g A chemist combines 59.9 mL of 0.282 M potassium bromide with 15.4 mL of 0.512 M silver nitrate. (a) How many grams of silver bromide will precipitate

Answer:

Explanation:

In all calorimetric experiment , the calorimeter must be isolated from the surrounding . Otherwise the heat change in the experiment can not be determined with precision .

The reaction is endothermic . Hence, there is lowering of temperature due to absorption of heat in the reaction equal to ΔH°. The value of ΔH° can be calculated by measuring fall in the temperature of the content . The fall in the temperature will be less when heat is allowed to come from the surrounding . Less fall of temperature will result in less ΔH° to be calculated .

Hence in the given experiment , if the student neglects to put lid on the cup , the experiment will give less value of ΔH°.

Answer:

HBr, CH3C(OH)2 and CH3COOH, Br-

Explanation:

The conjugate acid-base pairs acid reacts with base to form a conjugate acid and conjugate base.

Conjugate acid is formed when a bases receives a proton  (H+) and a conjugate base is formed when an acid losses a proton (H+).

From the given equation:

HBr, CH3C(OH)2 and CH3COOH, Br- are conjugate acid-base pair, where HBr is an acid and CH3C(OH)2 is a conjugate acid while CH3COOH and Br- is the conjugate base.

Answer:

The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0.  It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it

Thats all i know

Answer:

2,2,3,3-tetrapropyloxirane

Explanation:

In this case, we have to know first the alkene that will react with the peroxyacid. So:

What do we know about the unknown alkene?

We know the product of the ozonolysis reaction (see figure 1). This reaction is an oxidative rupture reaction. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If [tex](CH_3CH_2CH_2)_2C=O[/tex] is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.

What is the product with the peroxyacid?

This compound in the presence of alkenes will produce peroxides. Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product 2,2,3,3-tetrapropyloxirane. (see figure 2)

Answer:

A) CH3CH2CH2CH2CH2CH2OH

Explanation:

For this question, we have the following answer options:

A) CH3CH2CH2CH2CH2CH2OH

B) (CH3CH2)2CH(OH)CH2CH3

C) (CH3CH2)2CHOHCH3

D) (CH3CH2)3COH

E) (CH3CH2)2C(CH3)OH

We have to remember the reaction mechanism of the substitution reaction with [tex]PBr_3[/tex]. The idea is to generate a better leaving group in order to add a "Br" atom.

The [tex]PBr_3[/tex] attacks the "OH" generation new a bond to P (O-P bonds are very strong), due to this new bond we will have a better leaving group that can remove the oxygen an allow the attack of the Br atom to generating a new C-Br bond. This is made by an Sn2 reaction. Therefore we will have a faster reaction with primary substrates. In this case, the only primary substrate is molecule A. So, "CH3CH2CH2CH2CH2CH2OH" will react faster.

See figure 1

I hope it helps!

Answer:

AgI I the limiting reactant.

Explanation:

The balanced equation for the reaction is given below:

2AgI + HgI2 → Ag2HgI4

Next, we shall determine masses AgI and HgI2 that reacted from the balanced equation.

This is illustrated below:

Molar mass of Agl = 108 + 127 = 235 g/mol

Mass of AgI from the balanced equation = 2 x 235 = 470 g

Molar mass of HgI2 = 201 + (2x127) = 455 g/mol

Mass of HgI2 from the balanced equation = 1 x 455 = 455 g

From the balanced equation above,

470 g of AgI reacted with 455 g of HgI2.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above,

470 g of AgI reacted with 455 g of HgI2.

Therefore, 2 g of AgI will react with

= (2 x 455)/470 = 1.94 g of HgI2.

From the calculations made above, we can see that only 1.94 g out of 3.50 g of HgI2 given is needed to react completely with 2 g of AgI.

Therefore, AgI I the limiting reactant.

Answer:

a. the Kb of NO₂⁻ to find the hydroxide concentration.

Explanation:

When sodium nitrite is dissolved in water, it dissociates in sodium cation and nitrite anion according to the following equation.

NaNO₂(s) ⇒ Na⁺(aq) + NO₂⁻(aq)

Na⁺ comes from NaOH (strong base) so it doesn't react with water.

NO₂⁻ comes from HNO₂ (weak acid) so it reacts with water according to the following equation.

NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)

This is the basic reaction of nitrite ion, so we need the Kb of NO₂⁻ to  find the hydroxide concentration.

Answer:

The answer is "[tex]\bold{\log \frac{[0] mole}{[R]mole}}[/tex]"

Explanation:

[tex]E_{cell} =E_{cell}^{\circ} - \frac{0.0591}{n}= \log\frac{[0]}{[R]}\\[/tex]

In the above-given equation, we can see from [tex]E_{ceu}[/tex], of both oxidant [tex]conc^n[/tex]as well as the reactant were connected. however, weight decreases oxidant and reduction component concentration only with volume and the both of the half cells by the very same factor  and each other suspend

[tex]\to \log \frac{\frac{\text{oxidating moles}}{25 \ ml}}{\frac{\text{moles of reduction}}{25 ml}} \ \ = \ \ \log \frac{\frac{\text{oxidating moles}}{30 \ ml}}{\frac{\text{moles of reduction}}{30 ml}} \\\\\\[/tex]

[tex]\to {\log \frac{[0] mole}{[R]mole}}[/tex]

The cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.

The cell potential has been given as the difference in the potential of the two half cells in the electrochemical reaction.

The two cells has been set with the concentration of solutions in the oxidation and reduction half cells.

The cell potential has been changed when there has been a change in the potential of the half cells.

The volume of 30 mL to the solution has been, resulting in the cell potential difference of x.

With the volume of 25 mL, there has been the difference in the potential being similar to the 30 mL solution, i.e. x.

Thus, the cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.

Learn more about cell potential, here:

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Answer:

Sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon are the elements of third period.

Explanation:

There are three energy levels in sodium atom. It has 11 electrons revolving around the nucleus. the atomic radius of sodium atom is 227 ppm. Magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon has also three energy levels like sodium because all these elements belongs to third period. There are 12 electrons present in magnesium, 13 in aluminium, 14 in silicon, 15 in phosphorus, 16 in sulfur, 17 in chlorine, and 18 electrons in argon. The atomic radius of magnesium atom is 173 ppm.  The atomic radius of aluminium atom is 143 ppm.  The atomic radius of silicon atom is 111 ppm.  The atomic radius of phosphorus atom is 98 ppm.  The atomic radius of sulfur atom is 87 ppm. The atomic radius of chlorine atom is 79 ppm and the atomic radius of argon atom is 71 ppm.

Answer:

466 torr

Explanation:

Step 1: Given data

Step 2: Calculate the final pressure

Since we have a gas changing at a constant temperature, we can calculate the final pressure using Boyle's law.

P₁ × V₁ = P₂ × V₂

P₂ = P₁ × V₁ / V₂

P₂ = 626 torr × 5.00 L / 6.72 L

P₂ = 466 torr

Answer:

-2, -2 and 0.

- Respiration is a process in which energy is produced and photosynthesis is a process in which energy is used.

Explanation:

Hello,

In this case, oxygen is a substance that is used for animals and us to acquire the energy necessary for several functions by the cellular respiration (we also need glucose), besides, it is a product of the photosynthesis carried out by vegetable cells (plants). Moreover, carbon dioxide and water are used by the plants to produce oxygen we need as well as glucose via the aforementioned photosynthesis, thus, both chemical reactions are shown below:

[tex]C_6H_{12}O_6+6O_2\rightarrow 6CO_2+6H_2O\ \ \ respiration\\\\6CO_2+6H_2O\rightarrow C_6H_{12}O_6+6O_2\ \ \ photosynthesis[/tex]

In such a way, since the oxygen in carbon dioxide and water has an oxidation state of -2 (reduced form) we can say that the respiration is a process in which energy is produced and since the oxygen yielded during the photosynthesis has an oxydation state of 0, we can say that photosynthesis is a process in which energy is used.

Best regards.

Answer:

A (Atomic mass number or Nucleon number)

Explanation:

The mass number is the total number of protons and nucleons in an atomic nucleus.

Hope this helps.

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Answer:

I would say c) a closed circuit.

Hope I was right.

Answer:

That means that if you are calculating entropy change, you must multiply the enthalpy change value by 1000. So if, say, you have an enthalpy change of -92.2 kJ mol-1, the value you must put into the equation is -92200 J mol-1

The entropy change in the surroundings associated with this reaction occurring at 25 degree C is calculated as ΔS = -ΔH/T J/K.

Entropy is a quantity which gives idea about the randomness or arrangement of atoms or molecules present in any sample.

Entropy change will be calculated as:
ΔS = -ΔH/T, where

ΔH = chnage in enthalpy (J/mole)

T = temperature (K)

So to calculate the entropy change first we have to know about the value of enthalpy in joules and then divide it by the temperature.

Hence the unit of entropy is joule per kelvin.

To know more about entropy, visit the below link:
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Answer:

Chemistry is one of the causes for global warming, and in some cases it can even cause certain illnesses.

Answer:

Chemists have both hurt the environment and helped the environment by their actions.

Explanation:

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Answer:

PLEASE LOOK INN TO THE FILE YOU WILL GET ANSWER AND ALSO SUMMARY THANKS FOR ASKING QUESTION.

Explanation:

Answer:

[tex]C_{C_4H_6}=0.179M[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2C_4H_6\rightarrow C_8H_{12}[/tex]

And the rate law is:

[tex]\frac{dC_{C_4H_6}}{dt}=kC_{C_4H_6}^2[/tex]

Which integrated is:

[tex]\frac{1}{C_{C_4H_6}} =\frac{1}{C_{C_4H_6}^0}+kt[/tex]

In such a way, the concentration after 10.0 min is:

[tex]\frac{1}{C_{C_4H_6}} =\frac{1}{0.200M}}+5.76x10^{-2}\frac{L}{mol*min}*10.0min\\ \\\frac{1}{C_{C_4H_6}}=5.58\frac{L}{mol} \\\\C_{C_4H_6}=\frac{1}{5.58\frac{L}{mol} } \\\\C_{C_4H_6}=0.179M[/tex]

Regards.

Answer:

[tex]T_b=-88.48\°C[/tex]

Explanation:

Hello,

In this case, since the entropy of vaporization is defined in terms of the enthalpy of vaporization and the boiling point of the given substance, nitrous oxide, as shown below:

[tex]\Delta _{vap}S=\frac{\Delta _{vap}}{T_b}[/tex]

Solving for the boiling point of nitrous oxide, we obtain:

[tex]T_b=\frac{\Delta _{vap}H}{\Delta _{vap}S}=\frac{16.53\frac{kJ}{mol}*\frac{1000J}{1kJ} }{89.51\frac{J}{mol} } \\ \\T_b=184.67K[/tex]

Which in degree Celsius is also:

[tex]Tb=184.67-273.15\\\\T_b=-88.48\°C[/tex]

Best regards.

Answer:

Kindly check the explanation section.

Explanation:

PS: kindly check the attachment below for the required diagram that is the diagram showing solid sodium chloride looks like at the atomic level.

The chemical compound known as sodium chloride, NaCl has Molar mass: 58.44 g/mol, Melting point: 801 °C and

Boiling point: 1,465 °C. The structure of the solid sodium chloride is FACE CENTRED CUBIC STRUCTURE. Also, solid sodium chloride has a coordination number of 6: 6.

In the diagram below, the positive sign shows the sodium ion while the thick full stop sign represent the chlorine ion.

The NaCl has been the ionic structure with an equal number of sodium and chlorine ions bonded.

In the structure, there has been each Na ion bonded with the Cl ions. There has been the transfer of electrons between the structure in order to attain a stable configuration.

The expected structure of the NaCl would be the image attached below.

The image has been the cubic structure of NaCl. With the presence of Na ions at the vertex of the structure, there has been the presence of the Cl ion with every Na ion for the electron transfer.

For more information about the structure of NaCl, refer to the link:

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Answer:

CO3^2-

Explanation:

In qualitative analysis, we try to use chemical reactions to determine the composition of an unknown substance. The addition of certain reagents to the unknown solution gives certain results that show the presence or absence of certain species from the unknown sample.

When dilute HCl is added to an unknown sample and effervescence is observed, then the unknown sample must contain CO3^2- or HCO3^-. The presence of these species is confirmed if the gas evolved is passed through limewater and the gas turns limewater milky.

Answer:

All chemical reactions involve energy. Energy is used to break bonds in reactants, and energy is released when new bonds form in products. Endothermic reactions absorb energy, and exothermic reactions release energy. The law of conservation of energy states that matter cannot be created or destroyed.

Answer:

4.46 mol of NH3

Explanation:

The equation of he reaction is given as;

2N + 3H2 --> 2NH3

From the stochiometry of the reaction, 1 mol of Nitrogen produces 2 mol of Ammonia.

Mass of Nitrogen = 31.2g

Molar mass of Nitrogen = 14g/mol

Number of moles = Mass / Molar mass = 31.2 / 14 = 2.23 mol

Since 1 mol of N = 2 mol of NH3;

2.23 mol of N2 would produce x

x = 2.23 * 2 = 4.46 mol of NH3

Answer:

condensation

Explanation:

What led to the formation of oceans after the water on the Earth's surface evaporated is a condensation reaction of water vapour.

The water present on the surface of the earth was able to evaporate as a result of the hot condition of the primitive earth. As the earth cools down the water vapour present in the atmosphere began to condense, gradually forming puddles of water and eventually leading to the formation of various oceans as we currently have it on the earth.

Answer:

as earth cooled, water in the atmosphere condensed.

Explanation:

Answer:

\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}  

Explanation:

0.030 cm³ × ? = x m³

You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.

For example, you know that centi means "× 10⁻²", so  

1 cm = 10⁻² m

If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).

If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.

So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.

We choose the former because it has the desired units on top.

The "cm" is cubed, so we must cube the conversion factor.

The calculation becomes

[tex]\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 0.30 \times 10^{-6}\text{ m}^{3} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}\\\\\textbf{0.30 cm}^{\mathbf{3}} \times \left (\dfrac{\mathbf{10^{-2}}\textbf{ m}}{\textbf{1 cm}}\right )^{\mathbf{3}} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}[/tex]

Answer:

Explanation:

Given that:

The argon atoms are excited into an excited state before emitting the 488.0 nm laser.

the energy of the first ionization energy of argon is 1520 kJ mol-1.

SInce 1 eV = 96.49 kJ/mol

Therefore, the energy of the first ionization energy of argon in eV is = ( 1520/ 96.49) eV

=  15.75 eV

To find where  the energy level of the excited state lies below the vacuum energy level, let's first determine, the energy liberated by using planck expression.

[tex]E = \dfrac{hc}{\lambda}[/tex]

[tex]E = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}[/tex]

[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]

[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]

[tex]E =4.057 \times 10^{-19} \ J[/tex]

Converting Joules (J) to eV ; we get,

[tex]E =\dfrac{4.057 \times 10^{-19}}{1.6 \times 10^{-19}}[/tex]

E = 2.53 eV

The energy levels of the first exited state = -13.223 eV

The question is incomplete,the complete question is as follows:

A piece of solid Fe metal is put into an aqueous solution of Cu(NO3)2. Write the net ionic equation for any single-replacement redox reaction that may be predicted. Assume that the oxidation state of in the resulted solution is 2 . (Use the lowest possible coefficients for the reaction. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank. If no reaction occurs, leave all boxes blank and click on Submit.)

Answer:

Fe(s) + Cu^2+(aq) => Fe^2+(aq) + Cu(s)

Explanation:

An ionic equation is a chemical equation which shows clear image of reactions of the electrolytes in aqueous solution.

Molecular reaction equation for the reaction between iron and copper II nitrate is as follows:

Fe(s) + Cu(NO3)2(aq) => Fe(NO3)2(aq) +Cu(s)

The net ionic equation for any single-replacement redox reaction is as follows:

Fe(s) + Cu^2+(aq) => Fe^2+(aq) + Cu(s)

Answer:

an increase in 1-butene was observed when t-butoxide was used

Explanation:

When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.

Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.

The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.

The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.

Answer:

(a) [tex]M=0.257M[/tex]

(b) [tex]M=0.0510M[/tex]

(c) [tex]M =0.500M[/tex]

(d) [tex]M= 0.391M[/tex]

Explanation:

Hello,

In this case, since the molarity or molar concentration of a solution is computed by dividing the moles of solute by the volume of solution in liters, we proceed as follows:

(a) The molar mass of sodium chloride is 58.45 g/mol and the volume in liters is 0.100 L, therefore, the molarity is:

[tex]M=\frac{1.50gNaCl}{0.100L} *\frac{1molNaCl}{58.45gNaCl} =0.257M[/tex]

(b) The molar of potassium dichromate is 294.2 g/mol and the volume in liters is 0.100 L, therefore, the molarity is:

[tex]M=\frac{1.50gK_2Cr_2O_7}{0.100L} *\frac{1molK_2Cr_2O_7}{294.2gK_2Cr_2O_7} =0.0510M[/tex]

(c) The molar of calcium chloride is 111 g/mol and the volume in liters is 0.125 L, therefore, the molarity is:

[tex]M=\frac{5.55gCaCl_2}{0.100L} *\frac{1molCaCl_2}{111gCaCl_2} =0.500M[/tex]

(d) The molar of sodium sulfate is 142 g/mol and the volume in liters is 0.125 L, therefore, the molarity is:

[tex]M=\frac{5.55gNa_2SO_4}{0.100L} *\frac{1molNa_2SO_4}{142gNa_2SO_4} = 0.391M[/tex]

Best regards.

Answer:

105/22*(1.251-0.620)

105/22*0.631

4.772*0.631

3.011132

Hope it helps

Answer:

3.0

Explanation:

First, complete the operations inside the parenthesis according to the normal rules for significant figures. Because there are subsequent calculations, keep at least one extra significant figure when possible: (4.7727) × (0.631).

The final product will be rounded to two significant figures because it can’t be more precise than the least precise number in the problem, 22. The final product is 3.0.

Answer:

123.96 g Na₂O

Explanation:

4 Na  +  O₂  ⇒  2 Na₂O

You first need to find the limiting reagent.  Convert the reactants to moles and see which produces the least amount of product using the mole ratios in the chemical equation.

(91.96 g Na)/(22.99 g/mol Na) = 4 mol Na

(4 mol Na) × (2 mol Na₂O/4 mol Na) = 2 mol Na₂O

(32.0 g O₂)/(32.0 g/mol) = 1 mol O₂

(1 mol O₂) × (2 mol Na₂O/1 mol O₂) = 2 mol Na₂O

Since they both produce the same amount of product, you don't need to pick a limiting reagent.  Now, convert moles of Na₂O to grams.

(2 mol Na₂O) × (61.98 g/mol Na₂O) = 123.96 g Na₂O