use the references to access important values if needed for this question. hybrid orbitals are formed by combining the valence orbitals on an atom. a molecule has hybridization with 1 lone pair. the electron pair geometry of this molecule is: . the geometry of this molecule is: . this molecule will have approximate bond angles of (choose all that apply.):

Answer:

28.05316 grams

Explanation:

Answer:

25

Explanation:

25 because when you divide 50/2 it will be 25m/s

The best assessment of Gina's prediction is that the several medium-sized cubes will dissolve faster than the large cube of sugar.

The surface area of a substance refers to the total area on the surface of that object.

Surface area plays a vital role in the dissolution of a substance. When a solute dissolves, the action takes place only at the surface of each particle.

This suggests that when the total surface area of the solute particles is increased, the solute dissolves more rapidly, hence, the best assessment of Gina's prediction is that the several medium-sized cubes will dissolve faster.

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The mass of \(Al_{2} S_{3}\) produced is 15.616 grams.

The reaction taking place is as follows:

\(2Al+ 3S\) →\(Al_{2} S_{3} (s)\)

Moles of Al = mass/molar mass = 10.0g/27.0g/mol

= 0.370 mol

Moles of S = mass/molar mass = 15.0g/(32.065g/mol)

= 0.468 mol

Al and S reacts in the molar ratio of 2:3.

2 moles of Al reacts with 3 moles of S

0.370 moles of Al will react with S = (3/2)*0.370mol

= 0.555 mol

Similarly, 0.468 moles of S will react with Al = 2/3 *0.468mol

= 0.312 mol

Thus, Al is in excess and S is the limiting reactant (some of Al will be left over ,S will completely react)

So, moles of \(Al_{2} S_{3}\) produced=1/3*0.312 mol of S

= 0.104 mol

Mass of \(Al_{2} S_{3}\) produced = moles*molar mass of \(Al_{2} S_{3}\)

= 0.104mol*150.158g/mol

= 15.616 grams

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Answer:

56 %

Explanation:

The percent dissociation of a weak acid can be defined as:

Moles of weak acid that produced H₃O⁺ species / Total number of weak acid moles * 100%

We are given all the required data to calculate the percent dissociation:

0.42 mol / 0.75 mol * 100 % = 56 %

The percent dissociation is 56%.

Answer:

Explanation:

25.8 ml of .095 N NaOH is needed to neutralise the remaining acid

equivalent of NaOH used = 25.8 x .095 / 1000 = .002451 gm equivalent .

acid remaining = .002451 gm equivalent .

acid initially taken = 100 ml of .1 N  /  1000  = . 01 gm equivalent

acid reacted with metal = .01  -.002451 = .007549 gm equivalent

This must have reacted with same gram equivalent of metal oxide

.007549  gm equivalent = .15 gm of metal oxide

1 gm equivalent = 19.87 gm

equivalent weight of metal = 19.87 - equivalent weight of oxygen

= 19.87 - 8 = 11.87 .

1

Total pressure is defined as the sum of all partial pressures in a container, therefore in our question we have:

Ptotal = 5.76 atm

PO2 = 1.31 atm

PN = 2.22 atm

PAr = 0.77 atm

PCO2 = ?

Since the total pressure is the sum of all partial pressures, if we subtract the partial pressures that we know from the total pressure we will have the value of partial pressure for CO2

5.76 - 1.31 = 4.45

4.45 - 2.22 = 2.23

2.23 - 0.77 = 1.46 atm, this is the Partial Pressure for CO2

The third law of thermodynamics describes the entropy of a: solid.

The third law of thermodynamics states that the entropy of a pure crystalline substance approaches zero as the temperature approaches absolute zero (0 Kelvin or -273.15 degrees Celsius). This law implies that at absolute zero, a perfectly ordered and pure crystalline solid will have zero entropy.

The third law of thermodynamics is not specific to liquids or gases but applies to solids. In a solid, the molecules are highly ordered and have fixed positions in a regular lattice structure. As the temperature decreases towards absolute zero, the thermal motion of the molecules reduces, and the system becomes more ordered, resulting in a decrease in entropy.

In contrast, liquids and gases have higher entropy compared to solids at absolute zero because their molecules have more freedom of movement and are not as tightly arranged. Therefore, the third law of thermodynamics specifically addresses the entropy of solids and does not apply to liquids or gases.

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Answer:

- 10.6° C

Explanation:

Given that,

In the aqueous sucrose solution,

Molality of the solution (m) = 5.7

Freezing point constant (K) = 1.86

To find,

The freezing point of the solution = ?

Method:

To find its freezing point, we will first find out if the solute demonstrates an electrolyte or not.

In case it is not, the van't Hoff factor i.e. ration of particles' concentration denoted by (i) would be taken as 1. So,

ΔT(change in temperature) = i * m * K              (putting the given values)

= 1 * 5.7 * 1.86

= 10.6° C

Thus, the freezing point equals to 0 - 10.6° C = -10.6° C.

Explanation:

could you post the text? its inconvenient to go back and forth to the picture.

Answer:

eyes hair skin

Explanation:

eyes you can get from parents skin and hair

Answer:

H2C2O2 + 2KOH ----- K2C2O4 + 2H2O

Answer: False

Explanation:

Answer:

Carnivore, as they obtain food from other animals

Answer:

carnivores

Explanation:

Explanation:

2 KOH (aq) + H2SO4 (aq) →K2SO4 (aq) + 2 H2O (l)

Answer:

7.4 moles solute

Explanation:

From definition of Molarity (M) = moles solute (n)/volume of solution in Liters(V)

=> moles solute (n) = Molarity (M) x Volume (L) = (3.7 moles/L)(0.500L) = 7.4 moles solute

Answer:

65.2%

Explanation:

It is given that :

\($2NO + O_2 \rightarrow 2NO_2$\)

Therefore, 60 g of nitrogen oxide will produce 92 g of nitrogen dioxide

or 60 kg of nitrogen oxide will produce 92 kg of nitrogen dioxide

or 1500 kg of nitrogen oxide will produce \(2300\ kg\) of nitrogen dioxide

Therefore the percentage yield = (Actual yield / Expected yield) x 100

percentage yield = \($\frac{1500}{2300} \times 100$\)

Percentage yield = 65.217 %

Answer:

the person above me is correct, the answer for this is 65.2% or option D!

Explanation:

thank you person above me! i got a 100% on the test! :D

Answer:

Ar

Explanation:

Electron affinity and electron negativity both have the same trend of decreasing down a period. If the electron negativity is 0 than the electron affinity is also 0

Argon(Ar) has an electron affinity measuring closest to zero.

Electron affinity It is the degree at which an electron is added to a  neutral atom or molecule. That is, the ability of an atom or molecule to accept electrons

Argon is a group 0 element with a complete octet structure. It rarely combines with other atoms or elements. That is, it rarely gains or loses electrons.

This means that argon has a very low electron affinity because of its unreactive nature.

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Answer:

it is 52.7

Explanation:

so when it is 30 the degree is 5°

when it is 60, the degree is 10°

when it is 90, the degree is 15° and so on...

Answer:

T₂ = 695.33 K

Explanation:

Given data:

Initial volume = 150 mL

Initial temperature = 25°C (25+273 = 298 K)

Final temperature = ?

Final volume = 350 mL

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ /V₁

T₂ = 350 mL × 298 K / 150 mL

T₂ = 104300 mL.K / 150 mL

T₂ = 695.33 K

Answer:

H3PO4 + 3KOH -> K3PO4 + 3H20

Answer:

hope its helpful to uh....

The category of wastes that are not subject to regulation as RCRA hazardous wastes is option b, secure chemical landfill.

Radioactive and brake fern wastes are hazardous wastes that are subject to regulation under RCRA. However, secure chemical landfills are designed to safely manage and dispose of hazardous wastes, so they are exempt from RCRA regulations as long as they meet certain criteria.

These criteria include using a liner system to prevent leakage, installing a leachate collection system, and monitoring groundwater for contamination. It is important to note that even though secure chemical landfills are exempt from RCRA regulation, they are still subject to other federal and state regulations to ensure they are operating safely and minimizing environmental impacts.

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The true statements are:1. The addition of concentrated nitric acid to each standard solution results in a relatively constant ionic strength across the standard solutions.
2. The addition of concentrated nitric acid to each standard solution results in the required amount of excess nitrate ion.

These two statements are true because adding concentrated nitric acid to each standard solution maintains consistent ionic strength and provides the necessary excess nitrate ions for the reactions or analysis being performed. The other options do not accurately describe the effects of adding concentrated nitric acid to standard solutions.

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Answer: Hydropower plants can negatively affect the fish population in an ecosystem.

Boyles law shows the relation between effect of pressure on volume at constant temperature.

In boyle's law mass is independent variable and volume is dependent variable. The equation of boyle's law is

pV= constant.

Because the gas particles are compressed closer together as the pressure on a gas rises, the volume of the gas decreases.

In contrast, as a gas's pressure falls, its volume rises as a result of the gas's ability to spread its particles farther apart. Because the volume of the gas has risen, weather balloons become larger as they ascend through the sky to lower pressure areas.

Therefore, Boyles law shows the relation between effect of pressure on volume at constant temperature.

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The law described is Gay-Lussac's law. According to Gay-Lussac's law, the pressure of a gas is directly proportional to its absolute temperature when the volume and amount of gas are held constant. In other words, if the absolute temperature of a gas sample in a rigid container is doubled, its pressure will also double.

Gay-Lussac's law is one of the fundamental gas laws in thermodynamics. It is named after the French chemist Joseph Louis Gay-Lussac, who formulated this law in the early 19th century. The law can be mathematically expressed as P1/T1 = P2/T2, where P1 and P2 represent the initial and final pressures, and T1 and T2 represent the initial and final absolute temperatures of the gas.

This law is applicable when the volume of the gas remains constant. It provides a relationship between the pressure and temperature of a gas, illustrating that as the temperature increases, the gas molecules move with higher kinetic energy, resulting in increased collisions with the container walls, hence raising the pressure.

Conversely, if the temperature decreases, the pressure of the gas will decrease as well. Gay-Lussac's law is essential in understanding the behavior of gases under different temperature conditions and has practical applications in various fields, including chemistry, physics, and engineering.

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Answer:

Visible light waves let you see the world around you. The different frequencies of visible light are experienced by people as the colors of the rainbow.

Explanation:

Answer:

12ml per hour and

69% LOL

1) The rate of percolation would be 0.33 ml/s

Percolation rate = volume of water/time

                             =  600/1800s

                              = 0.33 ml/s

2) The percentage of water absorbed would be 40%

Percentage of water absorbed = amount absorbed/total water x 100

                 amount of water absorbed = 100 - 60

                                                      = 40 ml

                  total amount of water = 100 ml

   Percentage of water absorbed = 40/100 x 100

                                                       = 40%

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The correct algebraic equation for determining the units of the rate constant k is p = M-x+y/s.

For the general reaction aA + bB →cC + dD, the general rate law, rate = k[A]x[B]y is given.

The units of the rate constant k can be determined using the algebraic equation derived from the given rate law. Let's look at the steps to derive the algebraic equation:

Step 1: Writing the units of rate

Rate = k[A]x[B]y, where the units of rate = M/s (Molarity per second), since concentration is in M and time is in seconds.

Step 2: Writing the units of concentration

Concentration is given in M, which is moles of solute per liter of solution.

Therefore, the units of [A] and [B] are M.

Step 3: Writing the units of the rate constant, k

Let's assume the units of k are p.

So, the units of rate constant, k can be determined using the following algebraic equation:

rate = k[A]x[B]yM/s

= p[M]x[M]y or pMx+y/s ... (Equation 1)

Equation 1 shows the units of the rate constant k when the concentration is in M and time is in seconds.

Therefore, the correct algebraic equation for determining the units of the rate constant, k, is p = M-x+y/s.

Reaction is a process that involves the rearrangement of atoms, ions, or molecules in either single or multiple steps.

A chemical reaction can be defined as a process where two or more reactants are converted into a single or multiple products.

A chemical reaction can be classified as physical, chemical, endothermic, exothermic, or reversible. The rate of a chemical reaction can be defined as the change in concentration of the reactants or products over time.

The rate of a reaction depends on various factors like temperature, pressure, surface area, and concentration.

The general rate law for a chemical reaction is given by rate = k[A]x[B]y where k is the rate constant and x and y are the orders of reaction with respect to A and B, respectively.

The rate constant k is a proportionality constant that relates the rate of reaction to the concentration of reactants raised to their respective orders. The units of the rate constant k depend on the units of concentration and time.

When the concentration is in M (molarity) and time is in seconds, the units of k can be calculated using the algebraic equation derived from the rate law.

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The units of the rate constant, k, in a rate law equation depend on the reaction order. For a zero-order reaction, k is in M/s. A first order reaction, k units are s^-1, and for a second order reaction, units are M^-1s^-1.

The algebraic equation for determining the units of the rate constant, k, when concentration is in moles per liter (M) and time is in seconds, depends on the order of the reaction represented by x and y in the rate law equation (rate = k[A]x[B]y)

The units for k vary because the rate of the reaction depends on the molar concentration of the reactants to the power of their respective reaction order. This changes the dimensions associated with the rate constant in the rate equation.

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