What is the name of the ionic compound CuBr

15.00g of hydrated zinc sulfate loses 6.579 H₂O during heating what is the formula for the hydrate is ZnSO₄.7H₂O

Mass of anhydrous ZnSO₄ = 15.00 g - 6.579 g = 8.421 g

8.412 g of anhydrous ZnSO₄ = 6.579 g

molar mass of anhydrous ZnSO₄ = 161.47 g/mol

161.47 g of  ZnSO₄  = (6.579 × 161.47) / 8.421

                                 = 126 g

no. of moles of H₂O = mass / molar mass

                                 = 126 / 18

                                 = 7 molecules of H₂O

the formula for hydrate is ZnSO₄.7H₂O

Thus, 15.00g of hydrated zinc sulfate loses 6.579 H₂O during heating what is the formula for the hydrate is ZnSO₄.7H₂O

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Answer:

1.25 M

Explanation:

(Step 1)

Convert grams to moles using the molar mass of Na₂O.

Atomic Mass (Na): 22.990 g/mol

Atomic Mass (O): 15.999 g/mol

Molar Mass (Na₂O): 2(22.990 g/mol) + 15.999 g/mol

Molar Mass (Na₂O): 61.979 g/mol

 34.8 g Na₂O               1 mole
----------------------  x  ---------------------  =  0.561 moles Na₂O
                                    61.979 g

(Step 2)

Convert milliliters to liters.

1,000 mL = 1 L

 450.0 mL                 1 L
------------------  x  -------------------  =  0.4500 L
                             1,000 mL

(Step 3)

Calculate the molarity using the molarity ratio.

Molarity = moles / volume (L)

Molairty = 0.561 moles / 0.4500 L

Molarity = 1.25 M

The partial pressure is the individual pressure of a gas. The partial pressure of gas A is 1.11 × 10⁻⁴atm.

What is partial pressure?

In a mixture of gases, the partial pressure of a gas is the individual pressure of that gas in the mixture of gases.

Given the reaction; A (g) → 3 B (g)

Kp = P(B)³/PA

Where Kp  = 67900

PB = 2.00 atm

67900 = (2)³/PA

PA =  (2)³/67900

PA = 1.11 × 10⁻⁴atm

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33.136 grams of ethylene glycol should be mixed with  375 ml of Water to make a 7.50% by Mixture.

Density is the measure of how much “stuff” is in a given amount of space.

DENSITY = MASS / VOLUME

We have :-

→ Density of ethylene glycol = 1.09 g/mL

→ Volume of water = 375 mL

→ Concentration (v/v %) = 7.50 %

Let the volume of ethylene glycol (solute) be 'x' mL .

So, volume of the solution = Volume of solute + Volume of solvent

                                              = (x + 375) mL

Concentration (v/v %) = Vol. of solute/Vol. of solution × 100

⇒ x/(x + 375) × 100 = 7.5

⇒ 100x = 7.5(x + 375)

⇒ 100x = 7.5x + 2812.5

⇒ 100x - 7.5x = 2812.5

⇒ 92.5x = 2812.5

⇒ x = 2812.5/92.5

⇒ x = 30.4 mL

Mass of ethylene glycol = Volume × Density

                                          = 30.4 × 1.09 = 33.136 g

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As per the octet rule, the element tellurium will make 2 covalent bonds to complete it's octet.

What is octet rule?

The octet rule describes an atom's propensity to favor eight electrons in its valence shell. Atoms with fewer than eight electrons in the outermost shell are more likely to interact with one another and create better stable molecules. We ignore d or f electrons while considering the octet rule. The octet rule is useful for main group elements (those not in the transition metal or inner-transition metal blocks) since it only involves the s and p electrons. An octet in these atoms corresponds to an electron configuration ending in s2p6.

According to octet rule, an element require to complete it's octet (i.e. 8 electrons in the outermost shell). So, in the case of tellurium there are 6 electrons in the outermost shell therefore, we require two more electrons to complete the octet for which the tellurium would be expected to make 2 covalent bonds.

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The IUPAC name for the compound shown is 3-ethyl-2,2-dimethylpentane.

International Union of Pure and Applied Chemistry is referred to as IUPAC. The terminology for naming organic compounds has been provided by IUPAC. The root name, prefix, and suffix are the three components that make up an IUPAC name.

There are five carbon atoms in the longest chain. Consequently, pent is the structure's root name. Choose the longest chain where the substituents are represented by the fewest numbers.

On the longest chain, three substituents are present. It consists of one ethyl group and two methyl groups. One ethyl group and two methyl groups are substituted at C-2 and C-3, respectively. Therefore, 3-ethyl-2,2-dimethyl will be the prefix. Alkane makes up the functional group. Therefore, the suffix is ane.

This ends up naming the compound as 3-ethyl-2,2-dimethylpentane.

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Answer:

Explanation:

To find the number of entities in a given substance we use the formula

N = n × L

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

However we weren't given the number of moles only the mass of oxygen was given. we can find the number of moles from that by using the formula

[tex]n = \frac{m}{M} \\ [/tex]

m is the mass

M is the molar mass

n is the number of moles

Molar mass of oxygen = 16 g/mol

mass in question = 25 g

We have

[tex]n = \frac{25}{16} = 1.56 \\ [/tex]

number of moles = 1.56 mol

The number of oxygen atoms is equal to

N = 1.56 × 6.02 × 10²³ = 9.3912 × 10²³

We have the final answer as

Hope this helps you

Mole of sulfur hexafluoride gas that were collected is 20.081 mol

Mass of  sulfur hexafluoride gas that were collected is 0.137 gram

Sulphur hexafluoride gas is used as the electrical insulating material in circuit and breaker, cables and capacitor and sulfur hexafluoride gas is the nontoxic gas and it is an inorganic compound it is colorless and odorless and non flammable

Here given data is

Pressure = 0.220 atm.

Temprature = -4.0 °C = -4.0 °C +273 .15 K = 269.15 K

Volume =  5.0 L

Using ideal gas quation

PV = nRT

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R = gas constant =  0.0821 L.atm/K.mol

Applying the equation as

0.220 atm ×  5.0 L = n × 0.0821 L.atm/K.mol ×  269.15 K

n = 22.09/1.1

n = 20.081

Molar mass of sulfur hexafluoride gas =  146.06 g/mol

The formula for calculations of mole

Moles = mass taken/molar mass

20.081 = mass/146.06 g/mol

Mass = 0.137 gram

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Answer:

Explanation:

Here, we want to get the metric prefix m value

This means we want to get power to which it would be raised

Mathematically,we have this as the milli

The milli refers to thousandth

From what we have here, this is the power of -3

So the prefix m represents :

The molar mass of the gas in grams per mole if 6.2 g of gas occupies 5.4L is 336.97 grams.

Molar mass is defined as the mass of a sample of a certain chemical divided by the quantity of the material, expressed as the number of moles in the sample.

It can also be defined as the product of the mass of a specific substance and the amount of that substance in the sample.

Given Pressure = 1.3 atm

Temperature = 285 K

Volume = 5.4 L

Gas content = 8.3

So, PV = nRT

n = RT / PV

n = 8.3 x 285 / 1.3 x 5.4

n = 336.97 grams

Thus, the molar mass of the gas in grams per mole if 6.2 g of gas occupies 5.4L is 336.97 grams.

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To solve the problem we will assume the following:

1. Air behaves as an ideal gas during all the process.

2. The initial air equivalent to 8200L is at atmospheric pressure. It means 1 atm.

3. The temperature remains constant.

Taking into account the above, we can apply the Boyle-Marriote Law that relates the change in pressure and volume at constant temperature. The equation that we will use will be:

Where,

P1 is the atmospheric pressure. 1atm

V1 is the initial volume of air required, 8200L

P2 is the final pressure we want to find

V2 is the final volume, it means the volume of the spherical air tank. We will calculate this volume using the volume equation for a sphere:

r is the radius of the sphere, r=73cm/2=36.5cm

So, the volume of the spherical air tank will be:

No, we clear P2 from the first equation and replace known data:

The pressure of the gas must be 40.3 atm

Answer: 40.3

Explanation:

We have to find the molar mass of Au(OH)₃. To do that we have to look for the atomic masses of each element. Also we have to consider that one molecule of

Au(OH)₃ has 1 atom of Au, 3 atoms of O and 3 atoms H.

atomic mass of Au = 196.97 amu

atomic mass of O = 16.00 amu

atomic mass of H = 1.01 amu

molar mass of Au = 196.97 g/mol

molar mass of O = 16.00 g/mol

molar mass of H = 1.01 g/mol

molar mass of Au(OH)₃ = 1 * molar mass of Au + 3 molar mass of O + 3 molar mass of H

molar mass of Au(OH)₃ = 1 * 196.97 g/mol + 3 * 16.00 g/mol + 3 * 1.01 g/mol

molar mass of Au(OH)₃ = 248.0 g/mol

Answer: the molar mass of Au(OH)₃ is 248.0 g/mol.

Cation

Type 1

Type 2

Anion

The Roman numeral represents  charge and the oxidation state of the transition metal ion.

One of the example is, iron that forms two ions, Fe2+ and Fe3+. To differentiate, Fe2+ is named iron (II) and Fe3+ is named iron (III).

Those compounds in which the cation has only one charge are Type 1 binary ionic compounds whereas compounds in which the cation can have multiple forms are Type 2 binary ionic compounds.

Type II Binary Ionic Compounds contain Transition metals with non-metal ions.

A monatomic cation derives its name from the name of the element. For example, Na+ is called sodium  whereas monatomic anion is named by taking the root of the element and then adding -ide.

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Ans6

wer:

b

Explanation:

For a species to survive it must be within its geographical zone for all the genetic factors.

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Will Argon, Neon, and Krypton react same

Reactivity is the relative capacity of an atom or molecule or radical to undergo a chemical reaction with another atom or molecule or compound called as reactivity

In the noble gases only helium and neon are inert and the other noble gases will react with a limited scale under very specific conditions and krypton will form solid with fluorine and xenon will form a variety of compounds with oxygen and fluorine and the name comes from the fact that these elements are virtually unreactive towards other elements and they do not react with other element

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Answer ans explanation

2.51x10^4 - 1.5x10^3 = 2.36^4

Answer

2.36^4

When calcium nitrate and potassium fluoride solutions react to form a precipitate, the type of reaction involved is the process is double replacement

Calcium nitrate and potassium fluoride solutions react to form calcium fluoride and potassium nitrate. Calcium fluoride would precipitate out whereas potassium nitrate would be in aqueous form.

Ca (NO₃)₂ (aq) + 2 KF (aq) → CaF₂ (s) + 2 KNO₃ (aq)

In double replacement reaction, the ionic compounds would exchange their respective ions to form a new compound. Here two ionic compounds, nitrate and fluoride from calcium and potassium respectively are exchanged to form calcium fluoride and potassium nitrate.

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Yes, the experimentation with corn and other crops led to the development of new fuels called biofuels.

Biofuel is a biodegradable,  inexhaustible, fuel that is produced from Biomass. Biofuel is considered the easiest available and pure fuel on the earth. Biofuels are manufactured from biomass such as wood and straw, which are liberated by direct combustion of dry matter and converted into a gaseous and liquid fuel.

The organic matter such as sludge, sewage, and vegetable oils, can be changed into biofuels by a wet process such as fermentation and digestion. Biofuel is available in all regions of the world, and mainly includes fuels such as Biodiesel, Bioethanol, and Bio methanol.

The two types of biofuels are bioethanol and biodiesel most commonly used these days. Both of these biofuels are the first generation of biofuel technology.

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2.67 grams of Glycerin should be added to water of mass 169.8 grams to give a solution with freezing point -3.81°.

When glycerin will be added to water, its freezing point will decrease. This phenomena is given a name "Depression in Freezing Point". This is called a colligative property.

We can find the depression in freezing point by using the formula,

ΔT = i.m.K

Where,

i is the Vant Hoff's factor.

Kf is cryoscopic constant

m is the molality of the solution,

Molality m can be defined as,

m = moles of solute/mass of solvent(in KG)

First, let s find molality of solution,

m = Moles of Glycerin/mass of water

m = (Added mass of glycerin(W)/Molecular mass of glycerin)/mass of water.

Molecular weight of glycerin = 92 g/mole

Mass of water = 169.8 g

In kilograms,

mass of water = 0.1698 Kg.

Now,

m = W/92 x 0.1698

Now, putting all the values in the formula,

ΔT = i.m.Kf

Assuming 100% disassociation, we can take i = 1,

But first, ΔT = 0-(-3.81) °C

ΔT  = 3.81  °C.

So, now we can write,

3.81 = 1.86 x W/(92 x 0.1698)

W = 3.81 x 92 x .01698 / 1.86

W = 2.67 grams.

Hence, of is required to add 2.67 grams of glycerin in water.

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Answer:

In order to understand the Maxwell equations and relations derived from Maxwell relations, you will have to know Euler's Reciprocity.

 It states that for any state thermodynamic quantity (let x and y) having a state function phi, it must satisfy the following condition.

[tex](\frac{ {∂}^{2}\phi }{∂x \cdot∂y}) = (\frac{ {∂}^{2}\phi }{∂y \cdot∂x})[/tex]

Maxwell equations: These are a set of equations derived from the application of Euler's Reciprocity. The four Maxwell equations are as follows:

[tex]dH = TdS + VdP[/tex]

[tex]dG = VdP - SdT[/tex]

[tex]dA = -PdV - SdT[/tex]

[tex]dU = TdS - PdV[/tex]

Let's derive each Maxwell relations step by step,

1) dH = TdS + VdP

Enthalpy is a function of Entropy & Pressure

[tex] \sf \qquad \qquad H = f(S,P)[/tex]

[tex]dH = TdS + VdP[/tex]

[tex]dH = (\frac{∂H}{∂S})_{_P} dS + (\frac{∂H}{∂P})_{_S}dP[/tex]

Comparing both the above equation,

[tex]T = (\frac{∂H}{∂S})_{_P}[/tex]

[tex]V =(\frac{∂H}{∂P})_{_S}[/tex]

now we know that Enthalpy is a state function hence applying cross reciprocality,

[tex](\frac{∂V}{∂S})_{_P}= (\frac{∂T}{∂P})_{_S}[/tex]

This is called the first Maxwell relation,

Similarly,

2) dG = -SdT + VdP

Free energy is a function of Temperature & Pressure,

[tex] \sf \qquad \qquad G = f(T,P)[/tex]

[tex]dG = -SdT + VdP[/tex]

[tex]dG = (\frac{∂G}{∂P})_{_T} dP + (\frac{∂G}{∂T})_{_P}dT[/tex]

Comparing both the above equation,

[tex](\frac{∂G}{∂P})_{_T} = V [/tex]

[tex] (\frac{∂G}{∂T})_{_P}= -S[/tex]

now we know that Free energy is a state function hence applying cross reciprocality,

[tex]-(\frac{∂S}{∂T})_{_P}= (\frac{∂V}{∂P})_{_T}[/tex]

This is called the second Maxwell relation,

3) dA = -PdV - SdT

helmholtz free energy is a function of Temperature & Volume,

[tex] \sf \qquad \qquad A = f(T,V)[/tex]

[tex]dA = -PdV - SdT[/tex]

[tex]dA = (\frac{∂A}{∂V})_{_T} dV + (\frac{∂A}{∂T})_{_V}dT[/tex]

Comparing both the above equation,

[tex](\frac{∂A}{∂V})_{_T} = -P [/tex]

[tex] (\frac{∂A}{∂T})_{_V}= -S[/tex]

now we know that helmholtz free energy is a state function hence applying cross reciprocality,

[tex]-(\frac{∂S}{∂V})_{_T}= -(\frac{∂P}{∂T})_{_V}[/tex]

[tex](\frac{∂S}{∂V})_{_T}= (\frac{∂P}{∂T})_{_V}[/tex]

This is called the third Maxwell relation,

4) dU = TdS - PdV

Internal energy is a function of Entropy & Volume,

[tex] \sf \qquad \qquad A = f(S,V)[/tex]

[tex]dU = TdS - PdV [/tex]

[tex]dU = (\frac{∂U}{∂S})_{_V} dS + (\frac{∂U}{∂V})_{_S}dV[/tex]

Comparing both the above equation,

[tex](\frac{∂U}{∂V})_{_S} = -P [/tex]

[tex] (\frac{∂U}{∂S})_{_V}= T[/tex]

now we know that Internal energy is a state function hence applying cross reciprocality,

[tex](\frac{∂T}{∂V})_{_S}= -(\frac{∂P}{∂S})_{_V}[/tex]

This is called the fourth Maxwell relation,

The main significance of the Maxwell relation is that those thermodynamic quantities which are unmeasurable can be replaced with measurable quantities with the help of the Maxwell relation.

The derivative of the extensive asset in relation to the extensive asset gives the intensive asset; with respect to the intensive asset, the derivative of the extensive asset gives the extensive asset. This is the result of the overall Maxwell relations.

The coefficient of expansion and compression of a gas in thermodynamics is the application of the Maxwell relations.

There are 3 coefficients introduced,

Coefficient of thermal expansion (expansivity) α

[tex]α= \frac{1}{V} (\frac{∂V}{∂T} )_{_P}[/tex]

Coefficient of isothermal compressibility β

[tex]β = -\frac{1}{V} (\frac{∂V}{∂P} )_{_T}[/tex]

Isochoric thermal expansion coefficient γ

[tex] γ= \frac{1}{P} (\frac{∂P}{∂T} )_{_V}[/tex]

For ideal gases,

PV = nRT

For one mole ideal gas (n=1),

PV = RT

Taking derivative with respect to T at constant pressure,

[tex]V(\frac{∂P}{∂T} )_{_P}+ P(\frac{∂V}{∂T} )_{_P}= R(\frac{∂T}{∂T} )_{_P}[/tex]

At constant pressure ∂P = 0, & R.H.S = 1, hence

[tex](\frac{∂V}{∂T} )_{_P}= \frac{R}{P}[/tex]

[tex]α= \frac{1}{V} (\frac{∂V}{∂T} )_{_P}[/tex]

[tex]α= \frac{1}{V} \cdot \frac{R}{P}[/tex]

[tex]\sf Also, PV=RT\\ \frac{R}{PV} = \frac{1}{T}[/tex]

[tex]\boxed{α= \frac{1}{T}}[/tex]

Following the same procedure, by taking derivating w.r.t. pressure at constant temperature.

[tex]V(\frac{∂P}{∂P} )_{_T}+ P(\frac{∂V}{∂P} )_{_T}= R(\frac{∂T}{∂P} )_{_T}[/tex]

[tex]V+ P(\frac{∂V}{∂P} )_{_T}= 0[/tex]

[tex](\frac{∂V}{∂P} )_{_T}= \frac{-V}{P}[/tex]

Substituting the above value in,

[tex]β = -\frac{1}{V} (\frac{∂V}{∂P} )_{_T}[/tex]

[tex]β = -\frac{1}{V} \cdot \frac{-V}{P}[/tex]

[tex] \boxed{β = \frac{1}{P}}[/tex]

Repeating the same procedure again, i.e. derivative w.r.t. T at constant volume

[tex]V(\frac{∂P}{∂T} )_{_V}+ P(\frac{∂V}{∂T} )_{_V}= R(\frac{∂T}{∂T} )_{_V}[/tex]

At constant Volume ∂V = 0, and R.H.S = 1, hence overall equation becomes,

[tex]V(\frac{∂P}{∂T} )_{_V} = R \\ (\frac{∂P}{∂T} )_{_V} = \frac{R}{V}[/tex]

Substituting above value in,

[tex]γ= \frac{1}{P} (\frac{∂P}{∂T} )_{_V}[/tex]

[tex]γ= \frac{1}{P} \cdot \frac{R}{V}[/tex]

[tex] \boxed{γ= \frac{1}{T}}[/tex]

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The Maxwell equation in thermodynamics is very important and useful because the set of relations allows the scientists to change specific unknown quantities.

The Maxwell equation in thermodynamics is very useful because this is the set of relations that allows physicists to change certain unknown quantities that are hard to measure in the real world. These quantities need to be replaced by many easily measured quantities. Maxwell's relations are a set of equations in thermodynamics that are derivable from the second derivatives and from the definitions of the thermodynamic potentials. These relations are named for the nineteenth-century physicist James Clerk Maxwell. So entropy and pressure are the natural variables of enthalpy. Maxwell relations are thermodynamic equations that establish the relations between various thermodynamic quantities in equilibrium and other fundamental quantities known as thermodynamical potentials

So we can conclude that Maxwell's thermodynamics are the set of relations allows the scientists to change unknown quantities.

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The heat of reaction is obtained from the calculation as -0.65kJ/mol.

We already know that the heat of the reaction is computed by the use of the information that have been presented in the question. We know that this is a 1:1 reaction thus;

Number of moles of HCl = 0.1102 M * 25/1000 = 2.755 * 10^-3 moles

Number of moles of NaOH = 0.1024 M * 50/1000 L = 5.12 * 10^-3 moles

We can see from above that the HCl is the limiting reactant.

The hat evolved is obtained from;

0.001 J/°C. * (25.93°C - 25.14°C) = 1.79 * 10^-3 J

The heat of the reaction is then;

-( 1.79 * 10^-3 J) * 10^-3 * 1/2.755 * 10^-3 moles

-0.65kJ/mol

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In this question, we need to analyze the solubility of some substances and the changes that can occur to each one upon heating, at 20°C, every substance listed is soluble, NaCl, Na2SO4, Li2CO3 and Sucrose, but when we have a change in temperature, it will affect directly the solubility of the solution, for example, if you increase the temperature, Lithium carbonate and Sodium sulfate, will have a lower solubility in water, therefore if we have a certain amount of these two substances at 20°C and in 100g of water, the solution will be soluble, but if we increase the temperature, the solubility will change and these two compounds will start to precipitate, as the solubility will be lowering down.

Therefore the answers are Na2SO4 and Li2CO3

BaSO₄. Option C is correct

The substances that dissolve readily in water are ionic compounds and polar covalent compounds. Examples of ionic compounds that dissolve in water are salts, oxides, hydroxides, sulfides, and the majority of inorganic compounds.

The molecules in a polar solvent have a dipole, like water, one side is more negative and one is more positive.

Ionic compounds are composed of a positive ion, normally a metal, and a negative ion, normally a nonmetal, so their forces are attracted to their charge difference.

Thus, a polar solvent dissolves each ion with its corresponding parts, dissociating the two ions of the ionic compound.

Since sulfides are ionic compounds hence the substance that will dissolve most readily in water is BaSO₄. The molecule is formed by one barium cation Ba2+ and one sulfide anion S2-. The two ions are bound through an ionic bond.

The following are the examples of a chemical property of a substance;

1) A banana reacts with orange and turns brown

2) Iron will rust when exposed to oxygen and water

3) Iron will react with acid to form iron chloride.

The term chemical reaction has to do with the change in the properties of substances that are combined together in order to yield a product. Recall that the properties of the reactant change as the product is formed.

When a chemical reaction occurs

1: A new color may appear

2) A new gas may be seen

3) The temperature may be changed

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When the partial pressure of oxygen is 156 torr and the atmospheric pressure is 743 torr, the mole fraction of oxygen is 0.210.

Partial pressure is the pressure exerted by an individual gas in a mixture of gases. The partial pressure of a gas depends on its mole fraction.

The relationship between the partial pressure of a gas and the total pressure is given by Dalton's law, which states that the sum of the partial pressures is equal to the total pressure.

We can calculate the partial pressure of oxygen using the mathematical expression of Dalton's law:

pO₂ = P × X(O₂)

X(O₂) = pO₂ / P

X(O₂) = 156 torr / 743 torr = 0.210

where,

The mole fraction of oxygen is 0.210 when its partial pressure is 156 torr and the atmospheric pressure is 743 torr.

The complete question is:

The partial pressure of oxygen was observed To be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of oxygen present.

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The number of molecules in 59.73 grams of the theoretical acid borofuric acid, H₂B₂O₂ is 6.47 × 10²³ molecules.

The number of molecules in a substance can be estimated by multiplying the number of moles in the substance by Avogadro's number (6.02 × 10²³) as follows:

no of molecules = no of moles × 6.02 × 10²³

According to this question, there are 59.73 grams of the theoretical acid borofuric acid. The molar mass of this acid is as follows:

H₂B₂O₂ = 1(2) + 10.8(2) + 16(2) = 55.6g/mol

moles = 59.73g ÷ 55.6g/mol = 1.07mol

no of molecules = 1.07 × 6.02 × 10²³

no of molecules = 6.47 × 10²³ molecules

Therefore, 6.47 × 10²³ molecules is the number of molecules in the acid.

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