A reaction occurs when solid X is placed into solution Y. As a result, the temperature of the new solution increases by 3°C. The temperature rises another 2°C when more of solid X is added to the solution. What does this indicate about the effect of adding more solid? It increased the number of molecular collisions. It decreased the number of molecular collisions. It decreased the space between the molecules and lowered the reaction rate. It increased the space between the molecules and increased the reaction rate

Answer: The given statement is false.

Explanation:

A single bond is defined as the bond through which two same or different atoms are combined to each other by a single line. It is represented by a hyphen.

For example, C-H shows that both carbon and hydrogen atoms are attached to each other through single bond.

It is not necessary that all the molecules contain a single bond. This is because there are molecules which contain a double or triple bond also.

For example, [tex]H_{2}C=CH_{2}[/tex] is ethene molecule and it contains a double bond.

Therefore, we can conclude that the statement all molecules contain only single bond, is false.

Answer: Option (A) is the correct answer.

Explanation:

A transition element is defined as the element which has incompletely filled d-orbital.

For example, atomic number of chromium is 24 and its electronic configuration is [tex][Ar]4s^{1}3d^{5}[/tex].

Also, sometimes the s-orbital is completely filled and sometimes it is half-filled.

Hence, we can conclude that a single factor that classifies an element as a transition metal is that the  highest occupied orbital is a "d" orbital.

After 30 years, 12.5 grams will remain from a 100 grams sample of a radioactive material with a 10-year half-life. This is because the half-life is the period of time it takes for a substance undergoing decay to decrease by half.

The concept in question is related to radioactive decay and the half-life of a substance, both core concepts in physics and nuclear chemistry. For a substance with a half-life of 10 years, after each period of 10 years, it will be reduced to half of its previous amount. So, in a timeline, starting with 100 grams of this substance, after 10 years, you'll have 50 grams left. 10 years later (i.e., 20 years total), half of these 50 grams decay, leaving you with 25 grams. After another 10 years (i.e., 30 years total), half of this 25 grams decays, leaving you with 12.5 grams. Thus, in answer to your question, after 30 years of a 100 g sample of radioactive material with a 10-year half-life, 12.5 grams of it will remain.

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The atom with 7 valence electrons gains one electron from  the second atom with one valence electrons will bond together through ionic bonding.

Ionic compounds are formed between metals and non-metals. The metals are electropositive and easily loss electrons to electron deficient non metals to achieve octet.

When the non-metals gain electrons they acquire a negative charge and metals gain a positive charge by losing electron. These opposite charges electrostatically attracts and form the ionic bond.

The atom with one valence electron will donate it to the atom with 7 valence electrons to achieve octet. The donated atom acquires a positive charge and the received atom gains a negative charge. These ions attracts electrostatically to form the bond.

For example, in NaCl, sodium chloride is an ionic compound formed by donating an electrons from the sodium metal to the Cl atom. Thus, Na becomes Na + and Cl becomes Cl-. Therefore, option c is correct.

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Final answer:

To calculate ΔG, we use the equation ΔG = ΔH - TΔS, converting temperature to Kelvin and substituting the given values. The calculation gives a ΔG of approximately -136.47945 kJ/mol, so the closest answer is -136 kJ.

Explanation:

To calculate the change in Gibbs free energy (ΔG), we can use the Gibbs free energy equation:

ΔG = ΔH - TΔS

First, convert the temperature from Celsius to Kelvin:

T(K) = T(°C) + 273.15 = 110 °C + 273.15 = 383.15 K

Then substitute the given values into the equation:

ΔH = -85.5 kJ/mol

ΔS = 0.133 kJ/(mol)·K

T = 383.15 K

ΔG = (-85.5 kJ/mol) - (383.15 K)(0.133 kJ/(mol)·K)

ΔG = -85.5 kJ/mol - (50.97945 kJ/mol)

ΔG = -136.47945 kJ/mol

The closest answer to ΔG is option 4, -136 kJ.

Answer:

51 grams of ammonia would be formed.

Explanation:

[tex]3H_2+N_2\rightarrow 2NH_3[/tex]

Moles of hydrogen gas = 4.50 moles

According to reaction, 3 moles of hydrogen gas gives 2 moles of ammonia gas.

Then 4.50 moles of hydrogen gas will give:

[tex]\frac{2}{3}\times 4.50 mol=3.00 mol[/tex]of ammonia

Mass of 3.00 moles of ammonia:

[tex]17 g/mol\times mol=51 g[/tex]

51 grams of ammonia would be formed.

Final answer:

To find out how many grams of ammonia are produced from 4.50 moles of hydrogen, the balanced chemical equation for the synthesis of ammonia and the 3:2 mole ratio between hydrogen and ammonia are used. It is calculated that 4.50 moles of hydrogen will produce 3.00 moles of ammonia which is equivalent to 51.093 grams of ammonia.

Explanation:

To determine how many grams of ammonia would be formed from the complete reaction of 4.50 moles of hydrogen, we must use the balanced chemical equation for the synthesis of ammonia:

N2(g) + 3H2(g) → 2NH3(g)

From the equation, we see that 3 moles of hydrogen gas (H2) produce 2 moles of ammonia (NH3). Hence, the mole ratio is 3 moles H2 to 2 moles NH3. To find the amount of ammonia produced from 4.50 moles of hydrogen, we can set up a ratio:

3 moles H2 : 2 moles NH3 = 4.50 moles H2 : x moles NH3

Calculating this gives us:

4.50 moles H2 / 3 moles H2 * 2 moles NH3 = 3.00 moles NH3

To convert moles of NH3 to grams, we need to know the molar mass of NH3, which is approximately 17.031 g/mol:

3.00 moles NH3 * 17.031 g/mol = 51.093 grams of ammonia.

Answer:

B is correcta-moundo

Explanation:

Final answer:

According to Boyle's Law, when the temperature and amount of gas are kept constant, the product of the initial pressure and initial volume is equal to the product of the final pressure and final volume. By plugging in the given values and solving the equation, the new volume is determined to be 1.29 liters.

Explanation:

To solve this problem, we can use the relationship between pressure and volume described by Boyle's Law. According to Boyle's Law, when the temperature and amount of gas are kept constant, the product of the initial pressure and initial volume is equal to the product of the final pressure and final volume.

Mathematically, this can be represented as P1 * V1 = P2 * V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

In this case, the initial pressure is 1.1 atm and the initial volume is 4.0 L. If we increase the pressure to 3.4 atm, we can plug these values into the equation to find the new volume:

(1.1 atm) * (4.0 L) = (3.4 atm) * (V2)

Solving for V2, we divide both sides of the equation by 3.4 atm:

V2 = (1.1 atm * 4.0 L) / 3.4 atm = 1.29 L

Therefore, the new volume will be 1.29 liters.

Answer : The balanced chemical reaction will be,

[tex]HNO_3(aq)+NH_3(aq)\rightarrow NH_4NO_3(aq)[/tex]

Explanation :

Balanced chemical reaction : It is a reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

As per question.

When aqueous nitric acid react with aqueous ammonia then it gives ammonium nitrate as a product.

The balanced chemical reaction will be,

[tex]HNO_3(aq)+NH_3(aq)\rightarrow NH_4NO_3(aq)[/tex]

This reaction is an acid-base reaction in which an acid react with a base to give salt and water as a product.

Answer is: volume is 20 mL.
c₁(CH₃COOH) = 2,5 M.
c₂(CH₃COOH) = 0,5 M.
V₂(CH₃COOH) = 100 mL.
V₁(CH₃COOH) = ?
c₁(CH₃COOH) · V₁(CH₃COOH) = c₂(CH₃COOH) · V₂(CH₃COOH).
2,5 M · V₁(CH₃COOH) = 0,5 M · 100 mL.
V₁(CH₃COOH) = 0,5 M · 100 mL ÷ 2,5 M.
V₁(CH₃COOH) = 20 mL ÷ 1000 mL/L =0,02 L.

Answer:

20 milliliters

Explanation:

I don't cap

Answer:

2.31 molar

Explanation:

175 g/101 g/mol=1.73 mol

1.73mol/.750 L = 2.31 mol/L

Final answer:

To estimate the concentration of hydroxide ions ([OH-]) in a 0.125 M carbonate (CO3²-) solution, we would simply use the stoichiometry assuming that each carbonate ion becomes one hydroxide ion, giving an approximate [OH-] value of 0.250 M. This is a simplification and actual values may vary according to equilibrium dynamics.

Explanation:

To determine the concentration of hydroxide ions ([OH-]) in a solution that is 0.125 M in carbonate ions (CO32-), we need to understand the relationship between carbonate and hydroxide ions in an aqueous solution. Carbonate ions can react with water in a basic reaction to form bicarbonate ions (HCO3-) and hydroxide ions (OH-).

In this reaction, each carbonate ion can produce two hydroxide ions. Therefore, if we have a 0.125 M solution of carbonate ions, we could potentially have a hydroxide concentration of 0.125 M × 2 = 0.250 M. But, since this is a basic reaction and not all carbonate will turn into hydroxide ions (due to equilibrium with bicarbonate), we need additional information to calculate the exact hydroxide concentration, such as the equilibrium constant for the reaction.

However, without this information, we can assume a 1:1 stoichiometry for simplicity, giving us an initial approximate [OH-] concentration of 0.250 M. This is a simplification and the actual concentration could be less due to the reasons stated above.

For more accurate results, we would use the equilibrium constant for the reaction between carbonate and water to calculate the exact [OH-].

Did you ever get the correct answer?

Final answer:

To find the mass of HCl produced, use stoichiometry based on the balanced chemical equation.

Explanation:

To find the number of grams of HCl that can be produced, we need to use the stoichiometry of the balanced chemical equation. From the equation, we can see that for every 2 moles of AlCl3, 6 moles of HCl are produced. Therefore, we can set up a proportion:

2 moles AlCl3 / 6 moles HCl = 249 g AlCl3 / x g HCl

Solving for x, we find that x = 747 g HCl. Therefore, 747 grams of HCl can be produced when 249 g of AlCl3 is reacted.

Answer: The gas formed by the reaction of sodium sulfite and hydrochloric acid is [tex]SO_2[/tex]

Explanation:

When sodium sulfite reacts with hydrochloric acid, it leads to the formation of many products.

The chemical equation for the reaction of two reacts are:

[tex]Na_2SO_3(s)+2HCl(aq.)\rightarrow 2NaCl(aq.)+SO_2(g)+H_2O(l)[/tex]

By Stoichiometry of the reaction:

1 mole of solid sodium sulfite reacts with 2 moles of aqueous hydrochloric acid to produce 2 moles of aqueous sodium chloride, 1 mole of sulfur dioxide gas and 1 mole of water molecule.

Hence, the gas formed by the reaction of sodium sulfite and hydrochloric acid is [tex]SO_2[/tex]

The gas formed by the reaction of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_3}[/tex] with HCl is [tex]\boxed{{\text{S}}{{\text{O}}_2}}[/tex]

Further Explanation:

The five types of chemical reactions are as follows:

1. Combination reactions:

These reactions are also known as a synthesis reaction. These are the reaction in which two or more reactants combine to form single product. These are generally accompanied by the release of heat so they are exothermic reactions.

Examples of combination reactions are as follows:

(a) [tex]{\text{Ba}}+{{\text{F}}_2}\to{\text{Ba}}{{\text{F}}_2}[/tex]

(b) [tex]{\text{CaO}}+{{\text{H}}_2}{\text{O}}\to{\text{Ca}}{\left({{\text{OH}}}\right)_2}[/tex]

2. Decomposition reactions:

The opposite of combination reactions is called a decomposition reaction. Here, a single reactant gets broken into two or more products. Such reactions are usually endothermic because energy is required to break the existing bonds between the reactant molecules.

Examples of decomposition reactions are as follows:

(a) [tex]2{{\text{H}}_2}{{\text{O}}_2}\to2{{\text{H}}_2}{\text{O}}+{{\text{O}}_2}[/tex]

(b) [tex]2{\text{NaCl}} \to{\text{2Na+C}}{{\text{l}}_2}[/tex]

3. Displacement reactions

Also known as replacement or metathesis reactions. Here, one of the reactants gets replaced by the other one. Generally, the more reactive element displaces the less reactive element. Both metals and non-metals can take part in displacement reactions.

Examples of displacement reactions are as follows:

(a) [tex]{\text{Cu}}+{\text{AgN}}{{\text{O}}_3}\to{\text{Ag}}+{\text{Cu}}{\left({{\text{N}}{{\text{O}}_3}}\right)_2}[/tex]

(b) [tex]{\text{C}}{{\text{l}}_2}+{\text{KBr}}\to{\text{B}}{{\text{r}}_2}+{\text{KCl}}[/tex]

4. Double displacement reactions

These are the reaction in which ions of two compound interchange with each other to form the product. For example, the general double displacement reaction between two compound AX and BY is as follows:

[tex]{\text{AX}}+{\text{BY}}\to{\text{AY}}+{\text{BX}}[/tex]

Examples of double displacement reactions are as follows:

(a) [tex]{\text{N}}{{\text{a}}_2}{\text{S}}+{\text{HCl}}\to{\text{NaCl}}+{{\text{H}}_2}{\text{S}}[/tex]

(b) [tex]2{\text{KOH}}+{\text{Cu}}{\left({{\text{N}}{{\text{O}}_{\text{3}}}}\right)_2}\to2{\text{KN}}{{\text{O}}_3}+{\text{Cu}}{\left({{\text{OH}}}\right)_2}[/tex]

5. Combustion reactions:

These are the reactions that take place when hydrocarbons are burnt in the presence of oxygen to form carbon dioxide and water. These are also referred to burning.  

Example of combustion reactions are as follows:

(a) [tex]{\text{C}}{{\text{H}}_4}+{{\text{O}}_2}\to{\text{C}}{{\text{O}}_2}+{{\text{H}}_2}{\text{O}}[/tex]

(b) [tex]{{\text{C}}_{10}}{{\text{H}}_{14}}+12{{\text{O}}_2}\to10{\text{C}}{{\text{O}}_2}+4{{\text{H}}_2}{\text{O}}[/tex]

The given reaction occurs as follows:

[tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{3}}}\left(s\right)+{\text{2HCl}}\left({aq}\right)\to{\text{2NaCl}}\left({aq}\right)+{\text{S}}{{\text{O}}_{\text{2}}}\left(g\right)+{{\text{H}}_{\text{2}}}{\text{O}}\left(l\right)[/tex]

This is an example of a double displacement reaction. Here [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_3}[/tex] reacts with HCl to form NaCl, [tex]{\text{S}}{{\text{O}}_2}[/tex] and [tex]{{\text{H}}_{\text{2}}}{\text{O}}[/tex] . Of all the products formed, NaCl is present in the aqueous phase, [tex]{{\text{H}}_{\text{2}}}{\text{O}}[/tex] is in liquid state and [tex]{\text{S}}{{\text{O}}_2}[/tex] is in gaseous state.

So the gas formed in the above reaction is [tex]{\mathbf{S}}{{\mathbf{O}}_{\mathbf{2}}}[/tex] .

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Answer details:

Grade: High School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: SO2, Na2SO3, HCl, NaCl, H2O, double displacement reaction, combination reaction, combustion reaction, displacement reaction, decomposition reaction, aqueous, liquid, gaseous state.

Answer:

2.03125g of acetylene

Explanation:

First thing's first, we have to write out the balanced chemical equation;

CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g)

Water is in excess, so CAC2 is our limiting reactant. i.e it determines the amount of product that would be formed.

1 mol of CaC2 produces 1 mol of C2H2

In terms of mass;

Mass = Number of moles * Molar mass

where the molar mass of the elements are;

Ca = 40g/mol

C = 12g/mol

H = 1g/mol

CaC2 = 40+ (2*12) = 64g/mol

C2H2 =( 2 * 12) + ( 2 * 1) = 26g/mol

64g (1 * 64g/mol) of CaC2 produces 26g ( 1mol * 26g/mol) of C2H2

5g would produce x?

64 = 26

5 = x

Upon solving for x we have;

x = (5 * 26) / 64

x = 2.03125g

By using a balanced chemical equation and molar conversions, we can calculate that adding water to 5.00 g of CaC2 will produce approximately 2.03 grams of acetylene.

In order to find out how many grams of acetylene are produced by adding water to 5.00 g of CaC2, we need to start by writing a balanced chemical equation for the reaction: CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(aq). From this equation, we can see that one molecule of calcium carbide reacts with two molecules of water to produce one molecule of acetylene and one molecule of calcium hydroxide.

Next, we calculate the molar mass of CaC2 which is approximately 64 g/mol. Therefore, 5.00 g of CaC2 is approximately 0.0781 moles. According to the chemical equation, one mole of CaC2 produces one mole of C2H2, so 5.00 g of CaC2 would also produce 0.0781 moles of acetylene.

Last, we convert the moles of acetylene to grams by multiplying by its molar mass (26.04 g/mol), which gives us approximately 2.03 g of acetylene. Therefore, approximately 2.03 grams of acetylene are produced by adding water to 5.00 g of CaC2.

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Explanation :

(a) The full form of IMA is Ideal Mechanical Advantage. It is given as the ratio of the radius of the wheel and the radius of the axle.

From the figure,

The radius of the wheel is 6 cm

and the radius of the axle is 1 cm.    

[tex]IMA=\dfrac{6\ cm}{1\ cm}=6\ cm[/tex]

So, the IMA of the doorknob is 6.

(b) The IMA of the inclined plane is given by the ratio of the distance covered by the effort (L) and the distance covered by the resistance (h).

So, [tex]IMA=\dfrac{4.5\ m}{1.5\ m}=3[/tex]

So, the IMA of the inclined plane is 3.

(c) Work done is the product of the force acting and the displacement caused.

Mathematically, it can be written as :

[tex]W=F\times d[/tex]

F is applied force

d is the distance or displacement.

Combustion- is the correct answer on E.D.G

Answer:

B. COMBUSTION

Explanation:

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