I am so lost, and suggestions to how I can solve this?

A sample of nickel ( specific heat capacity 0.446J/goC) is heated to 255.5oC and placed in a coffee cup calorimeter containing 250.0 g of water at a temperature of 20.0oC. Calculate the mass of the nickel sample in grams, if the final temperature of the system is 35.5oC. The heat capacity of H2O is 4.18 J/gC.

Answer:

C) 14.1 moles

Explanation:

1 mole = 6.02 *10^ 23

8.50 * 10^24 molecules *(1 mole/6.02*10^23 molecules) ≈ 14.1 moles

The number of moles there are in 8.50 X [tex]10^{24}[/tex] molecules of sodium sulfate is 14.1 moles (option C).

The number of moles in a molecule of a substance can be calculated by dividing the number of molecules by Avogadro's number as follows:

no of moles = no of molecules ÷ 6.02 × [tex]10^{23}[/tex] molecules

According to this question, there are in 8.50 X [tex]10^{24}[/tex] molecules of sodium sulfate in a substance. The number of moles in this substance can be calculated as follows:

no of moles = 8.50 X [tex]10^{24}[/tex] ÷ 6.02 × [tex]10^{23}[/tex]

no of moles = 14.1 moles

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Answer:

The elements in the periodic table are arranged in order of increasing atomic number.

Explanation:

The elements are arranged based on their atomic numbers in a periodic table, not based on their atomic masses.  

• The chemical elements in the periodic table are demonstrated in order of their atomic number.  

• In the modern periodic table, the arrangement of elements is done based on their atomic number, not their relative atomic masses.  

• In the periodic table, the horizontal rows, known as periods, show the arrangement of elements in order of increasing atomic number.  

Thus, based on atomic numbers, elements are arranged in a periodic table.

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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]

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:

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:

As Bob add more solute to dissolves, the rate of solution will be decreases

Explanation:

At certain temperature the maximum amount of solute dissolve in the specific amount of solution or solvent.

There are many factors that affect solubility of the solute in solvent. Nature of the solute also affects the rate of solubility.

Temperature also affects the rate of solubility. Solubility increase when the temperature increases.

Pressure: Change in pressure have no effect on the rate of solubility

The rate of solution:  is a measure that how fast a substance dissolves in solvent.

Amount of solute already dissolved affect the rate of solution. If a solvent have little amount of solute in the solution it dissolve quickly. But when you have more solute in the solution the process was slower.

1 g of lead dissolves in 100 g of water at room temperature.  Similarly 5 mg of sodium hydroxide pellet dissolve in 100ml of water at room temperature.

So if Bob add 1 mg more of NaOH in the same 100ml of water the rate of the solubility decreases.

As there are some factors that decrease the solubility rate of NaOH pellet as follow:

• large size of the solute dissolve slowly as the solubility process take on      the surface of the solute particles, powders dissolve fast.

• less amount of solvent with ratio to solute

• saturated solution

So due to all above reasons the rate of the solubility decreases upon adding a little more amount of NaOH pellet.

Answer:

a

Explanation:

:)

Answer:

1 grams Strontium Fluoride to mol = 0.00796 mol

10 grams Strontium Fluoride to mol = 0.07961 mol

50 grams Strontium Fluoride to mol = 0.39804 mol

100 grams Strontium Fluoride to mol = 0.79607 mol

200 grams Strontium Fluoride to mol = 1.59214 mol

500 grams Strontium Fluoride to mol = 3.98036 mol

1000 grams Strontium Fluoride to mol = 7.96072 mol

- Hopefully this helps a bit :)

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:

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: 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:

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

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:

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.

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.

1.88 g

We are given;

We are required to determine the amount of uranium left after 2 days

N = N₀ × 0.5^n

Where, N is the remaining mass, N₀ is the original mass and n is the number of half lives.

n = (2 × 24) ÷ 12 hours

  = 4

Therefore;

Remaining mass, N = 30.0 g × 0.5^4

                                = 1.875 g

                                = 1.88 g

Therefore, the mass of the remaining sample after decay is 1.88 g

Answer:

Concentration = 0.14 M

Explanation:

Given data:

Volume of base = 19.62 mL

Molarity of base = 0.341 M

Volume of acid = 25.00 mL

Molarity of acids = ?

Solution:

Number of moles of base = Molarity × volume in litter

Number of moles = 0.341 mol/L × 0.02 L

Number of moles =  0.007 mol

Concentration of acid:

Concentration = 1/2 (0.007 mol) / 25 mL × 10⁻³ L/mL

Concentration = 0.0035 mol / 25 × 10⁻³ L

Concentration = 1.4  × 10⁻¹ M

Concentration = 0.14 M

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:

[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]

Answer:

negatively-charged particles.(ELECTRON) B

Explanation:

Electrons are the negatively charged particles of atom. Together, all of the electrons of an atom create a negative charge that balances the positive charge of the protons in the atomic nucleus

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:same

Explanation:

Final answer:

Physical weathering refers to the process of breaking down rocks without changing their chemical composition. Examples include grinding away of rock, flaking due to pressure release, and water freezing and thawing.

Explanation:

Physical weathering refers to the process of breaking down rocks into smaller pieces without changing their chemical composition. Examples of physical weathering include the grinding away of a rock surface, the top layers of rock flaking off due to the release of pressure, and the freezing and thawing of water, which causes it to expand and contract and break the rock apart.

Other examples include tree roots growing in the cracks of rocks and animals digging burrows. On the other hand, rust forming due to oxidation and acid rain weathering a statue are examples of chemical weathering.

Final answer:

Physical weathering includes processes like abrasion, exfoliation, temperature fluctuations, freeze-thaw cycles, animal activities, and root wedging, all of which disintegrate rock mechanically.

Explanation:

The examples of physical weathering are those processes that result in the breakdown of rock without altering its chemical composition. Here are the examples that correctly apply to the concepts of physical weathering:

On the other hand, the following are not examples of physical weathering because they involve chemical changes to the rock: water dissolving soft rock, rust forming due to oxidation, and acid rain weathering a statue.

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:

Volume percent of a solution that has 10.0 g of ethanol= 12.37%

Volume percent of a solution that has 90.0 g of water = 87.69%

Explanation:

Volume Percent:

The volume percent helps us  to indicate the concentration of the solution when the volume of the solute and volume of the solution is given by

[tex]Volume Percent=\frac{\text{Volume of the Solute}}{\text{Volume of Solution}}\times100%[/tex]

With respect to density 1mL of ethanol

=> 0.7893g or 10g of methanol contains

=> [tex]\frac{1}{0.7893}\times10[/tex]

=> 12.67mL

And for water = [tex]\frac{1}{0.9987} \times 90.0[/tex]

=> 90.12mL

Now total volume = 90.12+ 12.67 = 102.79mL

%(v/v) for water = [tex]\frac{90.12}{102.79}\times 100[/tex]

 = 87.69%

%(v/v) for ethanol = [tex]\frac{12.67}{102.79}\times 100[/tex]

 = 12.37%

The volume percent of ethanol in the solution is 12.3%, and the volume percent of water is 87.7%. This is calculated by determining the volume of each substance using their respective densities and the mass provided and then finding the percentage of each substance's volume in the total solution volume.

To find the volume percent of a solution that has 10.0 g of ethanol with a density (D) of 0.7893 g/mL, and 90.0 g of water with a density (D) of 0.9987 g/mL, we first calculate the volumes of each component. For ethanol, divide 10.0 g by 0.7893 g/mL to get the volume in milliliters. For water, divide 90.0 g by 0.9987 g/mL to get the volume in milliliters.

To calculate the volume of ethanol: 10.0 g / 0.7893 g/mL = 12.7 mL
To calculate the volume of water: 90.0 g / 0.9987 g/mL = 90.2 mL
Assuming volumes are additive: Total volume of the solution = 12.7 mL (ethanol) + 90.2 mL (water) = 102.9 mL

Now, we can find the volume percent of ethanol: (volume of ethanol / total volume of solution) × 100% = (12.7 mL / 102.9 mL) × 100% = 12.3%

The volume percent of water is similarly calculated: (volume of water / total volume of solution) × 100% = (90.2 mL / 102.9 mL) × 100% = 87.7%

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Answer:

Mass = 1.76 g

Explanation:

Given data:

Volume of solution = 25 mL

Molarity = 0.45 M

Amount of CuSO₄.5H₂O = ?

Solution:

Molarity = number of moles / volume in litter

Number of moles = Molarity × Volume in L

Number of moles = 0.45 M  × 0.025 L

Number of moles = 0.011 mol

Mass of f CuSO₄.5H₂O:

Mass = Number of moles ×  molar mass

Mass = 0.011 mol × 159.6 g/mol

Mass = 1.76 g

Answer:

Explanation:

Polyatomic ions are ions which consist of more than one atom. For example, nitrate ion, NO3-, contains one nitrogen atom and three oxygen atoms. The atoms in a polyatomic ion are usually covalently bonded to one another, and therefore stay together as a single, charged unit.

To write the formula for a compound with a polyatomic ion, consider that the compound should be electrically neutral. Balance the charges from different ions, treating polyatomic ions as discrete units. For instance, calcium phosphate (Ca3(PO4)2) illustrates this with three calcium ions and two phosphate ions.

When writing the formula for a compound containing a polyatomic ion, you must consider that the compound has to be electrically neutral overall. This often involves balancing the number of positive and negative charges coming from different ions. The polyatomic ions are usually treated as discrete units in the formula.

For example, consider calcium phosphate, which has the formula Ca3(PO4)2. Here, there are three calcium ions (Ca²+), and two phosphate groups (PO4³-). Each PO4³- group comprises one phosphorus atom and four oxygen atoms. The overall compound is electrically neutral because the positive charges from the three calcium ions balance out the negative charges from the two phosphate ions.

Another example can be seen in the ionic compound sodium oxalate, which has the formula Na₂C₂O₄. Here, the polyatomic ion is C₂O₄²-, and the compound has two sodium ions (Na⁺) for each oxalate ion. The formula of this compound is not empirical because it represents the number of atoms in the discrete unit of the oxalate ion.

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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:

Answer:

option b : O₉S₁₀

Explanation:

There are some points to name a covalent compound, according to that

number of atoms numbered as

**mono for one, di - for two tri for three and so on.

the name of the atom as

**carbon as carbide, sulfur as sulfide, oxygen as oxide, foulrine as flouro etc.

So,

Formula of Nonoxide decasulfide

Now Naming of the Chemical compound

Following are the explanations of  terms

So keeping the above points in mind

The right answer is

Option b : O₉S₁₀

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:

12.04 × 10²³ atoms

Explanation:

Given data:

Mass of cobalt = 117.86 g

Number of atoms = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

Number of moles of cobalt = 117.86 g/ 58.9 g/mol

Number of moles of cobalt = 2 mol

one moles of cobalt =  6.022 × 10²³ atoms of cobalt

2 moles × 6.022 × 10²³ atoms / 1 mol

12.04 × 10²³ atoms