Which are produced when HCl reacts with Ca(OH)2?
Cl2, H3, and CaO
CaCl and H3O
CaO, Cl2, and H2O
CaCl2 and H2O

Answer:

Insulators

Explanation:

They could work as Insulators

Hope it Helps

One of the most significant characteristics of the Dead Sea is its salinity level. The images show the rock salt deposits on the shore.

The dead dea is the most salty landlocked lake on the Earth surface. It is located in the middle East, between Israel and Jordania. Its nort-south length is about 80 km and its east-west wide varies between 7 and 20 km.

The Jordan river is the most significan afluent on the Dead Sea, from which it receives most of its water.

The Dead Sea is the deepest and lowest water body on the Earth, being its surface level 400 mt below the sea level.

Due to different reasons, during the last decades, there was a sharp and continuous drop in the water level. For this reason the lake surface is decreasing year after year.

Among its many characteristics, one of them -and probably the most notorious one- is the salinity.  The percentage of dissolved salts in the lake is much higher than the one recorded in the oceans, seas or lakes of the globe. Its value is between 8 and 10 times higher the salt level of oceans.

The salts present in the lake are chlorine, magnesium, sodium, calcium, potassium, bromine and sulfur. Their concentration vary according to the strate and deph, but it tends to increase toward the lake’s bottom. The high salinity level provides an oily aspect, bleach taste, and extreme density.

Several rock salt deposits can be seen on the shore and near it. Salt columns can also be seen in deepest areas. These structures are becomming more common with time, as water evaporates, water provision from afluent sources decreases, and salinity near the surface increases.

Mainly due to the extreme salt level, which turns to be lethal, almost no forms of life inhabit these waters. Only some bacteria species and one algae species can be found in the water, and some plant species near the coasts.

What we can observe in these images are the rock salt deposits on the shore.

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The moles of NaOH used in the titration would be 0.00177 moles while the molarity of the vinegar will be 0.885 M

Recall that: mole = molarity x volume

In this case, the molarity of the NaOH= 0.1 M

Volume of NaOH = 17.7 - 0.0 = 17.7 mL

Mole of NaOH used = 0.1 x 17.7/1000 = 0.00177 moles.

Since NaOH and vinegar have 1:1 mole ratio, the mole of vinegar will also be 0.00177 moles.

Molarity of vinegar = mole/volume = 0.00177/0.002 = 0.885 M

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[tex]12 \times 2 + 2 \times 2 + 16 \times 2[/tex]

[tex] = 24 + 2 + 32 [/tex]

[tex] = 58 \: gram \: per \: mole[/tex]

[tex]58 = 100\% \\ 24 = x\%[/tex]

[tex]x = \frac{24 \times 100}{58} = 41.38\%[/tex]

The energy required to completely boil a 5.05 g-sample of liquid water at its boiling point is 2.73 kcal.

Boiling is the process in which, upon absorbing energy in the form of heat, a substance goes from the liquid state to the gaseous state.

We have a 5.05 g-sample of liquid water. We can calculate the heat required to completely boil it using the following expression.

Q = ΔH°vap × m

Q = 0.540 kcal/g × 5.05 g = 2.73 kcal

where,

The energy required to completely boil a 5.05 g-sample of liquid water at its boiling point is 2.73 kcal.

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Answer:

S.........................................

Answer:

456 K

Explanation:

PV = nRT

PV/ T = nR       since nR is constant for this scenario

P1 V1 /T1   =  P2 V2 / T2

500 * 1 / 200 =  (760*.75)  *2 /T2    Solve for T2 =456 K

(Note that .750 atm =   760 mmHg * .75 )

The new volume of the balloon assuming the temperature remains constant is 0.5 L.

To calculate the new volume of the balloon, we apply Boyle's law

Boyle's law state that the volume of a fixed mass gas is inversly proportional to its pressure at constant temperature.

Formula:

Where:

Make V' the subject of the equation

From the question,

Given:

Substitute these values into equaion 2

Hence, The new volume of the balloon assuming the temperature remains constant is 0.5 L.

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Explanation:

density= mass × volume

3.35×5

=. 16 .75

Water has a high specific heat capacity hence, water can keep the  large bodies of water cooler in the summer months and warmer in the winter months.

The specific heat capacity of a substance refers to the heat that must be added to the substance in order to increase the temperature of the substance by 1K.

It must be noted that water has a high specific heat capacity hence, water can keep the  large bodies of water cooler in the summer months and warmer in the winter months. Hence, the statement is false

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Answer:

False

Explanation:

I took the Chemistry exam

Explanation:

Cytoplasm is a semi-solid substances inside cells.

Answer:

Because the non-metal comes from the right hand side of the Periodic Table as we face it, its nuclear charge is very poorly shielded by its INCOMPLETE valence electronic shell, and this results in the well-known contraction of atomic radii across the Table from left to right.

Explanation:

Non metals usually gain electrons so they have more electron in the shells than protons in nucleus and hence they form negative ions. However, metals usually lose electron so they have more protons in nucleus than electrons in shells and hence they form positive ions.

Answer:

HEBW EISJS A A RID RO J KNE SINS DMO C EIS

Answer:

The correct formula for dinitrogen hexafluoride is Memorize the prefixes for number of atoms, Write the chemical symbol for the first element, add the number of atoms present as a subscript, Write the chemical symbol for the second element, Add the number of atoms present as a subscript, Practice with some examples.

Explanation:

1: Memorize the prefixes for number of atoms. In naming compounds, Greek prefixes are used to indicate the number of atoms present for each element. Covalent compounds are written out as molecular formulas due to the fact that each compound is a distinct, separate molecule.[1] Covalent compounds have the first element written out completely while the second element is named with the suffix "ide." For example, diphosphorus trisulfide has a chemical formula of P2S3. Below are the prefixes for 1-10:

1: Mono-

2: Di-

3: Tri-

4: Tetra-

5: Penta-

6: Hexa-

7: Hepta-

8: Octa-

9: Nona-

10: Deca-

2: Write the chemical symbol for the first element. When a compound has been written out, you must identify the elements and know their chemical symbols. The first element written is "first name" of the compound. Use the periodic table to find the chemical symbol for the element. For example: Dinitrogen hexafluoride. The first element is nitrogen and the chemical symbol for nitrogen is N.

3: Add the number of atoms as a subscript. To identify the number of atoms present for each element, you simply need to look at the prefix of the element. Memorizing the Greek prefixes will help you to be able to write chemical formulas quickly without looking anything up. For example: Dinitrogen has a the prefix "di-" which means 2; therefore, there are 2 atoms of nitrogen present. Write dinitrogen as N2.

4: Write the chemical symbol for the second element. The second element is the "last name" of the compound and will follow the first element. For covalent compounds, the element name will have a suffix of "-ide" instead of the normal ending of the element. For example: Dinitrogen hexafluoride. The second element is fluorine. Simply replace the "ide" ending with the actual element name. The chemical symbol for fluorine is F.

5: Add the number of atoms present as a subscript. As you did with the first element, identify the number of atoms present in the second element by reading the prefix. Using this prefix, write the number of atoms as a subscript to the right of the chemical symbol. For example: Hexafluoride has a prefix of "hexa-" which means 6; therefore, there are 6 atoms of fluorine present. Write hexafluoride as F6. The final chemical formula for dinitrogen hexafluoride is N2F6.

6: Practice with some examples. When first learning chemistry, there is a lot of memorization involved. It is kind of like learning a new language. The more examples you practice with, the easier it will be to decipher chemical formulas in the future and learn the language of chemistry. Sulfur dioxide: SO2 Carbon tetrabromide: CBr4 Diphosphorus pentoxide: P2O5

Answer:

"The Circuit" by author Francisco Jimenez, tells the story of Panchito and his experiences on a new country, having to deal with learning a second language and adapting to a new culture and environment. Even if you had not been in a similar situation as Panchito, you could connect his stories to your personal experiences at school in many ways. Maybe, the first classes of spanish or french were difficult to you as you could not understand all the words the teacher was saying, but as you were learning things were easier for you. Or you can connect Panchito's experiences by writing about that time in school you were not afraid of asking for help when you had to deal with a difficult situation 

Antifreeze is the coolant used in car engines that regulates the temperature. The cooling system turns over around 40 times to regulate the temperature.

Heat is the product of the mass, temperature change, and the specific heat of the substance.

Given,

The heat of surroundings = 44000 kJ

Mass of the substance = 8719 gm

Specific heat = 8.37 J/g degrees celsius

Change in temperature = 15 degrees celsius

Q system is calculated as:

8719 × 8.37 × 15

= 1095 kJ

The number of antifreeze is calculated as:

Δ H ÷ Q system = 44000 ÷ 1095

= 40

Therefore, 40 times the cooling system was turned over.

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Answer:

The R-O-R Chain

Explanation:

Please give me brainliest! Im so sorry if this is incorrect I really tried... -_- ;-;

Mechanical Energy to Thermal Energy

hope it helps...!!!

Answer:

B

Explanation:

Write the dissociation reaction of solid CoS:
[tex]\displaystyle \text{CoS}_\text{(s)} \rightleftharpoons \text{Co}_\text{(aq)}^{2+} + \text{S}_\text{(aq)}^{2-}[/tex]

The Ksp expression is hence:
[tex]\displaystyle K_{sp} = [\text{Co}^{2+}][\text{S}^{2-}][/tex]

If the molar solubility of CoS is 5.0 × 10⁻²² mol L⁻¹ (M), then 5.0 × 10⁻²² M of Co²⁺ and S²⁻ also forms. Hence, the value of Ksp is:
[tex]\displaystyle \begin{aligned} K_{sp} & = [\text{Co}^{2+}][\text{S}^{2-}] \\ \\ & = (5.0\times 10^{-22})^2 \\ \\ & = 2.5\times 10^{-43}\end{aligned}[/tex]

In conclusion, the answer is B.

When 6.1 moles of Cu is reacted with excess S 3.05 number of mole of Cu2S are produced.

The balance equation of the above reaction is:

2Cu +S --> Cu2S

This means that 2 moles of Cu = 1 mole of Cu2S

Therefore 6.1 moles of Cu = X mole of Cu2S

Make X the subject of formula,

X = 6.1 ×1/2

X = 3.05 moles

Therefore, when 6.1 moles of Cu is reacted with excess S 3.05 number of mole of Cu2S are produced.

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Taking into account the reaction stoichiometry, 0.886 grams of O₂ can be prepared by the decomposition of 12 grams of mercury oxide.

In first place, the balanced reaction of the decomposition of mercury oxide is:

2 HgO  → 2 Hg + O₂

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:

HgO: 2 moles ×216.59 g/mole= 433.18 grams

Hg: 2 moles ×200.59 g/mole= 401.18 grams

O₂: 1 mole ×32 g/mole= 32 grams

The following rule of three can be applied: if by reaction stoichiometry 433.18 grams of HgO form 32 grams of O₂, 12 grams of HgO form how much mass of O₂?

[tex]mass of O_{2} =\frac{12 grams of HgOx 32 grams of O_{2}}{433.18 grams of HgO}[/tex]

mass of O₂= 0.886 grams

Then, 0.886 grams of O₂ can be prepared by the decomposition of 12 grams of mercury oxide.

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Answer:

b

Explanation:

cause it stops at 8 and it starts going up until it hits 14

Answer:

[tex]\boxed{O_2 \: ^{2-} (\sigma2s)^2(\sigma∗2s)^2(\sigma2pz)^2(π2py,π2px)^4(π∗2py,π∗2px)^4}[/tex]

[tex]\boxed{O_2 \: ^{1-}(\sigma2s)^2(\sigma∗2s)^2(\sigma2pz)^2(π2py,π2px)^4{ (π∗2py)}^{2} {(π∗2px)}^{1}}[/tex]

Explanation:

[tex]\sf \: Electronic \: configuration \: of \: O_2 \: ^{2-} and \: O_2 \: ^{1-}[/tex]

Electronic configuration of any diatomic molecule can be determined using MOT(molecular orbital theory).

To know the electronic configuration of above two molecules, draw the MOT of O2 molecule.(refer the diagram)

There are 16 electrons in O2 molecule out of which Four electrons are in non bonded state. Remaining 12 electrons of 16 will be filled according to MOT, since O2 is not a S-P mixing case, in without S-P mixing case the sigma orbital(σ2pz) lies below the Pi orbital(π2px,π2py) in bonding state as you can see in the diagram, on other hand the other diatomic molecules like B2, C2 & N2 are S-P mixing case where Pi orbital lies below and sigma orbital lies above.

Now, the electronic configuration of O2 molecule is

[tex](\sigma2s)^2(\sigma∗2s)^2(\sigma2pz)^2(π2py,π2px)^4(π∗2py,π∗2px)^2 [/tex]

There are two electrons unpaired electrons in MOT hence the O2 molecule is paramagnetic which contradicts VBT, Thats why MOT dominant VBT.

in case of O2^2-(peroxide linkage) the two electrons enters in the Anti Bonding of pi orbital. hence the electronic configuration of O2^2- is

[tex](\sigma2s)^2(\sigma∗2s)^2(\sigma2pz)^2(π2py,π2px)^4(π∗2py,π∗2px)^4[/tex]

since the HOMO (highest occupied molecular orbital) is paired the O2^2- is diamagnetic in nature.

Coming to O2^1- (superoxide linkage) the one electron enters in the Anti Bonding of pi orbital hence the electronic configuration comes out,

[tex](\sigma2s)^2(\sigma∗2s)^2(\sigma2pz)^2(π2py,π2px)^4{ (π∗2py)}^{2} {(π∗2px)}^{1} [/tex]

The HOMO of O2^1- is unpaired hence it is Paramagnetic in nature.

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Answer: Heyo Kenji Here! Here's your answer- In a fat molecule, the fatty acids are attached to each of the three carbons of the glycerol molecule with an ester bond through the oxygen atom. During the ester bond formation, three molecules are released. Since fats consist of three fatty acids and a glycerol, they are also called triacylglycerols or triglycerides.

Explanation: Hope this helps!

Have a nice day!

- Kenji ^^

Answer: D

Explanation:

Based on the atomic masses of the two isotopes of boron, B-11 is more than twice as abundant as B-10. The correct option is option D.

The reason why some isotopes have a greater prevalence naturally common on Earth then others has something to do with the simple fact that various isotopes have varying relative abundances.

The proportion of a given isotope with a particular atomic mass that exists in nature is known as its relative abundance. Isotopes are different chemical elements that differ in nucleon number and neutron number, respectively. The average atomic mass of boron is 10.81 amu. Based on the atomic masses of the two isotopes of boron, B-11 is more than twice as abundant as B-10.

Therefore,  B-11 is more than twice as abundant as B-10. The correct option is option D.

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For a carbonate containing a kind of cation, we have that Option 1 is the correct answer: ammonium.

Generally, Any of two families of chemical substances generated from carbonic acid or carbon dioxide is referred to as carbonate.

In conclusion, Sodium, potassium, and ammonium are the three carbonates that are soluble.

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When the two vehicles collide, then vehicle with slow speed will pushes back.

Speed of any moving body is define as the distance travelled by that body in per unit of time.

As it is mentioned that one car has a speed of 30 mph and another one has a speed of 60mph, if both of these cars get collide then the car with high speed pushed back the car with low speed.

Hence the car with slow speed pushed back.

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Answer:

the answer is (A). .A nuclear reaction releases more energy per gram and appears to violate the law of conservation of mass.

Explanation:

Most mass is converted into energy so it violates the law.

Following the quantic theory, the energy of a photon equals the radiation frequency multiplied by the universal constant. ν = 2.923x10¹⁵ Hz. E = 3.09x10¹⁵Hz.

It is the branch of physics that studies objects and forces at a very low scale, at atoms, subatoms, and particles levels.

Quantum mechanics states that the elemental particles that constitute matter -electrons, neutrons, protons- have the properties of a wave and a particle.

It emerges from the quantic theory exposed by Max Planck (1922), in which he affirmed that light propagates in energy packages or photons.

He discovered the Universal Planck constant, h, used to calculate the energy of a photon.

He stated that the energy of a photon (E) equals the radiation frequency (ν) multiplied by the universal constant (h).

E = νh

In the exposed example, we need to calculate the energy required to change from the 3rd shell to the first shell.

To do it, we should know that the energy in a level (Eₙ) equals the energy associated to an electron in the most inferior energy level (E₁) divided by the square of the shell number (n²).

Eₙ = E₁ / n²

E₁ is a constant. We can express it in Joules or electroVolts

So, let us calculate the energy at level 1 and 3

Eₙ = E₁ / n²

        E₁ =  -13.6 eV / 1² =  -13.6 eV

        E₃ =  -13.6 eV / 3² =  -13.6 eV / 9 = - 1.51 eV

The change of energy can be calculated in two ways,

Option 1

ΔE = E₁ - E₃ = 2.18x10⁻¹⁸ - 2.42x10⁻¹⁹ = 1.93x10⁻¹⁸J

ΔE = E₁ - E₃ = 13.6 - 1.51 = 12.09 eV

Option 2

ΔE = -2.18x10⁻¹⁸ J (1/nf² - 1/ni²)

ΔE =-13.6 eV (1/nf² - 1/ni²)

Where nf is the final level and ni is the initial level. When the electron passes from its initial level to its final level it is called electronic transition.

ΔE = -2.18x10⁻¹⁸ J (1/nf² - 1/ni²)

ΔE = -2.18x10⁻¹⁸ J (1/1² - 1/3²)

ΔE = -2.18x10⁻¹⁸ J (1 - 0.111)

ΔE = -2.18x10⁻¹⁸ J (0.888)

ΔE = - 1.937x10⁻¹⁸ J

or

ΔE = -13.6 eV  (1/nf² - 1/ni²)

ΔE = -13.6 eV  (1/1² - 1/3²)

ΔE = -13.6 eV  (1 - 0.111)

ΔE = -13.6 eV  (0.888)

ΔE = -12.08 eV

This is the energy required for the electron to go from n= 3 to n = 1. The negative sign (-) means energy (as light or photons) released or emitted.

If we want to express the result in Hz, we just need to make a conversion.

1Hz ⇔ 6.626x10⁻³⁴J ⇔ 4.136x10¹⁵ eV.

The energy required for the electron to go from n= 3 to n = 1 is 3.09x10¹⁵ Hz.

Now, we need to calculate the frequency, ν. This is, how many times the wave oscillates back and foward per second.

To do it, we will use the universal Planck constant, h, and the absolute value of the energy, E.

ν = E/h = 1.937x10⁻¹⁸ J / 6.626x10⁻³⁴ Js = 2.923x10¹⁵ 1/s =  2.923x10¹⁵ Hz.

Answer:

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Answer:

4.3 %

Explanation:

Trust me.