a molecular weight higher than the MWCO of a membrane, it will be retained and not pass through the membrane. If the molecular weight of a solute is lower than the MWCO, it will pass through the membrane.

In this case, glucose has a molecular weight of 180 g/mol (6 carbons x 12 g/mol per carbon + 6 oxygens x 16 g/mol per oxygen) and albumin has a molecular weight of approximately 81,942 g/mol (607 amino acids x 135 g/mol per amino acid). Both of these solutes have molecular weights that are much lower than 20,000 g/mol, which is a typical MWCO for ultrafiltration or dialysis membranes.

They would both easily pass through the membrane and be lost during the process. Instead, a membrane with a much lower MWCO would be needed if we wanted to retain these solutes during a process such as ultrafiltration or dialysis.

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