A small, jagged particle with a density of 1.2 g/cm3 deposits on the ground before a large, round particle with a density of 0.82 g/cm3.

Which most likely influences the rate of deposition of the jagged particle?

large size
low density
high density
irregular shape

Answer:

what is the English translation, id like to help

Explanation:

Answer:

596 g of CO₂ is the mass formed

Explanation:

Combustion reaction:

C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)

We determine moles of ethylene that has reacted:

189.6 g . 1mol / 28g = 6.77 moles

We assume the oxygen is in excess so the limiting reagent will be the ethylene.

1 mol of ethylene produce 2 moles of CO₂ then,

6.77 moles will produce the double of CO₂, 13.5 moles.

We convert the moles to mass: 13.5 mol . 44 g /1mol = 596 g

Answer:

5 electrons

Explanation:

Valence electrons are electrons located in the outermost energy shell, so you want to count the number of electrons in the last energy shell.

You can divide the configuration into 1s2 / 2s22p6 / 3s23p3 to see the energy shells in this atom. There are 3 shells occupied by the atom's electrons, so you need to count the electrons in the third shell as those are its valence electrons.

2 + 3 = 5 valence electrons total

Note: you don't count the 3 before the letter because that only indicates the shell level, not the number of electrons. Count only the exponents.

An atom that has the following electron configuration.

1s22s22p63s23p3 contains only 5 valence electrons in its outermost shell.

Electronic configuration of chemical elements is the classification and ordering of electrons around the nucleus of an atom in reference to their energy levels.

Electrons can be ordered and classified using Aufbau's principle.

Aufbau's principle posits that in the ground state of an atom, electrons are ordered to the subshell of the lowest energy levels before filling the higher energy level subshell.

The valence electron of a chemical element is the outermost electron participating actively in a chemical bonding.

From the given electronic configuration, we will first determine the total atomic number (total electrons) before we determine the valence electrons in the outermost shell.

Given that

1s² 2s² 2p⁶ 3s² 3p³

There are (2 + 2 + 6 + 2 + 3) total electrons in the elctronic configuration.

= 15 total electrons

The maximum number of electrons in the energy levels is determined by using the formula is  [tex]\mathbf{2n^2}[/tex]

where;

n = 1, 2, 3 ....

In the first energy level, we have the maximum number of electrons to be:

= 2(1)²

= 2 electrons

In the second energy level, we have:

=  2(2)²

= 2(4) electrons

= 8 electrons

Since we have only 15 electrons, the number of the valence electrons in the atom is:

= 15- (2 + 8 ) electrons

= 15 electrons - 10 electrons

= 5 valence electrons left in the outermost shell.

In conclusion, an atom that has the following electron configuration 1s22s22p63s23p3 contains only 5 valence electrons in its outermost shell.

Learn more about electronic configuration here:

https://brainly.com/question/16762037?referrer=searchResults

Answer:  2.34 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

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

where,

[tex]P_1[/tex] = initial pressure of gas = 0.850 atm

[tex]P_2[/tex] = final pressure of gas = 1.50 atm

[tex]V_1[/tex] = initial volume of gas = 4.25 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]23.0^oC=273+23.0=293.0K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]11.5^oC=273+11.5=284.5K[/tex]

Now put all the given values in the above equation, we get:

[tex]\frac{0.850\times 4.25}{293.0K}=\frac{1.50\times V_2}{284.5}[/tex]

[tex]V_2=2.34L[/tex]

Thus the final volume will be 2.34 L

Hey there!

C₃H₆O(l) + O₂(g) => CO₂(g) + H₂O(g)

First thing I want to do is balance the equation. Balance C:

C₃H₆O(l) + O₂(g) => 3CO₂(g) + H₂O(g)

Balance H:

C₃H₆O(l) + O₂(g) => 3CO₂(g) + 3H₂O(g)

Balance O:

C₃H₆O(l) + 4O₂(g) => 3CO₂(g) + 3H₂O(g)

Now we can properly solve the problem.

a.)

Density of acetone is 0.800 g/mL, we need to form 67.2 L of CO₂ at STP.

For every one mole of acetone reacted, 3 moles of CO₂ is produced.

Let's convert 67.2 L to moles: At STP, one mole of a gas takes up 22.4 liters.

67.2 ÷ 22.4 = 3 moles

So turns out we want to produce 3 moles of CO₂, which means we need one mole of acetone.

The molar mass of acetone is 58.08 g/mol. The density of acetone is 0.800 g/mL. We need the volume of one mole.

58.08 ÷ 0.800 = 72.6

72.6 mL of acetone is needed.

b.)

For every one molecule of acetone reacted, 3 molecules of water is produced.

We are combusting 3.011 x 10²² molecules of acetone.

3.011 x 10²² x 3 = 9.033 x 10²²

Find the number of moles:

(9.033 x 10²²) ÷ (6.022 x 10²³) = 0.15 moles

The molar mass of water is 18.015 g/mol.

0.15 x 18.015 = 2.7 grams

2.7 grams of water vapor is formed.

Hope this helps!

Answer:

(D.) Nitrifiers are bacteria that generate nitrites or nitrates.

Explanation:

In the nitrogen cycle which occurs in nature, ammonia and ammonium compounds in the soil from organic sources and are converted to nitrites and nitrates by aerobic microorganisms.

Nitrifiers, as the name implies, are these such aerobic bacteria which oxidize inorganic constituents in the soil to generate energy. Examples of these nitrifiers are nitrobacter and nitrosomonas.

Answer:

1.0 *10^(-4) mol

Explanation:

For gases:

n1/n2 = V1/V2

n1/3.8*10^(-4) mol = 230 mL/ 860 mL

n1 = 3.8*10^(-4)*230/860 = 1.0 *10^(-4) mol

A balloon containing gas expands from 230 mL to 860 mL as more helium is added. 1.0 × 10⁻⁴ mole was the initial quantity of helium present if the expanded balloon contains 3.8 × 10⁻⁴ mole assuming constant temperature and pressure.

The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.

PV = nRT

where,

P = Pressure

V = Volume

n = number of moles  

R = Ideal gas constant

T = Temperature

Now, calculating the ratio between the initial and the final numbers of moles of gas

PV = nRT

or, [tex]\frac{V}{n} = \frac{(RT)}{P}[/tex]

or, [tex]\frac{V}{n} = k[/tex]

or, [tex]\frac{V_1}{n_1} = \frac{V_2}{n_2}[/tex]

or, [tex]\frac{n_1}{n_2} = \frac{V_1}{V_2}[/tex]        [Avogadro's Law]

Here, Volume and number of moles are the variables which are known.

Now put the values in above formula we get

⇒ [tex]\frac{n_1}{n_2} = \frac{V_1}{V_2}[/tex]

⇒ [tex]\frac{n_1}{3.8 \times 10^{-4}} = \frac{230}{860}[/tex]

⇒ [tex]n_{1} = \frac{3.8 \times 10^{-4} \times 230}{860}[/tex]

⇒ n₁ = 1.0 × 10⁻⁴ mole

Thus, we can say that 1.0 × 10⁻⁴ mole was the initial quantity of helium present if the expanded balloon contains 3.8 × 10⁻⁴ mole assuming constant temperature and pressure. Volume and number of moles are the variables which are known.

Learn more about the Ideal Gas here: https://brainly.com/question/1063475

#SPJ2

Answer:

32 grams

Explanation:

To find the molar mass of oxygen gas, you need to take a look at it's molecular formula. Oxygen is O2, and since each oxygen molecule has a molar mass of 16, 2 of them together have a molar mass of 32 grams. Hope this helps!

Answer:

3

Explanation:

Answer:

3

Explanation:

the 3 elements are carbon hydrogen and oxygen

Answer:

89.57 %

Explanation:

Don't have one just thought id help <3

Answer: 89.6

Explanation:

Answer:

-63.79 kJ/g

Explanation:

According to the law of conservation of energy, the sum of the heat released by the combustion of the new organic material (Qcomb) and the heat absorbed by the bomb calorimeter (Qbc) is zero.

Qcomb + Qbc = 0

Qcomb = -Qbc   [1]

We can calculate the heat absorbed by the bomb calorimeter using the following expression.

Qbc = C × ΔT

where,

Qbc = C × ΔT

Qbc = 28.81 kJ/°C × (30.28°C - 24.91°C) = 154.7 kJ

From [1],

Qcomb = -154.7 kJ

The heat of combustion per gram of the material is:

-154.7 kJ / 2.425 g = -63.79 kJ/g

Answer:

The heat of combustion per gram of the material is -63.8 kJ/ gram

Explanation:

Step 1: Data given

Mass of a ew organic material = 2.425 grams

The initial temperature of the calorimeter = 24.91 °C

The final temperature of the calorimeter = 30.28 °C

The heat capacity (calorimeter constant) of the calorimeter is 28.81 kJ /°C

Step 2: Calculate heat

Q = c*ΔT

⇒with c = the heat capacity (calorimeter constant) of the calorimeter is 28.81 kJ /°C

⇒with ΔT = The change of temperature = T2 - T1 = 30.28 - 24.91 = 5.37 °C

Q = 28.81 kJ/ °C * 5.37 °C

Q = 154.7 kJ

Step 3: Calculate the heat of combustion per gram of the material

heat of combustion per gram = -Q / mass   (negative since it's exothermic)

heat of combustion per gram = -154.7 kJ / 2.425 grams

heat of combustion per gram = -63.8 kJ/ gram

The heat of combustion per gram of the material is -63.8 kJ/ gram

Answer: False.

The final pressure in the system is 4.14 atm

Explanation: Please see the attachments below

Answer:

I hope you can get a good answer

Explanation:

What class is this for?

Answer:

The vibrations are then passed to 3 tiny bones in the middle ear called the ossicles. The ossicles amplify the sound. They send the sound waves to the inner ear and into the fluid-filled hearing organ (cochlea). Once the sound waves reach the inner ear, they are converted into electrical impulses.

Answer:

The beach ball will get larger at the beach because the molecules are moving faster.

Explanation:

We can answer this problem by using Charle's Law, which states that:

"For a constant mass of an ideal gas kept at constant pressure, the volume of the gas is directly proportional to its temperature"

Mathematically:

[tex]V\propto T[/tex]

where

V is the volume of the gas

T is the absolute temperature of the gas

Here, we are analyzing the air inside the ball (the gas). We are also told that when the ball is brought to the beach, the pressure does not change: therefore, we can apply Charle's Law.

We are told that when the ball is brought to the beach, the temperature increases from 18℃ to 32℃: therefore, since the volume of the air (and the ball) is proportional to the temperature, this means that the volume of the ball will increase as well.

The reason for this is that the ball is not thermally isolated, so the molecules of the air inside the ball reach soon the same temperature of the surroundings, and so they will move faster (higher temperature means higher average kinetic energy of the molecules, so the molecules move faster, and therefore the ball will expand).

So the correct option is

The beach ball will get larger at the beach because the molecules are moving faster.

Jorge's beach ball when he is at the beach should be beach ball will get larger at the beach because the molecules are moving faster.

Here we applied Charle law, in this the volume of the gas is directly proportional to its temperature. Here the ball should not be isolated due to this it moved faster i.e. if there is high temperature that means there should be high kinetic energy so the molecules should be moved quicker due to this the ball should be expanded.

Learn more about molecule here: https://brainly.com/question/19531144

Answer: Cell A performs all of the organism's necessary functions, while cell B performs only one or a few functions.

Answer:

C

Explanation:

C is the only one that is true. Because sugar comes before juice in the ingredients listed, there is more of it.

Answer:

3041.5L

Explanation:

Data obtained from the question include:

V1 (initial volume) = 2785 L

P1 (initial pressure) = 830mmHg

P2 (final pressure) = stp = 760mmHg

V2 (final volume) =?

Applying the Boyle's law equation P1V1 = P2V2, the final volume propane can be achieved by as illustrated below:

P1V1 = P2V2

830 x 2785 = 760 x V2

Divide both side by 760

V2 = (830 x 2785)/760

V2 = 3041.5L

Therefore, the volume of the propane at standard pressure is 3041.5L

By applying Boyle's Law and using the given initial conditions, the volume of propane at standard pressure is calculated to be 3041.71 L.

The question involves applying the gas law to determine the volume of propane at standard pressure. The given conditions are 2785 L of propane at 830 mmHg, and we need to find the volume at standard pressure (760 mmHg). Using Boyle's Law, which states that P1V1 = P2V2 where P is pressure and V is volume, we can calculate the new volume at standard pressure.

First, convert the initial pressure to the same units as the standard pressure, which is already in mmHg. Then, set up the equation with P1 = 830 mmHg, V1 = 2785 L, P2 = 760 mmHg, and solve for V2:

V2 = (P1 × V1) / P2 = (830 mmHg × 2785 L) / 760 mmHg

V2 = 3041.71 L

So, the volume of propane at standard pressure is 3041.71 L.

Answer:

Concentration of solution is 402g/mL

Explanation:

The HCl solution has a purity of 35% w/w. That means there are 35g of HCl per 100g of solution. As density of solution is 1.15g/mL, volume of the solution is:

100g solution × (1mL / 1.15g) = 86.96mL. Thus concentration of solution in g/mL is:

35g of HCl / 86.96mL = 0.402g/mL. As 1L contains 1000mL:

[tex]0.402\frac{g}{mL} \frac{1000mL}{1L}[/tex] = 402g/mL

Concentration of solution is 402g/mL

Answer:

If 16 grams of sugar are in a spoonful, 2.83*10²² molecules of sugar are in the spoonful

Explanation:

Avogadro's Number or Avogadro's Constant is the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of that substance. Its value is 6.023 * 10²³ particles per mole. The Avogadro number applies to any substance.

So you need to know how many moles are 16 grams of sugar, C₁₂H₂₂O₁₁. For that you must know the molar mass of sugar. Being:

the molar mass of sugar is:

C₁₂H₂₂O₁₁= 12 *12 g/mole + 22 * 1 g/mole + 11* 16 g/mole= 342 g/mole

Then the following rule of three applies: if 342 grams of sugar are present in 1 mole of substance, 16 grams in how many moles will they be?

[tex]moles=\frac{16 grams*1 mole}{342 grams}[/tex]

moles= 0.047

Now, knowing Avogadro's number, you can apply the following rule of three: if 1 mole of sugar contains 6.023 * 10²³ molecules, 0.047 moles how many molecules does it have?

[tex]molecules=\frac{0.047 moles*6.023*10^{23} }{1 mole}[/tex]

molecules= 2.83*10²²

If 16 grams of sugar are in a spoonful, 2.83*10²² molecules of sugar are in the spoonful

Answer:

2.82x10^22 molecules.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10^23 molecules.

The above hypothesis implies that 1 mole of sugar, C12H22O11 also contains 6.02x10^23 molecules.

1 mole of C12H22O11 = (12x12) + (22x1) + (16x11) = 144 + 22 + 176 = 342g

Now, if 342g of C12H22O11 contains 6.02x10^23 molecules,

Therefore 16g of C12H22O11 will contain = (16x6.02x10^23)/342 = 2.82x10^22 molecules.

Therefore, 2.82x10^22 molecules of the sugar are in the spoon

Answer:

Synthetic fiber fabric has good heat resistance and thermoplasticity.

Good light resistance, light resistance is second only to acrylic.

Good chemical resistance. ...

High strength and elastic recovery. ...

Good water absorption. ...

you can choose any three from here

hope that was helpful.Thank you!!!

Final answer:

Polyester fabric has three main advantages: it is highly stain-resistant, has excellent wrinkle resistance, and is used in various industrial applications.

Explanation:

Advantages and Uses of Polyester Fabric:

Answer:

Total pressure increased

Explanation:

When gas C is added in the vessel then number of mole increases and number of collision depends on the number of molecules present in the vessel and on adding gas C ,mole also increases hence number of collision increases therefore pressure also increases because number of collision increases.

Total pressure increases.

Answer: The empirical formula for the given compound is [tex]Cr_2S_3[/tex]

Explanation : Given,

Mass of product = 0.963 g

Mass of Cr = 0.500 g

Mass of S = 0.963 g  - 0.500 g = 0.463 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Cr =[tex]\frac{\text{Given mass of Cr}}{\text{Molar mass of Cr}}=\frac{0.500g}{52g/mole}=0.00962moles[/tex]

Moles of S = [tex]\frac{\text{Given mass of S}}{\text{Molar mass of S}}=\frac{0.463g}{32g/mole}=0.0145moles[/tex]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00962 moles.

For Cr = [tex]\frac{0.00962}{0.00962}=1[/tex]

For S = [tex]\frac{0.0145}{0.00962}=1.5[/tex]

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of Cr : S = 1 : 1.5

The ratio of Cr : S = 2 : 3

Step 3: Taking the mole ratio as their subscripts.

The ratio of Cr : S = 2 : 3

Hence, the empirical formula for the given compound is [tex]Cr_2S_3[/tex]

Answer:

[tex]P_{2} = 3.259\,atm[/tex]

Explanation:

Let suppose that air inside the tire behaves ideally. The equation of state for ideal gases is:

[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex]

As tire can be modelled as a closed and rigid container, there are no changes in volume and number of moles. Hence, the following relationship is constructed:

[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}[/tex]

The final pressure is:

[tex]P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}[/tex]

[tex]P_{2} = \frac{298.15\,K}{278.15\,K}\cdot (3.04\,atm)[/tex]

[tex]P_{2} = 3.259\,atm[/tex]

Answer:

P2=3,26 atm

Explanation:

Considering that this gas behaves like an ideal gas, the equation used in noble gases is:

PxV = nxrxT

p is pressure, v is volume, n is number of moles, r is a constant of noble gases whose value is 0.082x10exp23 as long as you use it in the equations it has the same value (that is why it is called constant) and t would be the temperature.

Because v, r and n are values that did not change throughout the reaction of the gas, we are only going to take into account the p and t that did change throughout the reaction and that is why there is an initial and final pressure and a final and initial temperature.

So:

P1 / T1 = P2 / T2 (where value 1 refers to start, and 2 to end)

Regarding this equation, since we want to know the value of P2, which is the FINAL pressure of the gas, that is, what pressure did it achieve with the reaction, the equation that we would use would be solving for P2:

P2 = (T2 / T1) xP1 = 3.26 atm

Answer:

Ammonia

Explanation:

Answer:

ammonia NH3

Explanation:

pls mark brainliest

Answer:

.

Explanation:

Eardrum or tympanic membrane changes sound to vibration and pass through malleus, incus and stapes.

Answer:

d

Explanation:

Carbohydrates are compounds containing carbon, hydrogen, and oxygen. Therefore, a is true.

An empirical formula is the simplest ratio of atoms present in a compound. Therefore, C2H4O2 and C3H6O3, (if you simplified them like you would a fraction) would be CH2O. Therefore b is correct,

They also have the same % composition, with a ratio of 1 carbon : 2 hydrogen : 1 oxygen. Therefore, c is correct.

Since a, b and c are all correct, the answer is d, all of the above are true.

Final answer:

Methanol, ethanoic acid, and glucose all share the same empirical formula CH2O, despite having different molecular structures and functions.

Explanation:

What methanol (CH2O), ethanoic acid (C2H4O2), and glucose (C6H12O6) all have in common is that they all have the same empirical formula, which is CH2O. This simplified empirical formula indicates a molar ratio between carbon, hydrogen, and oxygen, which is 1:2:1. However, these compounds have different molecular formulas, meaning they contain different numbers of atoms, but when simplified to the lowest whole number ratio, they share the same empirical formula.

Methanol, a simple alcohol with the molecular formula CH4O, ethanoic acid (also known as acetic acid, found in vinegar), with a molecular formula of C2H4O2, and glucose, a simple sugar with the molecular formula C6H12O6, while all different compounds, share the empirical formula CH2O. This showcases an interesting aspect of organic chemistry where different compounds can have identical empirical formulas but vary greatly in structure and function.

Answer:

Boyels law

Explanation:

The general gas equation is p1v1/T1=p2v2/T2 and T1 and T2 are constant so all that is left is p1v2=p2v2

If temperatures T₁ and T₂ are constant in the combined gas law, the remaining relationship is Boyle's Law, defined as P₁V₁ = P₂V₂.

When we consider the equation for the combined gas law and keep the temperatures T₁ and T₂ constant, we are left with Boyle's Law. Boyle's Law states that the pressure of a gas varies inversely with its volume when temperature is held constant.

This can be represented mathematically as P₁V₁ = P₂V₂, assuming the number of moles (n) and the temperature (T) are constant. This law is named after the English scientist Robert Boyle who first described it in 1662.

Answer:

10grams

Explanation:

So it weighs 236 grams added with 25 grams. So it now weighs 261 grams so 10 grams of sugar remains in it.

Final answer:

Michelle added 10 grams of sugar to her coffee. The calculation involved subtracting the initial coffee weight and the weight of added milk from the final mixture weight.

Explanation:

To determine how many grams of sugar Michelle added to her coffee, we need to analyze the before and after weights of the mixture. The initial weight of the coffee was 236 grams, and after adding milk, the total weight became 271 grams. Given that Michelle also added 25 grams of milk, we can calculate the weight of the sugar.

The calculation to find the weight of the sugar added is as follows:

Subtract the initial weight of the coffee (236 grams) from the final weight of the mixture (271 grams) to account for the added ingredients: 271 grams - 236 grams = 35 grams.

Subtract the weight of the milk (25 grams) from the 35 grams to find the weight of the sugar: 35 grams - 25 grams = 10 grams.

Therefore, Michelle added 10 grams of sugar to her coffee.

Answer:

Si is the answer I hope this help