When running an experiment, it is essential to record data and observations within your lab notebook. The data and observations are then discussed and analyzed, results and discussion. However, as you write a conclusion, some data are more important to include than others. Which of the following data would you include if you had to write a conclusion in for outcome of this week's LLE experiment?
a. MP range of product
b. mass of product
c. mass of flask and cork support
d. instrument used to evaporate the solvent
e. volume of crude reaction mixture used
f. % recovery
g. volume of acid and base used
h. theoretical MP
i. mass of flask, cork support, and product

Answer:

1.) Yes, Cs₂S

2.) Yes, MgF₂

3.) Yes, MgS

4.) No

Explanation:

Nonmetals are located on the right side of the periodic table in the p-block. It is safe to assume that all the other elements are metals.

Ionic bonds form between a metal and nonmetal. When determining the empirical formula, the cation (the element losing electrons) is written before the anion (the element gaining electrons). The charges can be balanced by modifying the quantity of element.

1.) Cs₂S

  -----> Cesium = metal and Sulfur = nonmetal

  -----> Cesium = Cs⁺

  -----> Sulfur = S²⁻

2.) MgF₂

  -----> Magnesium = metal and Fluorine = nonmetal

  -----> Magnesium = Mg²⁺

  -----> Fluorine = F⁻

3.) MgS

  -----> Magnesium = metal and Sulfur = nonmetal

  -----> Magnesium = Mg²⁺

  -----> Sulfur = S²⁻

4.) Both cesium and strontium are metals

The freezing point of the sucrose solution is -0.435°C.

The freezing point of the solution is determined from the freezing point depression formula below:

Kf(H₂O) = 1.86 Cm

m is molality of solution = moles of solute/mass of solvent

moles of sucrose = 8.0/342.3 = 0.0233 moles

m = 0.0233/0.1 = 0.233 molal

ΔT = 0.233 m * 1.86°C/m.

ΔT = 0.435 °C.

Freezing point of sucrose solution = 0°C - 0.435°C

Freezing point of sucrose solution  = -0.435°C.

In conclusion, the freezing point of sucrose solution is determined from the freezing point depression.

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Answer:

Potassium chloride.

Answer:

11.9 moles Cl₂

Explanation:

To find the number of moles, you need to use the Ideal Gas Law. The equation looks like this:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = constant (0.0821 L*atm/mol*K)

-----> T = temperature (K)

Before you can plug the given values into the equation, you first need to convert Celsius to Kelvin.

P = 33.3 atm                         R = 0.0821 L*atm/mol*K

V = 11.5 L                              T = 120. °C + 273.15 = 393.15 K

n = ? moles

PV = nRT

(33.3 atm)(11.5 L) = n(0.0821 L*atm/mol*K)(393.15 K)

382.95 = n(0.0821 L*atm/mol*K)(393.15 K)

382.95 = (32.2776)n

11.9 = n

Answer:

O Kinetic energy does not change, but the potential energy does

Taking into account the reaction stoichiometry, 340.0 moles  of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

In first place, the balanced reaction is:

CO₂ + 4 H₄  → CH₄ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?

[tex]moles of CH_{4} =\frac{85.1 moles of CO_{2}x4 moles of CH_{4} }{1 moles of CO_{2}}[/tex]

moles of CH₄= 340.4 moles

Then, 340.0 moles  of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

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Answer:

a) Ca (s) + 2 H₂O (l) ---> Ca(OH)₂ (aq) + H₂ (g)

b) Al₂O₃ (s) + 3 C (s) + 3 Cl₂ (g) --->  2 AICI₃ (s) + 3 CO (g)

c) 2 FeCl₃ (aq) + 3 H₂S (g) ---> Fe₂S₃ (s) + 6 HCI (aq)

d) CaCO₃ (s) + 2 HCI (aq) ---> CaCl₂ (aq) + CO₂ (g) + H₂O (l)

Explanation:

For an equation to be balanced, there must be an equal amount of each type of atom on both sides of the reaction. The reactants are on the left side and the products are on the right side.

a) ------------------------------------------------------------------------------------------------------

The unbalanced equation:

Ca (s) + H₂O (l) ---> Ca(OH)₂ (aq) + H₂ (g)

Reactants: 1 calcium, 2 hydrogen, 1 oxygen

Products: 1 calcium, 4 hydrogen, 2 oxygen

The balanced equation:

Ca (s) + 2 H₂O (l) ---> Ca(OH)₂ (aq) + H₂ (g)

Reactants: 1 calcium, 4 hydrogen, 2 oxygen

Products: 1 calcium, 4 hydrogen, 2 oxygen

b) ------------------------------------------------------------------------------------------------------

The unbalanced equation:

Al₂O₃ (s) + C (s) + Cl₂ (g) --->  AICI₃ (s) + CO (g)

Reactants: 2 aluminum, 3 oxygen, 1 carbon, 2 chlorine

Products: 1 aluminum, 1 oxygen, 1 carbon, 3 chlorine

The balanced equation:

Al₂O₃ (s) + 3 C (s) + 3 Cl₂ (g) --->  2 AICI₃ (s) + 3 CO (g)

Reactants: 2 aluminum, 3 oxygen, 3 carbon, 6 chlorine

Products: 2 aluminum, 3 oxygen, 3 carbon, 6 chlorine

c) ------------------------------------------------------------------------------------------------------

The unbalanced equation:

FeCl₃ (aq) + H₂S (g) ---> Fe₂S₃ (s) + HCI (aq)

Reactants: 1 iron, 3 chlorine, 2 hydrogen, 1 sulfur

Products: 2 iron, 1 chlorine, 1 hydrogen, 3 sulfur

The balanced equation:

2 FeCl₃ (aq) + 3 H₂S (g) ---> Fe₂S₃ (s) + 6 HCI (aq)

Reactants: 2 iron, 6 chlorine, 6 hydrogen, 3 sulfur

Products: 2 iron, 6 chlorine, 6 chlorine, 3 sulfur

d) ------------------------------------------------------------------------------------------------------

The unbalanced equation:

CaCO₃ (s) + HCI (aq) ---> CaCl₂ (aq) + CO₂ (g) + H₂O (l)

Reactants: 1 calcium, 1 carbon, 3 oxygen, 1 hydrogen, 1 chlorine

Products: 1 calcium, 1 carbon, 3 oxygen, 2 hydrogen, 2 chlorine

The balanced equation:

CaCO₃ (s) + 2 HCI (aq) ---> CaCl₂ (aq) + CO₂ (g) + H₂O (l)

Reactants: 1 calcium, 1 carbon, 3 oxygen, 2 hydrogen, 2 chlorine

Products: 1 calcium, 1 carbon, 3 oxygen, 2 hydrogen, 2 chlorine

The periodic trend is very useful because it helps is understand the properties and reactivities of elements in chemical reactions

Some examples of periodic trends are:

Ionization energy can simply be defined as that energy required to remove a valence electron from an atom of an element in a gaseous state to form a gaseous ions l

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The theoretical yield of the CO2 produced is 4.22 g.

The balanced reaction equation is;

2C2H3Cl + 5O2 ------>2CO2 + 2H2O + 2HCl

Number of moles of  vinyl chloride =  9.10g /62 g/mol = 0.15 moles

Number of moles of oxygen = 7.80g/32 g/mol = 0.24 moles

Now;

2 moles of C2H3Cl reacts with 5 moles of oxygen

0.15 moles of C2H3Cl reacts with 0.15 moles * 5 moles / 2 moles

= 0.375 moles

Hence oxygen is the limiting reactant;

5 moles of oxygen produces 2 moles of CO2

0.24 moles of oxygen produces 0.24 moles * 2 moles /5 moles

= 0.096 moles

Mass of CO2 produced = 0.096 moles * 44 g/mol

= 4.22 g

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The density of the solid is =0.3076g/ml

The mass of solid in flask = 36.65g

The volume of both solid and Toluene = 50ml

The mass of both solid and Toluene = 58.58g

The density of toluene = 0.864g/ml

The density of the solid = X

Note: The density of the solid is calculated from the density of both the solid and toluene.

But the density of solid+ toluene;

=mass/volume

= 58.58g/50ml

= 1.1716g/ml

Therefore the density of solid alone = density of both solid +toluene - density of toluene.

= 1.1716g/ml - 0.864g/ml

= 0.3076g/ml

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Answer: 63.75 %

Explanation:

The balanced chemical equation for this reaction is:

NaHCO3 + HA = NaA + H2O + CO2

According to the chemical equation, one mole of NaHCO3 will react with one mole of HA to form one mole of NaA, H2O and CO2. All the components are in same mole ratio of 1:1:1:1

So,

Moles of CO2 = weight/molar mass = 0.561/44 = 0.01275 moles

Moles of CO2 will be equal to moles of NaHCO3 = 0.01275 moles

Moles = weight/mass

Weight = moles × mass

So the weight of NaHCO3 = 0.01275 × 84.01 = 1.071 g

The percentage mass of NaHCO3 in the original mixture =(1.071/1.68)× 100% = 63.75 %

Therefore, the percent by mass of NaHCO3 in the original mixture was 63.75 %

I think it should be suspension.

Answer:

H₂(g) = oxidized

N₂(g) = reduced

Explanation:

Oxidized substances lose electrons and increase in oxidation number. Reduced substances gain electrons and decrease in oxidation number.

Lone elements always have an oxidation number of 0. So, N₂(g) and H₂(g) have oxidation numbers of 0.

Within compounds, hydrogen generally always have an oxidation of +1. Therefore, the hydrogen in NH₃(g) has an oxidation number of +1. In order to make the overall compound neutral, the nitrogen in NH₃(g) must have an oxidation number of -3 (to balance the +3 from the 3 hydrogens).

Since H₂(s) is losing electrons (goes from 0 to +1), it is being oxidized. Since N₂(g) is gaining electrons (goes from 0 to -3), it is being reduced.

Answer:

49.8 g C₆H₅Br

Explanation:

The theoretical yield is the maximum calculated amount of product that can be made from a reaction. To find the theoretical yield, you need to (1) determine the chemical reaction which is taking place, then (2) calculate the molar masses of the relevant reactants and product, then (3) convert grams of each reactant to grams of bromobenzene (using molar masses and mole-to-mole ratio from reaction coefficients), and then (4) determine the limiting reagent and the theoretical yield.

(Step 1)

Bromine = Br₂

Benzene = C₆H₆

Bromobenzene = C₆H₅Br

The balanced equation:

Br₂ + 2 C₆H₆ ----> 2 C₆H₅Br + H₂

(Step 2)

Molar Mass (Br₂): 2(79.904 g/mol)

Molar Mass (Br₂): 159.808 g/mol

Molar Mass (C₆H₆): 6(12.011 g/mol) + 6(1.008 g/mol)

Molar Mass (C₆H₆): 78.114 g/mol

Molar Mass (C₆H₅Br): 6(12.011 g/mol) + 5(1.008 g/mol) + 79.904 g/mol

Molar Mass (C₆H₅Br): 157.01 g/mol

(Step 3)

52.3 g Br₂           1 mole             2 moles C₆H₅Br         157.01 g
----------------- x  ------------------  x  -------------------------  x  ---------------  =
                          159.808 g              1 mole Br₂               1 mole

=  103 g C₆H₅Br

24.8 g C₆H₆         1 mole            2 moles C₆H₅Br          157.01 g
-------------------  x  ---------------  x  --------------------------  x  ---------------  =
                             78.114 g           2 moles C₆H₆             1 mole

=  49.8 g C₆H₅Br

(Step 4)

Because C₆H₆ produces a smaller amount of product than Br₂, it must be the limiting reagent. This means that it will run out before all of the Br₂ is used up. Therefore, the actual theoretical yield of C₆H₅Br is 49.8 grams.

The rate of a reflection decreases when the temperature increases

This goes to say that the rate of reflection will go down more whenever there is an increase in temperature

Reflection can simply be defined as the the throwing back by a body of light without absorbing it.

reflection of

In conclusion, the rate of a reflection decreases when the temperature increases

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The element which is losing mass from the above reaction is Cu

Mg(s) + CuSO4(aq) -> MgSO4(aq) + Cu(s)

A compound simply refers to a chemical compound which contains two contains two or more elements which are chemically combined together.

So therefore, the element which is losing mass from the above reaction is Cu

Mg(s) + CuSO4(aq) -> MgSO4(aq) + Cu(s)

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The mass of melted gold to release the energy would be  3, 688. 8 Kg

It is important to note that the formula for  energy is

Q = n × HF

Where

n = number of moles

HF = heat of fusion

Let's find number of moles

235.0 = n × 12.550

number of moles = [tex]\frac{235}{12. 550}[/tex] = 18. 725 moles

The molar mass of Gold is 197g/ mol

We have that;

Number of moles = mass/ molar mass

Mass = number of moles × molar mass

Substitute the values

Mass = 18. 725 × 197

Mass = 3, 688. 8 Kg

Thus, the mass of melted gold to release the energy would be  3, 688. 8 Kg

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The true statements are:

Energy is the ability to do work.

Energy can exist in many form, two of which are kinetic and potential energy.

The worker is holding a filled gas cylinder still gives the gas potential energy by lifting it above the ground.

The molecules of the gas also has kinetic energy by virtue of their constant random motion resulting in thermal energy.

In conclusion, the energy of the filled gas is both potential and kinetic.

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The following applies to the above equation:

Brønsted-Lowry acid is the any chemical species that acts as a donor of protons while Brønsted-Lowry base is any chemical species that acts as a proton acceptor.

In the following equation: HCOOH(aq) + PO43-(aq) = HCOO-(aq) + HPO42-(aq)

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Answer:

4.30 moles

Explanation:

You need to use the Ideal Gas Law. The equation looks like this:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas Constant (0.08206 atm*L/mol*K)

-----> T = temperature (K)

Before you can plug the given values into the equation and simplify to find "n", you need to convert Celsius to Kelvin. The final answer should have 3 sig figs to match the sig figs of the given values.

P = 3.85 atm                        R = 0.08206 atm*L/mol*K

V = 32.4 L                            T = 80.4 °C + 273.15 = 353.55 K

n = ? moles

PV = nRT

(3.85 atm)(32.4 L) = n(0.08206 atm*L/mol*K)(353.55 K)

124.74 = n(29.012313)

4.30 = n

1. The resulting concentration will be 0.00044 mol/L

2. The minimum mass of sodium sulfite to add will be 0.4032 grams.

1. Using m1v1=m2v2

  m1 = 0.01 mol/L, v1 = 20 mL, v2 = 450 mL

       m2 = m1v1/v2 = 0.01 x 20/450 = 0.00044 mol/L

2. [tex]Na_2SO_3 + Ca(NO_3)_2 --- > 2NaNO_3 + CaSO_3[/tex]

Mole ratio of the reactants = 1:1

Mole of 80 mL, 0.0400 mol/L Ca(NO3)2 = 80/1000 x 0.0400 = 0.0032 mol

Equivalent mole of Na2SO3 = 0.0032 moles

Mass of 0.0032 moles Na2SO3 = 0.0032 x 126 = 0.4032 grams

Thus, the minimum mass of sodium sulfite to be added must be 0.4032 grams.

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Answer:

163 L

Explanation:

To find the volume, you need to use the Ideal Gas Law. The equation looks like this:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = constant (0.0821 L*atm/mol*K)

-----> T = temperature (K)

Before you can plug the given values into the equation, you first need to convert Celsius to Kelvin.

P = 3.50 atm                         R = 0.0821 L*atm/mol*K

V = ? L                                   T = 75.3 °C + 273.15 = 348.45 K

n = 20.0 moles

PV = nRT

(3.50 atm)V = (20.0 moles)(0.0821 L*atm/mol*K)(348.45 K)

(3.50 atm)V = 572.1549

V = 163 L

The number of moles of carbon dioxide produced will be 8.6 moles.

First, let's look at the balanced equation of the reaction. The complete combustion of ethanol gives carbon dioxide and water as follows:

[tex]C_2H_5OH + 3O_2 --- > 2CO_2 + 3H_2O[/tex]

We can see from the equation that the burning of 1 mole of ethanol produces 2 moles of carbon dioxide. In other words, the mole ratio is 1:2.

Thus, with 4.3 moles of ethanol, the amount, in moles, of carbon dioxide that will be produced will be:

                            4.3 x 2 = 8.6 moles

In other words, 8.6 moles of carbon dioxide will be produced from the complete combustion of 4.3 moles of ethanol.

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Reasoning includes one or more scientific principles in this context.

This involves the thought about something in a sensible and logical manner due to one or more scientific principles being applied.

It is very important to the claim and evidence as it helps ascertain their validity in science.

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The labelled diagram of Daniell cell is as shown below

The half-reactions of oxidation and reduction taking place on electrodes are shown:

Oxidation at anode: Zn(s)→Zn 2+(aq)+2e −

Reduction at cathode: Cu 2+ (aq)+2e − →Cu(s)

A Daniell cell is the best model of a galvanic cell which converts chemical energy into electrical energy. The Daniell cell consists of two electrodes of different metals, Zn and Cu; each electrode is in communication with a solution of its own ion; Zinc sulphate and copper sulphate respectively.

The key difference between Daniell cell and galvanic cell is that Daniell cell operates only copper and zinc as electrodes whereas a galvanic cell can have a mixture of metals as electrodes.

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The second reaction with the volume of 5L will occur faster as compared to the first reaction of volume 10L.

Volume is inversely proportional to the rate of reaction.

As volume increases rate of reaction decreases and as volume decreases rate of reaction increases.

Let's consider a reaction A   →  B.

r is rate of reaction and K is rate constant, A is the concentration of reaction.

r = k(A)

r is directly proportional to the A.

But the concentration of A is in Moles/liters (i.e. moles per unit volume)

r is inversely proportional to the volume of vessel or chamber.

 Hence as volume increases rate of reaction decreases.

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Answer:

Techniques and Tests

Qualitative analysis typically measures changes in color, melting point, odor, reactivity, radioactivity, boiling point, bubble production, and precipitation.