The energy of an orange lamp with a **frequency** of 5.10 x 10¹⁴ Hz is **3.38 x 10⁻¹⁹ J.**

**Frequency** is defined as the number occurrences of repeated events per unit of time. SI unit is **hertz (Hz)**.

Using the **formula**,

**E = hν**

where,

E = energy in joules (J)

h = Planck's constant = 6.63 x 10⁻³⁴ J

ν = frequency in hertz

Given :

**Frequency** = 5.10 x 10¹⁴ Hz

h = 6.63 x 10⁻³⁴ J

Putting the values,

E = (6.63 x 10⁻³⁴) x (5.10 x 10¹⁴) J

**E = 3.38 x 10⁻¹⁹ J**

The energy of an orange lamp with a **frequency** of 5.10 x 10¹⁴ is **3.38 x 10⁻¹⁹J.**

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The complete question is -

What is the energy of an orange lamp with a frequency of 5.10 x 10¹⁴ Hz ?

In the earth's mantle and core, how do the mass and density compare?

The **core **is **denser** than the **mantle**.

The** Earth** is partitioned into three primary layers. The **hot inner core,** the **molten outer core**, the** mantle**, and where the thin **crust**, support all life within the known universe. Most of the Earth's insides are made up of the mantle, the layer of molten rock underneath the strong outside, and the hot, dense core. As the **mass** and the **volume** of the mantle are greater than the core's mass which leads the mantle of having a low **density**. as a result the core is considered to be having more density than the mantle in spite of the mantle having more mass.

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Calculate the following round to the proper number of significant figures and write in standard scientific notation.2.51x104-1.5x103

Answer ans explanation

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

Answer

**2.36^4**

CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g) 13. What volume of O₂(g) is required to react with excess CS₂(l) to produce 4.0 L of CO₂(g)? (Assume all gases are measured at 0°C and 1 atm.)

The **volume** of a substance at STP for one mole is **22.4 L**. The volume of oxygen required to react with excess of carbon sulphide to produce 4 L of carbon dioxide 4.0 L is 12 L.

**Volume** of a substance is the space occupied by the particles of the substance. As the **temperature** increase, volume also increases. The volume of one mole of every substance at standard temperature and pressure is 22.41 L

According the **balanced reaction** given, 3 moles of O₂ is required to produce one mole **carbon dioxide**. One mole of O₂ is 22.4 L. Thus volume of 3 moles of oxygen is calculated as follows:

Volume of O₂= 3×22.41 =67.2 L.

Thus, 67.2 L of O₂ produces** one mole** or 22.4 L of carbon dioxide.

The volume **O₂ **required to produce 4 L of carbon dioxide is calculated as follows:

Volume of O₂ = (67.2 L × 4 L) / 22.41 L

= 12 L.

Therefore, the **volume** of **oxygen** molecule required to produce 4 L of carbon dioxide on the reaction with excess of carbon sulphide is 12 L.

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3. The density of C₂H4 (OH)₂ is 1.09 g/me. How Many gram of C₂H4 (OH)2 Should be Mixed With 375 ml of Water to make a 7.50% by Mixture?

**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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how many atoms are in 6 moles of oxygen

Each oxygen molecule has two atoms

6 * 2 = 12

12

6 * 2 = 12

12

Calculate the heat of reaction when 25.00 mL of 0.1102 M HCl(aq) at 25.14°C is added to 50.00 mL of 0.1024 M NaOH(aq) at the same temperature in a coffee-cup calorimeter that has a calorimeter constant of 0.001 J/°C. After mixing the temperature of the solution was observed to be 25.93°C.

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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What mass of glycerin (C3H8O3) must be dissolved in 169.8 g water to give a solution with a freezing point of -3.81°C? Kf for water = 1.86°C*kg/mol.

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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Determine the numerical ages of rock samples that contain a parent isotope with a half-life of 100 million

years and have the following percentages of original parent isotope:

50%: Age =

25%: Age =

6%: Age =

50% = 100 million years

25% = 200 million years

6% = 400 million years

I’m not really sure if this is right

25% = 200 million years

6% = 400 million years

I’m not really sure if this is right

An atom with 3 protons in the nucleus and 3 electrons in the orbitals would have what overall charge?

Ans6

wer:

b

Explanation:

The element tellurium would be expected to form covalent bond(s) in order to obey the octet rule.

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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Calculate the molar mass of gold (III) hydroxide, Au(OH)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.**

b) Given the following standard enthalpy changes at 298 K, calculate the standard enthalpy change for the reaction given below. S(s) + O2(g) → SO2 (g) AH = -296.8 kJ mol-1 SO2(g) + 3/2 O2(g) → SO3 (g)AH = -98.9 kJ mol-¹ S(s) + 3/2 O2(g) → SO3 (g) AHr = ?

−791.4 kJ is the **standard enthalpy** change for the reaction.

Given,

SO2(g) → S(s) + O2(g)

ΔH° = +296.8 kJ

2SO2(g) + O2(g) → 2SO3(g)

ΔH° = −197.8 kJ

Modified equation:

S(s) + 2O2(g) → 2SO2(g)

ΔH° = −593.6 kJ, multiply by 2 and flip

2SO2(g) + O2(g) → 2SO3(g)

ΔH° = −197.8 kJ here no change

The 2SO2 will balance out by canceling it when the equations are added. Add the **enthalpies** for the final answer:

−593.6 + (−197.8) = −791.4 kJ

The change in enthalpy of a compound when one mole of the compound is generated from all components of the same constituent is known as the **standard enthalpy** of formation, also known as the **standard heat** of formation.

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Saturated solutions of each of the following compounds are made at 20°C. Circle the letter(s) of the solution(s), which will form a precipitate upon heating.a) NaClb) Na2SO4c) Li2CO3d) Sucrose

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

calculate the percentage oxygen in aluminum phosphate

To determine the percentage of oxygen in the molecule we must first know its structure and how many oxygen atoms are contained in it. The structure of aluminum phosphate is AlPO4.

It means that there are 4 oxygen atoms, 1 Aluminum atom, and 1 phosphate atom.

Now we will determine the weight of the molecule AlPO4 by adding the atomic weights of the elements as follows:

Element Atomic mass # of atoms Mass

Al 26.9815 1 26.9815

P 30.9738 1 30.9738

O 15.999 4 63.996

Total mass of AlPO4 = 26.9815+30.9738+63.996 = 121.9513 g/mol

Now we will determine the mass percentage with the following equation:

[tex]\text{Mass percentage = }\frac{Oxygen\text{ mass}}{\text{Total mass of the molecule }}\times100[/tex]We replace the known terms:

[tex]\begin{gathered} \text{Mass percentage = }\frac{63.996}{\text{121.9513 }}\times100 \\ \text{Mass percentage = }52.477 \end{gathered}[/tex]So, the percentage of oxygen in aluminum phosphate is 52.477%

Given the decomposition reaction:2SI3(g)->2SO2(g) + O2(g)According to Le Châtelier’s principle, what will happen when the volume of the container is increased for the chemical reaction that had reached equilibrium? A)Increasing volume, increases pressure and favors the products. B) Increasing volume, decreases pressure and favors the products.C)Increasing volume, increases pressure and favors the reactants.D) Increasing volume, decreases pressure and favors the reactants.

To analyze the equilibrium and how it shifts aaccording to Le Châtelier’s principle, we have to see how the system reacts to the change.

The change is an increase in the volume of the container. Since all the compounds on equilibrium are in gaseous state, their volume is the same as its container, so an increase in the volume of the container increase the volume of the compounds.

The reactant is SI₃, but we have 2 of them for each reaction.

The products are 2 molecules of SO₂ and one molecule of O₂.

In total for each reaction, we have 2 molecules in the reactant part and 3 molecules in the product part.

Since they are all in gaseous form, this means that the products occupy more space than the reactants, that is, an increase in the volume will favor the products, because this increase will left more space for them to occupy.

Thinking in preassure, an increase in the volume will decrease pressure, because, by the Boyle's Law, they are inversely proportional (assuming ideal gas). Since there are more molecules per reaction on the products side, this will favor the products, since more molecules make more pressure and now it has been decreased.

Thus:

**Increase in volume -> decrease in pressure -> favors the products.**

This matches **alternative B**.

Sulfur hexafluoride gas is collected at -4.0 °C in an evacuated flask with a measured volume of 5.0 L. When all the gas has been collected, the pressure in the

flask is measured to be 0.220 atm.

Calculate the mass and number of moles of sulfur hexafluoride gas that were collected.

answer needs has the correct number of significant digits

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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What is the IUPAC name for the compound shown?

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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19) A sample of metal ore is reacted according to the following reaction:Fe(s) +2HCL (aq) --> FeCl2(aq) + H2(g)If 24.06 mL of 5.6 M HCL are used, what mass of Fe was in the ore? Keep the answer with 2 decimal places

Assuming all the HCl presend in the 24.06mL reacted, we can follow the steps:

1 - Use the concentration and the volume to calculate the number of moles of HCl that reacted

2 - Apply the stoichiometry ratios to calculate the number os moles of Fe that reacted

3 - Use the tomic weight of Fe to calculate the mass of that amount of number of moles of Fe.

1 - The concentration is given by the equation:

[tex]C=\frac{n_{\text{solute}}}{V_{\text{solution}}}[/tex]The number of moles of solute is the same as the number of moles o HCl, because it is the solute in this case:

[tex]\begin{gathered} C=\frac{n_{HCl}}{V_{\text{solution}}_{}} \\ n_{HCl}=C\cdot V_{\text{solution}} \end{gathered}[/tex]So, we have:

[tex]\begin{gathered} C=5.6mol/L \\ V_{\text{solution}}=24.06mL=24.06\times10^{-3}L \\ n_{HCl}=5.6mol/L\cdot24.06\times10^{-3}L=0.134736mol \end{gathered}[/tex]2 - The coefficients of Fe and HCl are 1 and 2, respectively, so we have the following relation between their number of moles:

Fe --- HCl

1 --- 2

[tex]\begin{gathered} \frac{n_{Fe}}{1}=\frac{n_{HCl}}{2} \\ n_{Fe}=\frac{n_{HCl}}{2}=\frac{0.134736mol}{2}=0.067368mol \end{gathered}[/tex]3 - The atomic weight of Fe can be checked on a periodic table:

[tex]M_{Fe}=55.845g/mol[/tex]So, we have:

[tex]\begin{gathered} M_{Fe}=\frac{m_{Fe}}{n_{Fe}} \\ m_{Fe}=n_{Fe}M_{Fe}=0.067368mol\cdot55.845g/mol=3.76216\ldots g\approx3.76g \end{gathered}[/tex]So, there was approximately **3.76 g of Fe**.

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

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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Why are Roman numerals needed in the names of ionic compounds containing a metal that forms more than one type of ion? Type one contains a metal with a invariant charge-One that does not vary from one compound to another. Type two contains a metal with a charge that can differ in different compounds. Match the items in the left column to the appropriate blanks in the sentences on the right

**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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The metric prefix m would be presented as 10 to the power of:

**Answer:**

[tex]-3[/tex]

**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 :

[tex]10^{-3}[/tex]How many molecules are in 59.73 grams of the theoretical acid borofuric acid, H2B2O2?

remember units and sig figs.

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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8. How many oxygen atoms are in 25 g of oxygen?

1.9 × 1024 atoms

2.4 × 1026 atoms

9.4 × 1023 atoms

1.5 × 1025 atoms

**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

9.4 × 10²³ oxygen atomsHope this helps you

Will Argon, Neon, and Krypton react the same or differently? Explain. (make it clear and simple. thank u)

-need help asap. thank u so much :)

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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a gaseous product of a chemical reaction is collected at 285k and 1.3atm. what was the molar mass of the gas in grams per mole if 6.2 g of gas occupies 5.4l

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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Calcium nitrate and potassium fluoride solutions react to form a precipitate. Classify this reaction.

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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Which of the following substances dissolves most readily in water?a. CHb. NH3c. BaSO4d. CaCO3

Answer:

**BaSO₄. **Option C is correct

The substances that dissolve readily in water are **ionic compounds** and** polar covalent** compounds. Examples of** ionic compound**s 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 s**ulfides **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 b**ond.

A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in aspherical air tank that measures 73.0 cm wide.The biologist estimates she will need 8200. L of air for the dive. Calculate the pressure to which this volume of air must becompressed in order to fit into the air tank. Write your answer in atmospheres. Round your answer to 3 significant digits.0atm0.0XS ?EoloPEBH

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:

[tex]P_1V_1=P_2V_2[/tex]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:

[tex]V_2=\frac{4}{3}\pi r^3[/tex]r is the radius of the sphere, r=73cm/2=36.5cm

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

[tex]\begin{gathered} V_2=\frac{4}{3}\pi\times(36.5cm)^3=20.4\times10^4cm^3 \\ V_2=20.4\times10^4cm^3\times\frac{1L}{1000cm^3}=204L \end{gathered}[/tex]No, we clear P2 from the first equation and replace known data:

[tex]\begin{gathered} P_2=\frac{P_1V_1}{V_2} \\ P_2=\frac{1atm\times8200L}{204L} \\ P_2=40.3atm \end{gathered}[/tex]**The pressure of the gas must be ****40.3 atm**

**Answer: ****40.3 **

34.8 g of Na₂O are used to form a solution with a volume of 450.0 mL L. What is the

molarity?

34.89

**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

For a species to survive it must be within its ________ for all ___________ factors.

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

To know more about **species **visit:

**https://brainly.com/question/13259455**

#SPJ13

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