Bonding between two elements of equal electronegativity would be

Answers

Answer 1

The given question is incomplete. The complete question is :

Bonding between two elements of equal electronegativity would be

a.100% covalent

b.50% ionic

c. metallic in character

d. primarily ionic

Answer: a. 100% covalent

Explanation:

An ionic bond is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element or the metal and the element which accepts the electrons is known as electronegative element or non metal.

A covalent bond is formed when an element shares its valence electron with another element. This bond is formed between two non metals. When the two non metals have equal electronegativities, the bond is called as non polar covalent.

Metallic bonding is the type of chemical bonding occurs between the atoms of a metals.In this type of bonding arises due to electrostatic interaction between the electron cloud of de-localized electrons with positively charges metal ions.

Answer 2

Final answer:

A bond between two elements with equal electronegativity is termed a pure or nonpolar covalent bond, characterized by equal sharing of electrons without charge separation on the molecule.

Explanation:

Bonding between two elements of equal electronegativity results in a pure covalent bond, also known as a nonpolar covalent bond. This type of bond forms when electrons are shared equally between the atoms. Since there is no significant difference in electronegativity, there are no stable regions of net negative or positive charge on the molecule's surface, meaning the bond is nonpolar.


Related Questions

Percent yield for sodium bromide if 3.87 moles of iron 3 bromide yielded 10.4 moles of sodium bromide

Answers

Answer:

89.57 %

Explanation:

Don't have one just thought id help <3

Answer: 89.6

Explanation:

An atom has the following electron configuration.

1s22s22p63s23p3

How many valence electrons does this atom have?

3
5
8
15

Answers

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:

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When 189.6 g of ethylene (C2H4) burns in oxygen to give carbon dioxide and water, how many grams of CO2 are formed? C2H4(g) + O2(g) → CO2(g) + H2O(g) (unbalanced)

Answers

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

The middle ear converts sound from what into vibrations that can be sent through what?

Answers

Answer:

.

Explanation:

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

Air was put into an automobile tire when the tire was cold, at 5.00 C. The tire's air pressure was 3.04 atm. Later, the weather warms up. Also, the automobile is driven, further warming the tire. What will the tire's pressure be when the temperature is 25.0 C?

Answers

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

What are the 3 advantages and uses of polyester fabric

Answers

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:

Polyester fabric is highly stain-resistant, making it a durable option for apparel and home furnishings. It can resist spills and stains, making it easier to clean and maintain.Polyester fabric has excellent wrinkle resistance due to the cross-linking of polymer strands. This makes it a popular choice for permanent-press clothing that requires minimal ironing or maintenance.Industrial polyester fibers, yarns, and ropes are used in various applications such as car tire reinforcements, conveyor belts, safety belts, and coated fabrics. Polyester's high-energy absorption properties make it suitable for these industrial uses.

The stopcock connecting a 1.30 L bulb containing xenon gas at a pressure of 6.15 atm, and a 4.25 L bulb containing oxygen gas at a pressure of 3.53 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is 8.33 atm. True or false?

Answers

Answer: False.

The final pressure in the system is 4.14 atm

Explanation: Please see the attachments below

Jorge inflates a beach ball to a volume of 4L in his air-conditioned house where the temperature is 18℃. That afternoon he takes the beach ball to the beach with some friends. The temperature at the beach is 32℃, and the air pressure at the beach is the same as it was at Jorge's house. What will happen to Jorge's beach ball when he is at the beach? Choose the claim that best answers the question: * 1 point The beach ball will get smaller at the beach because the molecules are moving slower. The beach ball will stay the same size at the beach because the pressure is constant. The beach ball will get larger at the beach because the molecules are moving faster. The beach ball will get larger at the beach because the pressure will cause it to expand. The beach ball will get smaller at the beach because the molecules will collide less often.

Answers

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.

What should be happening for beach ball?

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

A 2.425 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 24.91 ∘ C to 30.28 ∘ C. The heat capacity (calorimeter constant) of the calorimeter is 28.81 kJ / ∘ C, what is the heat of combustion per gram of the material?

Answers

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,

C: calorimeter constantΔT: change in the temperature

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

Which element in period 3 has both metallic and non metalik properties​

Answers

Answer:

Si is the answer I hope this help

Sample of gas initially occupies 4.25 L at a pressure of 0.850 atm at 23.0°C. What will the volume be if the temperature is changed to 11.5°C, and the pressure is changed to 1.50 atm?

Answers

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

If 16 grams of sugar, C12H22O11, are in a spoonful, how many molecules of sugar are in the spoonful?

Answers

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:

C: 12 g/moleH: 1 g/moleO: 16 g/mole

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

Stoichiometry!

Please note:

- Use 6.022x1023 for avogadro’s number

- Ignore sig figs and do not round the final answer.

- Keep it to 1 decimal place.

Answers

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!

El succinato de butilo es un repelente de insecto utilizado en la casa contra hormigas y cucarachas. Su composicion es de 65.28 por ciento de C, 9,63 por ciento de H y de 27,79 por ciento de O. Su masa molecular determinada experimentalmente es de 230 u, ¿cuales son las formulas empiricas y molecular del succinato de butilo?

Answers

Answer:

what is the English translation, id like to help

Explanation:


The ingredient label for a fruit drink reads: water, sugar, orange juice, high fructose corn syrup, vitamin C, sodium benzoate (preservative), Red 40, Blue 1.
What can you conclude about the fruit drink’s contents?

A) The product contains no preservatives
B) The product contains more juice than water.
C) The product contains more sugar than juice.
D) The product contains more sugar than any other ingredient.


Answers

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.

Nitrifiers are bacteria
a. capable of living in symbiosis in the roots of some plants.
b. capable of oxidizing NH3 using NO.
c. that convert N2 to NH3.
d. that generate nitrites or nitrates.
e. capable of using NH3 as an electron donor.

Answers

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.

Consider a mixture of two gases A and B, confined to a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the total pressure in the vessel?

Answers

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.

what is the mass of one mole of oxygen gas?

Answers

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!

it should be 32grams

A balloon containing helium gas expands from 230
mL to 860 mL as more helium is added. What was
the initial quantity of helium present if the
expanded balloon contains 3.8 x 10-4 mol,
assuming constant temperature and pressure?
Which of the variables are known?


Answers

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.

Ideal Gas Law

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

An ionic precipitation reaction takes place when aqueous solutions of potassium phosphate and aluminum nitrate are combined in a beaker.
K3PO4(aq) + Al(NO3)3(aq) →
i. Predict the products and write a balanced chemical equation for the reaction,
including state symbols.
K3PO4(aq) + Al(NO3)3(aq) →
ii. Write the complete balanced ionic equation, including state symbols.
iii. Write the complete balanced net ionic equation, including state symbols.​

Answers

Answer:

I hope you can get a good answer

Explanation:

What class is this for?

6) At constant temperature, the pressure of a gas is 780 mmHg and
volume of the gas is 2.0 L. Calculate the volume of the gas when
pressure decreases to 620 mmHg.

Answers

Answer:

The volume of the gas when  pressure decreases to 620 mmHg is 2.52 L

Explanation:

Boyle's law is a law related to gases that establish a relationship between the pressure and the volume of a given amount of gas, without variations in temperature, that is, at constant temperature. So, this law establishes that at constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

Having a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment, if you vary the volume of gas to a new value V2, then the pressure will change to P2, and it will be true:

P1 * V1 = P2 * V2

In this case:

P1: 780 mmHgV1: 2 LP2: 620 mmHgV2: ?

Replacing:

780 mmHg* 2 L= 620 mmHg* V2

Solving:

[tex]V2=\frac{780 mmHg* 2L}{620 mmHg}[/tex]

V2= 2.52 L

The volume of the gas when  pressure decreases to 620 mmHg is 2.52 L

Answer:

2.52 L

Explanation:

Step 1:

The following data were obtained from the question:

Initial pressure (P1) = 780 mmHg

Initial volume (V1) = 2.0 L.

Final pressure (P2) = 620 mmHg.

Final volume (V2) =?

Step 2:

Determination of the final volume of the gas.

From the question given, we discovered that the temperature is constant. Since the temperature is constant, the gas is simply obeying Boyle's law.

Applying the Boyle's law equation, the final volume of the gas can be obtained as follow:

P1V1 = P2V2

780 x 2 = 620 x V2

Divide both side by 620

V2 = (780 x 2) /620

V2 = 2.52 L

Therefore, the volume of the gas when the pressure decreases to 620 mmHg is 2.52 L

Se toman 200 ml de una disolución de un frasco de ácido clorhídrico ClH en cuya etiqueta se lee: Pureza: 35 % en peso; densidad = 1,15 g/ml. ¿Cuál es su concentración en g/l?

Answers

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

Consider the equation for the combined gas law. If T1, and T2, are constant, you are left with whose
law?​

Answers

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.

17. A gas tank holds 2785 L of propane, C3H8, at 830. mm
Hg. What is the
volume of the propane at standard pressure?​

Answers

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.

Michelle has a cup of hot coffee. The liquid coffee weighs 236 grams. Michelle adds a few teaspoons sugar and 25 grams of milk to the coffee. Michelle stirs the mixture until everything is combined. The mixture now weighs 271 grams. How many grams of sugar did Michelle add to the coffee?

Answers

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.

What do methanol (CH2O), ethanoic acid (C2H4O2), and glucose (C3H6O3) all have in common?
a. They are all carbohydrates
b. They all have the same empirical formula
c. They all have the same % composition
d. All of the above are true

Answers

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.

A 0.500-g sample of chromium metal reacted with sulfur powder to give 0.963 g of product. Calculate the empirical formula of the chromium sulfide.

Answers

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]

2. What is the percentage by mass of sodium (Na) in a formula unit of sodium
hydrogen carbonate (NaHCO3)?

Answers

Final answer:

The percentage by mass of sodium in sodium hydrogen carbonate (NaHCO3) is 27.4%, calculated by dividing the atomic mass of sodium by the molar mass of NaHCO3 and multiplying by 100%.

Explanation:

To calculate the percentage by mass of sodium (Na) in a unit of sodium hydrogen carbonate (NaHCO3), you first need to know the atomic masses of the elements involved. The atomic mass of Na is approximately 23 g/mol, H is about 1 g/mol, C is approximately 12 g/mol and O is approximately 16 g/mol.

Using these numbers, the molar mass of NaHCO3 is about 84 g/mol (23 for Na, 1 for H, 12 for C, and 48 for the three O atoms). To find the percentage of Na in NaHCO3, divide the atomic mass of Na by the molar mass of NaHCO3 and multiply by 100%:

(23 g/mol ÷ 84 g/mol) x 100% = 27.4%.

Therefore, the percentage by mass of sodium in a formula unit of sodium hydrogen carbonate is 27.4%.

Learn more about Percentage by Mass here:

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The mass percentage of sodium (Na) in a formula unit of sodium hydrogen carbonate (NaHCO₃) is 27.36%. The calculated molar mass of NaHCO₃ is 84.01 g/mol.

To determine the percentage by mass of sodium (Na) in a formula unit of sodium hydrogen carbonate (NaHCO₃), we need to follow these steps:

1. Find the atomic masses of each element:

 Sodium (Na) = 22.99 g/mol

 Hydrogen (H) = 1.01 g/mol

 Carbon (C) = 12.01 g/mol

 Oxygen (O) = 16.00 g/mol

2. Calculate the molar mass of NaHCO₃:

         NaHCO₃ = 22.99 g/mol (Na) + 1.01 g/mol (H) + 12.01 g/mol (C) + 3 x 16.00 g/mol (O)

         NaHCO₃ = 22.99 + 1.01 + 12.01 + 48.00 = 84.01 g/mol

3. Determine the mass percentage of sodium (Na):

          Mass % Na = (22.99 g / 84.01 g) x 100

          Mass % Na = 27.36%

Therefore, the mass percentage of sodium (Na) in sodium hydrogen carbonate (NaHCO₃) is 27.36%.

Will mark brainliest, please help!

Answers

Answer:

Ammonia

Explanation:

Answer:

ammonia NH3

Explanation:

pls mark brainliest

A scientist is comparing two cells.

One cell, cell A, is from a unicellular organism, and the other cell, cell B, is from a multicellular organism.
What functional differences will the scientist most likely see when comparing the two cells?

A. Cell A performs all of the organism's necessary functions, while cell B only performs one or a few functions.
B. Cell A is very large and has many organelles, while cell B is very small and has only a few organelles
C. Cell A performs only one or a few functions, while cell B performs all of the organism's necessary functions.
D. Cell A is very small and has only a few organelles, while cell B is large and has many organelles

Answers

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

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