As the ball falls, its energy will decrease and its energy will increase

Explanation:

Heterogeneous mixture is a mixture with non-uniform composition.

The properties of the mixture like concentration may change for different parts of the mixture.

Colloids contain solute particles of size [tex]2nm-500nm[/tex].The presence of these particles makes the mixture heterogeneous.

Suspensions contain solute particles of size [tex]500nm-1000nm[/tex].These particles settle to the bottom of the mixture which makes the composition of the bottom different from the top.

So,colloids and suspensions are two types of heterogeneous mixtures.

Final answer:

Yes, the statement is true; suspensions and colloids are both types of heterogeneous mixtures, differentiated by the size of their particles and their tendency to settle.

Explanation:

The statement is true: suspensions and colloids are indeed two types of heterogeneous mixtures. A suspension is a heterogeneous mixture where large particles are dispersed throughout a liquid or gas and can settle over time, examples include paint and mud. On the other hand, a colloid is also a heterogeneous mixture, but it consists of much finer particles that don't settle upon standing, such as fog, milk, and butter.

Colloids can be found in various states like sols (solid particles in a liquid), gels (solids absorbing liquids), aerosols (solid or liquid particles in a gas), or emulsions (liquids dispersed in other liquids). They exhibit the Tyndall effect, where light is scattered by the particles, distinguishing them from solutions, which are homogenous and clear. Suspensions, with their larger particle size, do not maintain their dispersion upon standing and are often cloudy.

It's important to note that while suspensions and colloids are heterogeneous due to the presence of dispersed particles, these particles differ significantly in size between the two types, which influences the stability and appearance of the mixture.

Answer:

A) 1.5 moles  B) 1.5 moles  C) 3.5 moles  D) 2/3 moles

Explanation:

A) 1 mole of CO₂ gives 1 mole of CH₄ and 2 moles of H₂O, i.e, 3 moles of products. So 0.5 moles of CO₂ gives 3 x 0.5 = 1.5 moles of products.

B) 1 mole of BaCl₂ gives 2 moles of AgCl and 1 mole of Ba(NO₃)₂, i.e, 3 moles of products. So 0.5 moles of BaCl₂ gives 3 x 0.5 = 1.5 moles of products.

C) 1 mole of C₃H₈ gives 4 moles of H₂O and 3 moles of CO₂, i.e, 7 moles of products. So 0.5 moles of C₃H₈ gives 7 x 0.5 = 3.5 moles of products.

D) 3 moles of H₂SO₄ gives 1 mole of Fe₂(SO₄)₃ and 3 moles of H₂, i.e, 4 moles of products. So 0.5 moles of H₂SO₄ gives 4/3 x 0.5 = 2/3 moles of products.

To calculate the number of moles of product produced when a given amount of reactant is completely reacted, use the stoichiometric coefficients of the balanced chemical equation.

To calculate the number of moles of product produced when a given amount of reactant is completely reacted, we need to use the stoichiometric coefficients of the balanced chemical equation. Each coefficient represents the ratio of moles between reactants and products.

For example, in the first equation, CO2 (g) + 4H2 (g) → CH4 (g) + 2H2O (l), the coefficient of CO2 is 1 and the coefficient of CH4 is also 1. This means that for every 1 mole of CO2 that reacts, 1 mole of CH4 is produced.

Therefore, if 0.500 mole of CO2 were to react completely, 0.500 mole of CH4 would be produced.

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Answer:

Total mass of 5 lead fishing weights= 1250 grams

Explanation:

Given:

Mass of a lead fishing weight =250 grams

To find total mass of 5 such fishing weights

We can apply unitary method to find total mass of 5 fishing weights.

If mass of 1 lead fishing weight = 250 grams

∴ Total mass of 5 lead fishing weights = [tex]250\times 5 =1250\ grams[/tex]

Answer:

All of these

option a to f are molecule

Explanation:

Molecule:

Molecule is formed when two or more than two atoms of an element combine chemically.

N₂O₄ is a molecule where Nitrogen combine with oxygen through covalent bond.

The name of this compound is Dinitrogen tetraoxide.

Na₂CO₃ is a Compound where sodium carbon and oxygen combined to form a Compound.

Its name is sodium carbonate.

CClF3 is a Compound: It is a polar Compound

It named as Chlorofloromethan.

OBr₂ is polar Molecule. This molecule consist of oxygen and Bromine atoms and a covalent compound.

It named as oxygen dibromide.

FeSO₄ is a compound . it consists of iron Oxygen and sulfate atoms.

It named as Iron Sulfate.

LiBr is molecule. it is ionic in nature and Known as molecular compound

It named as Lithium bromide.

All molecule can not be compound as some molecule have same type of atom such as ozone or hydrogen molecule., but all compounds are molecule, so with this concept all of these are molecule.

Answer: CH3COOH + CH3CHOHCH3 -> CH3COOCH(CH3)CH3 + H2O

Explanation: Estherification is the reaction between alkanoic acids and alkanols to porduce esters and what. ethanoic acid will react with 2-propanol to produce an ester and water.

Ethanoic acid reacts with 2-propanol to form an ester and water in the presence of sulfuric acid.



[tex]\bold {CH_3COOH + CH_3CHOHCH_3 \rightarrow CH_3COOCH(CH_3)CH_3 + H_2O}[/tex]

[tex]\bold {CH_3COOH + CH_3CHOHCH_3 \rightarrow CH_3COOCH(CH_3)CH_3 + H_2O}[/tex]

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Answer:

water is formed

Explanation:

acid and base maxing form salt and water by combining proton and hydroxyl ions

Answer:

See the answer below, please.

Explanation:

A precipitation reaction is defined as one that yields an insoluble compound (precipitate) as a product by mixing two different solutions. An example:

NaCl + AgN03 -> Agcl (precipitate) + NaN03

Answer:

K2CO3 + PbCl2 → 2KCl + PbCO3

Explanation:

This is correct because a percipitant solution is one that has an insoluble compound in it and PbCO3 is insoluble since the Pb is not amonium or an alkaline metal. The compound PbCO3 is unsoluable and 2KCl is soluable.  

SOLUBILITY RULES:

1.) Hydroxides (OH−), carbonates (CO32−), and phosphates (PO43−) are insoluble, except for compounds containing group 1 alkali metals and ammonium (NH4+).

2.) Chlorides (Cl−), bromides (Br−), and iodides (I−) are soluble, except for compounds containing silver (Ag+), mercury(I) (Hg22+), and lead (Pb2+).

Answer:

Explanation:

It could make burrows and homes in the dirt. It could hide food and itself in the soil. The Soil will grow food.

In an ecosystem , a mouse would make burrows in soil and soil would provide nutrients to the mouse.

Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.

Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.

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Answer:

34 gram of FeO produced 8 gram of oxygen.

Explanation:

Given data:

Mass of FeO = 34 g

Mass of oxygen = ?

Solution;

Chemical equation:

2FeO → 2Fe + O₂

Number of moles of FeO:

Number of moles = mass/ molar mass

Number of moles = 34 g /71.8 g/mol

Number of moles = 0.5 mol

Now we will compare the moles of FeO with oxygen:

             FeO       :       O₂

                2         :        1

                0.5      :      1/2 × 0.5 = 0.25

Mass of oxygen:

Mass = number of moles × molar mass

Mass =  0.25 mol × 32 g/mol

Mass = 8 g

So 34 gram of FeO produced 8 gram of oxygen.

Answer: Add more water to the solution and remove a small amount of the solution and mix it with water.

Answer:

Frequency = 1.25 ×10¹³ Hz

Explanation:

Given data:

Wavelength of light = 24.0  μm (2.4 ×10⁻⁵ m)

Frequency = ?

Solution:

Formula:

Speed of light = wavelength × frequency

Speed of light /wavelength = frequency

Frequency = 3×10⁸ m /s /2.4 ×10⁻⁵m

Frequency = 1.25 ×10¹³ s⁻¹

s⁻¹ = Hz

Frequency = 1.25 ×10¹³ Hz

34.73%

We are given;

A mixture that contains;

We are required to determine mass percent of NaCl in the mixture;

Mass of the mixture = Mass of NaCl + Mass of NH₄Cl + Mass of SiO₂

                                 = 0.639 g + 0.350 g + 0.851 g

                                 = 1.84 g

The mass % of NaCl = (Mass of NaCl ÷ Mass of the mixture) 100

                                  = (0.639 g ÷ 1.84 g) × 100

                                  = 34.73%

Therefore, the mass % of NaCl in the mixture is 34.73%

Answer:

199 g

Explanation:

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

Explanation:

This law was given by french chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

For example:

In given reaction there are two reactant one with a mass of 115 g and other with the mass of 84 g thus the resultant product must have a mass of 199 g.

Chemical equation:

A + B → AB

115 g + 84 g = 199 g

119 g = 199 g

Final answer:

The mass of the products after the reaction between 115 grams and 84 grams of two substances, according to the Law of Conservation of Mass, will be 199 grams, assuming no material is lost during the reaction.

Explanation:

If 115 grams of a substance reacts with 84 grams of another substance, according to the Law of Conservation of Mass, the mass of the products after the reaction will be the sum of the masses of the reactants. This is because in a chemical reaction, matter is neither created nor destroyed, so the total mass of the reactants must equal the total mass of the products.

In this case, adding the masses of the reactants gives us:

115 g + 84 g = 199 g

Therefore, the mass of the products after the reaction will be 199 grams. However, this assumes a perfectly efficient reaction with no losses to factors such as gas escaping the reaction vessel or side reactions. In a practical setting, a slight discrepancy might be observed due to such factors.

Answer:

Gases mix easily because of their high kinetic energy and low inter-molecular forces.

Explanation:

Answer:

Gases mix easily because of their high kinetic energy and low inter-molecular forces.

Explanation:

The relationship between the two reactions is correctly explained as chemical equation 1 is the opposite of Reaction 2.The reactant in Reaction 2 breaks apart, making it a decomposition reaction.

Chemical equation is a symbolic representation of a chemical reaction which is written in the form of symbols and chemical formulas.The reactants are present on the left hand side while the products are present on the right hand side.

A plus sign is present between reactants and products if they are more than one in any case and an arrow is present pointing towards the product side which indicates the direction of the reaction .There are coefficients present next to the chemical symbols and formulas .

The first chemical equation was put forth by Jean Beguin in 1615.By making use of chemical equations the direction of reaction ,state of reactants and products can be stated. In the chemical equations even the temperature to be maintained and catalyst can be mentioned.

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Answer:

[tex]1.6\times 103[/tex]g of K₂SO₄ will be produced

Explanation:

Given [tex]K_3PO_4[/tex] is available in excess

Reaction:

Cr₂(SO₄)₃ + 2K₃PO₄  ----------> 3K₂SO₄ + 2CrPO₄

From the above reaction, it is clear that the moles of (Cr₂(SO₄)₃), [tex]K_3PO_4[/tex] and  K₂SO₄  are 1, 2 and 3 respectively

We can say that 1 mole of chromium(iii) sulfate (Cr₂(SO₄)₃) react with 2 mole of potassium phosphate (K₃PO₄) to produce 3 mole of K₂SO₄

molar mass of Cr₂(SO₄)₃ = 147 g/mol

molar mass of K₃PO₄ = 212 g/mol

molar mass of K₂SO₄ = 174 g/mol

We can write as;

     Cr₂(SO₄)₃               +       2K₃PO₄             ----------> 3K₂SO₄  + 2CrPO₄

     1 mol (147 g/mol)         2 mol  (212 g/mol)           3 mol  (174g/mol)

Therefore, we have

         Cr₂(SO₄)₃  +  2K₃PO₄          ----------> 3K₂SO₄ + 2CrPO₄

             147 g         424 g                           522 g

So, we can see that 147 g of Cr₂(SO₄)₃ reacts with 424 g of 2K₃PO₄ to produce  522 g of K₂SO₄

             147 g of  Cr₂(SO₄)₃  = 522 g of K₂SO₄

             So, 450 g of  Cr₂(SO₄)₃ = [tex]\frac{(522\times 450)}{147}[/tex]g of K₂SO₄ = 1597.959 g = [tex]1.59\times 103[/tex] g = [tex]1.6\times 103[/tex] g

So,[tex]1.6\times 103[/tex]g of K₂SO₄ will be produced

Answer:

[tex]\large \boxed{\text{460 dg}}[/tex]

Explanation:

1. Convert decagrams to grams

[tex]\text{Mass} =  \text{4.6 dag} \times\dfrac{\text{10 g}}{\text{1 dag}} = \text{46 g}[/tex]

2. Convert grams to decigrams

\[tex]\text{Mass} =  \text{46 g} \times\dfrac{\text{10 g}}{\text{1 dg}} = \text{460 dg}\\\\\large \boxed{\textbf{4.6 dag = 460 dg}}[/tex]

Answer:

the correct answer would be c.

Answer:  The correct answer is: C. what you're going to do and how you're going to do it

Explanation:  A good experiment should follow a design with a few logical steps. An experiment must be carried out methodically in order to perform valid and reliable measurements. Experimenting is a procedure to check one or several hypotheses about a specific topic.

Answer:

One complete revolution around a circular path.

Explanation:

Let us take the case of a car moving in a circular track of radius r metres.

In one revolution, the car covers the length(distance) equal to the perimeter of the circle.

In this case, distance traveled = 2[tex]\pi[/tex]r metres

But after one complete revolution, the car reaches the same position as it was at the beginning of the motion.

Hence, the initial and final points coincide or the car hasn't changed it's position w.r.t the initial point.

So in this case, the displacement is zero.

Hence, revolution of a car around a circular path is an example of an object traveling a distance but having no displacement.

Answer:

Structures Attached

Explanation:

1. Total Number of e⁻  in NF₃

Number of e in Nitrogen = 5

Number of e in Flourine = 7

Total Number of e⁻ in NF₃ = 5 + 7(3)

Total Number of e⁻ in NF₃ = 5 + 21

Total Number of e⁻ in NF₃ = 26

2. Total Number of e⁻  in H₂S

Number of e in Hydrogen = 1

Number of e in sulphur = 6

Total Number of e⁻ in H₂S = 1(2) + 6

Total Number of e⁻ in H₂S = 2 + 6

Total Number of e⁻ in H₂S = 8

3. Total Number of e⁻  in F₂

Number of e in Flourine = 7

Total Number of e⁻ in F₂ = 7(2)

Total Number of e⁻ in F₂ = 14

4. Total Number of e⁻  in CO

Number of e in carbon = 4

Number of e in oxygen = 6

Total Number of e⁻ in CO = 4 + 6

Total Number of e⁻ in CO = 10

5. Total Number of e⁻  in SO₂

Number of e in Sulphur = 6

Number of e in oxygen = 6

Total Number of e⁻ in SO₂ = 6 + 6(2)

Total Number of e⁻ in SO₂ = 6 + 12

Total Number of e⁻ in SO₂ = 18

6. Total Number of e⁻  in O₂

Number of e in Oxygen = 6

Total Number of e⁻ in O₂ = 6(2)

Total Number of e⁻ in O₂ = 12

7. Total Number of e⁻  in SF₂

Number of e in Sulphur = 6

Number of e in Flourine = 7

Total Number of e⁻ in SF₂ = 6 + 7(2)

Total Number of e⁻ in SF₂ = 6 + 14

Total Number of e⁻ in SF₂ = 20

8. Total Number of e⁻  in BH₃

Number of e in Boron = 3

Number of e in Hydrogen = 1

Total Number of e⁻ in BH₃ = 3 + 1(3)

Total Number of e⁻ in BH₃ = 3 + 3

Total Number of e⁻ in BH₃ = 6

9. Total Number of e⁻  in CHCl₃

Number of e in Carbon = 4

Number of e in Hydrogen = 1

Number of e in chlorine = 7

Total Number of e⁻ in CHCl₃ = 4 + 1+ 7(3)

Total Number of e⁻ in CHCl₃ = 4 + 1 + 21

Total Number of e⁻ in CHCl₃ = 26

10. Total Number of e⁻  in CS₂

Number of e in Carbon = 4

Number of e in Sulphur = 6

Total Number of e⁻ in CS₂ = 4 + 6(2)

Total Number of e⁻ in CS₂ = 4 + 12

Total Number of e⁻ in CS₂ = 16

11. Total Number of e⁻  in BeCl₂

Number of e in Beryllium = 2

Number of e in Chlorine = 7

Total Number of e⁻ in BeCl₂ = 2 + 7(2)

Total Number of e⁻ in BeCl₂ = 2 + 14

Total Number of e⁻ in BeCl₂ = 16

12. Total Number of e⁻  in HCN

Number of e⁻ in Hydrogen = 1

Number of e⁻ in Carbon = 4

Number of e⁻ in Nitrogen = 5

Total Number of e⁻ in HCN = 1 + 4 + 5

Total Number of e⁻ in HCN = 10

13. Total Number of e⁻  in C₂H₂

Number of e⁻ in Carbon = 4

Number of e⁻ in Hydrogen = 1

Total Number of e⁻ in C₂H₂ = 4(2) + 1(2)

Total Number of e⁻ in C₂H₂ = 8 + 2

Total Number of e⁻ in C₂H₂ = 10

14. Total Number of e⁻  in SiO₂

Number of e⁻ in Silicon = 4

Number of e⁻ in Oxygen = 6

Total Number of e⁻ in SiO₂ = 4 + 6(2)

Total Number of e⁻ in SiO₂ = 4 + 12

Total Number of e⁻ in SiO₂ = 16

15. Total Number of e⁻  in H₂O₂

Number of e⁻ in Hydrogen = 1

Number of e⁻ in Oxygen = 6

Total Number of e⁻ in H₂O₂ = 1(2) + 6(2)

Total Number of e⁻ in H₂O₂ = 2 + 12

Total Number of e⁻ in H₂O₂ = 14

16. Total Number of e⁻  in SO₂⁻

Number of e in Sulphur = 6

Number of e in Oxygen = 6

Total Number of e⁻ in SO₂⁻ = 6 + 6(2)

Total Number of e⁻ in SO₂⁻ = 6 + 12

Total Number of e⁻ in SO₂⁻ = 18

17. Total Number of e⁻  in CH₃OH

Number of e⁻ in Carbon = 4

Number of e⁻ in Hydrogen = 1

Number of e⁻ in Oxygen = 6

Total Number of e⁻ in CH₃OH = 4 + 1(3) + 6 + 1

Total Number of e⁻ in CH₃OH = 4 + 3 + 6 + 1

Total Number of e⁻ in CH₃OH = 14

18. Total Number of e⁻  in NO₃

Number of e⁻ in Nitrogen = 5

Number of e⁻ in Oxygen = 6

Total Number of e⁻ in NO₃ = 5 + 6(3)

Total Number of e⁻ in NO₃ = 5 + 18

Total Number of e⁻ in NO₃ = 23

19. Total Number of e⁻  in ClO

Number of e⁻ in Chlorine = 7

Number of e⁻ in Oxygen = 6

Total Number of e⁻ in ClO = 7 + 6

Total Number of e⁻ in ClO = 7 + 6

Total Number of e⁻ in ClO = 13

20. Total Number of e⁻  in CH₂O

Number of e⁻ in Carbon = 4

Number of e⁻ in Hydrogen = 1

Number of e⁻ in Oxygen = 6

Total Number of e⁻ in CH₂O = 4 + 1(2) + 6

Total Number of e⁻ in CH₂O = 4 + 2 + 6

Total Number of e⁻ in CH₂O = 12

Lewis structures for different molecules are provided, including nitrogen trifluoride, hydrogen sulfide, fluorine, carbon monoxide, sulfur dioxide, oxygen, sulfur difluoride, boron trihydride, chloroform, carbon disulfide, beryllium chloride, hydrogen cyanide, acetylene, silicon dioxide, hydrogen peroxide, sulfate, methanol, nitrate, chlorite, and formic acid.

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Answer:

[tex]\large \boxed{\text{77.4 mL}}[/tex]

Explanation:

                Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

    V/mL:     249

c/mol·L⁻¹:  0.0443     0.285

1. Calculate the moles of Ba(OH)₂

[tex]\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}[/tex]

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

[tex]\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1  mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}[/tex]

3. Calculate the volume of HCl

[tex]V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}[/tex]

The empirical formula for [tex]C_2H_6O_2[/tex] is [tex]CH_3O[/tex], obtained by finding the simplest whole-number ratio of the elements after dividing the number of atoms in the molecular formula by the greatest common divisor.

To determine the empirical formula of [tex]C_2H_6O_2[/tex], we need to find the simplest whole-number ratio of the atoms of each element in the compound. We can start by noting that the number of carbon (C), hydrogen (H), and oxygen (O) atoms as represented in the molecular formula [tex]C_2H_6O_2[/tex].

Looking at the subscripts of the elements, we can divide each by the greatest common divisor of the subscripts, which is 2. Doing this, we get:

Carbon: 2 / 2 = 1

Hydrogen: 6 / 2 = 3

Oxygen: 2 / 2 = 1

Therefore, the empirical formula is [tex]CH_3O[/tex]. To clarify, this is the simplest form of a formula representing the chemical composition.

As a check, if we consider the molecular formula of a very different compound, glucose, which is [tex]C_6H_{12}O_6[/tex], the empirical formula is [tex]CH_2O[/tex], because it is the simplest whole-number ratio, which in glucose's case is 1:2:1 for carbon, hydrogen, and oxygen atoms, respectively.

Answer:

Density of  metal = 259 g/mL

Explanation:

Data Given:

Mass of metal = 518.0 g

Volume of Sugar =  2.00 mL

Density of = ? g/mL

The formula will be used is

                    d = m/v ........................................... (1)

where

d is density

m is the mass

v is the volume

So

put the given values in Equation (1)

d =  518.0 g / 2.00 mL

d = 259 g/mL

Density of  metal = 259 g/mL

Answer:

[tex]\large \boxed{\text{106 L}}[/tex]

Explanation:

We will need a balanced chemical equation with molar masses and volumes, so, let's gather all the information in one place.

MV/L:                                                                     22.71

M_r:         63.55

                  Cu + 4HNO₃ ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂

m/g:           149

(a) Moles of Cu

[tex]\text{Moles of Cu } =\text{149 g Cu } \times \dfrac{\text{1 mol Cu }}{\text{63.55 g Cu }} =\text{2.344 mol Cu}[/tex]

(b) Moles of NO₂

The molar ratio is 2 mol NO₂:1 mol Cu

[tex]\text{Moles of NO$_{2}$}= \text{2.344 mol Cu} \times \dfrac{\text{2 mol NO$_{2}$}}{ \text{1 mol Cu}} = \text{4.689 mol NO$_{2}$}[/tex]

(c) Volume of NO₂

The volume of 1 mol of an ideal gas at STP (0 °C and 1 bar) is 22.71 L.

[tex]\text{V} = \text{4.689 mol} \times \dfrac{\text{22.71 L}}{\text{1 mol}} = \textbf{106 L}\\\\\text{You can produce $\large \boxed{\textbf{106 L}} $ of NO$_{2}$.}[/tex]

Answer:

Gasoline

Explanation:

Pure Substance: are those having same type of atoms or molecules.

Compound: one or more atoms of the element combine to form a compound.

Mixture: is combination of one or more substances mix together.

Homogeneous Mixture: are those which are not distinguish by naked eye but can be separated into its components by physical means

Heterogeneous mixture: are those which can be distinguishing by naked eye and can be separated into its components by physical means.

So keeping in mind the above definitions

Table salt:

The chemical formula of table salt is NaCl and it a pure compound not mixture.

Gasoline:  

Gasoline is mixture of hydrocarbon, It contains small hydrocarbons ranging from 4 Carbon to 12 Carbon per molecule. it is a homogeneous mixture.

Aluminum:

Aluminum is a pure substance that is made up of same kind of atoms. so it is an element and not mixture. Its symbol is Al

Carbon dioxide:

Carbon dioxide is a pure compound and its chemical formula is CO₂ and not a mixture.

So, the write option is Gasoline.

Answer:

gasoline

Explanation:

Answer:

chemical weathering I think it is a right answers

Condensation is most likely to occur when air is saturated and contains condensation nuclei, as the saturation represents the maximum capacity for water vapor and the nuclei act as a base for the water vapor to condense into droplets.

Condensation can most likely occur in a given volume of air when the air is saturated and contains condensation nuclei. The saturation of air refers to its maximum capacity to hold water vapor. After reaching its saturation point, air cannot dissolve more water vapor, and any further addition of water vapor will lead to condensation. On the other hand, condensation nuclei are tiny particles that water vapor latches onto to form water droplets, facilitating the process of condensation. Therefore, air that is not only saturated with water vapor but also contains condensation nuclei has the strongest propensity for condensation to occur.

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Answer:

Molarity = 5.22 M

Explanation:

Given data:

Mass of sodium chloride = 7.0 g

Volume of solution = 23.0 mL ( 23.0/1000 = 0.023 L)

Molarity = ?

Solution;

Number of moles of NaCl = 7.0 g/ 58.4 g/mol

Number of moles of NaCl = 0.12 mol

Molarity = moles of solute / volume in litter

Molarity = 0.12 mol /  0.023 L

Molarity = 5.22 M

Answer:same

Explanation: