9) Describe a method that could be used to extract lead from lead oxide (11). You will need
to include charcoal, an ignition (or test) tube and a Bunsen burner. Make sure
you include
each step and what safety measures you would take?

Answers

Answer 1

Answer:

Explained below.

Explanation:

In order to extract lead from lead oxide, we need to add carbon from coke to the lead oxide.

We will need;

A bunsen burner

A test tube

Charcoal

First step is to heat charcoal in an oxygen free environment. The safety measure here to ensure an oxygen free environment is to add some zinc metal powder into the test tube in order to remove any atmospheric oxygen inside the container.

Secondly, we will now put charcoal inside the test tube and turn on the bunsen burner to heat it until coke is formed.

Thirdly, we now introduce lead oxide into the coke inside the test tube and the carbon in the coke reacts with the lead oxide to produce lead and carbon dioxide.

Throughout all this process, since we required an oxygen free environment inside the test tube, whenever the test tube is opened, its neck must immediately be warmed by heat with the tube placed in a horizontal direction or very close to being in that horizontal direction so that any movement of air moves outwards from the tube.


Related Questions

What is the first step in the curved arrow mechanism for the chlorination of benzene in the presence of FeCl3

Answers

Answer:

See explanation

Explanation:

The chlorination of benzene occurs in the presence of a Lewis acid. A Lewis acid is a compound that can accept a lone pair of electrons.

The first step in the chlorination of benzene is the formation of the ion Cl^+ which attacks the benzene ring.

This ion is formed when the Cl2 molecule undergoes heterolytic fission assisted by FeCl3 to yield FeCl4^- and Cl^+.

The Cl^+ electrophile now attacks the benzene ring to yield chlorobenzene.

Which equation was used by Albert Einstein to explain the photoelectric effect? [E = energy, h= planck's constan, and v = frequency]

Answers

Answer:

E = hv

Explanation:

Energy = planck constant × frequency

9. Which property is not important when selecting a material to use as a light bulb filament?
d
high melting point
high vapor pressure
high ductility

Answers

Answer:

high melting point

Explanation:

The filament of a bulb is often heated to very high temperatures as the bulb is in operation.

Many times, electric bulbs may have to be on for a whole day and they may reach temperatures that are outrageously high in the process.

The material of the filament must have a very high melting point so that it doesn't melt while the bulb is still in operation.

Answer:

Actually the answer is High Vapor pressure

Explanation:

The energy needed to collapse the H-bonding of ice is _____.

Answers

it depends on the widths of the recesses, and if it is causing the shelf to fracture and collapse into the sea, then a massive iceberg could be called from the life she,f and the ice shelf are way more important because it holds it up

For an ideal gas, classify the pairs of properties as directly or inversely proportional. You are currently in a sorting module. a. P and n b. V and n c. P and T d. T and V e. P and V1. directly proportional2. inversely proportional

Answers

Answer:

the result for the following are (a) P is directly proportional to n

(b) V is directly proportional to T (c) P is directly proportional to T (d) T is inversly proportional to V (e) P is inversely proportional to V

A 1.5-m 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa. Now paddle-wheel work is done on the system until the pressure in the tank rises to 150 kPa. Determine the entropy change of carbon dioxide during this process. Assume constant specific heats

Answers

Answer:

The entropy change of carbon dioxide = 0.719 kJ/k

Explanation:

Given:

1.5 m - 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa

The objective is to determine the entropy change of carbon dioxide

Formula used:

ΔS=

Solution:

On considering,

[tex]C_{P} =0.846 kJ/kg K\\C_V=0.657 kJ/kg k\\[/tex]

ΔS=[tex]mc_{v} lu\frac{p_{2} }{P_{1} }[/tex]

On substituting the values,

ΔS=[tex]2.7*0.657lu\frac{150}{100}[/tex]

ΔS=0.719 kJ/k

The entropy change is "0.719 kJ/K".

Given values are:

Mass of tank,

m = 2.7 kg

Pressure,

P₁ = 100 kPa

Rised pressure,

P₂ = 150 kPa

Assumption of constant specific heat is,

[tex]C_v = 0.657 \ kJ/kgK[/tex]

As we know the formula,

→ [tex]\Delta S = mC_v \ ln(\frac{P_2}{P_1} )[/tex]

         [tex]= (2.7)(0.657) \ ln (\frac{150}{100} )[/tex]

         [tex]= 1.7739\times 0.4055[/tex]

         [tex]= 0.7193 \ kJ/K[/tex]

Thus above answer is right.

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If I have 25g of Sodium, how much Sodium Chloride will I theoretically create?
O 50g NaCl
0 58.3g NaCl
O 63.7g Naci
0 35.4g NaCl

Answers

Answer:

64 g

Explanation:

Step 1: Write the balanced equation

2 Na + Cl₂ ⇒ 2 NaCl

Step 2: Calculate the moles corresponding to 25 g of Na

The molar mass of Na is 22.98 g/mol.

25 g × 1 mol/22.98 g = 1.1 mol

Step 3: Calculate the moles of NaCl formed from 1.1 moles of Na

The molar ratio of Na to NaCl is 2:2. The moles of NaCl formed are 2/2 × 1.1 mol = 1.1 mol.

Step 4: Calculate the mass corresponding to 1.1 moles of NaCl

The molar mass of NaCl is 58.44 g/mol.

1.1 mol × 58.44 g/mol = 64 g

Question 1
1 pts
How many mols of bromine are present in 35.7g of
Tin(IV) bromate?

Answers

Answer:

n = 0.0814 mol

Explanation:

Given mass, m = 35.7g

The molar mass of Tin(IV) bromate, M = 438.33 g/mol

We need to find the number of moles of bromine. We know that,

No. of moles = given mass/molar mass

So,

[tex]n=\dfrac{35.7}{438.33}\\\\n=0.0814\ mol[/tex]

So, there are 0.0814 moles of bromine in 35.7g of  Tin(IV) bromate.

Which safety feature works to slow down nuclear-fission chain reactions?
A. Proliferation
B. Control rods
C. Cooling rods
D. Fuel rods

Answers

Answer:

The answer is

-control rods

Answer:

B. Control Rods.... via A P E X

What should be done if a spectrophotometer reports an absorbance that is too high? Select one: Reposition the cuvette in the spectrophotometer. Pour out half the volume of the sample. Restart the spectrophotometer and try again. Dilute the sample. g

Answers

Answer:

The sample should be diluted

Explanation:

According to Beer Lambert's law, the absorbance of a sample depends on the concentration of the sample.

Hence, if the concentration of the sample is very high, the spectrophotometer will also report a very high value of absorbance.

When this is the case, the sample should simply be diluted and the readings are taken again using the spectrophotometer.

In an exothermic reaction, reactants are __ products.
a more stable than
b equal to
c equally stable as
d less stable than

Answers

Answer:

Less stable

Explanation:

When an exothermic reaction occurs, the reactants have a greater energy level than the outputs. The products, in other respects, are more stable than that of the reactants.

The outcomes of an exothermic reaction have a lower energy to react. The enthalpy of a process is the difference between some of the activation energy and the power of the products. 

For the molecular compound dioxygen difluoride , what would you multiply "grams of O2F2 " by to get the units "molecules of O2F2 "

Answers

Answer:

Molecules of O₂F₂ = mass of O₂F₂ × (1 mole O₂F₂ / 70 g O₂F₂) × (6.02 × 10²³ molecules / one mole of O₂F₂)

Explanation:

The Avogadros constant gives the the number of specified entities in one mole of a substance. One mole of any substance contains 6.02 × 10²³ particles. Therefore, one mole of O₂F₂ contains 6.02 × 10²³ molecules.

Also, the molar mass of a substance is the mass in grams of one mole of that substance. It is obtained by summing the relative atomic masses of all the atoms of the elements in the substance. For O₂F₂, the molar mass = (2 × 16 + 2 × 19) g/mol = 70 g/mol

Converting to molecules of O₂F₂;

To convert from grams of a substance to molecules of that substance, multiply by the ratio of one mole and mass of one mole, and then by the number of molecules per mole.

Molecules of A = mass of A × (1 mole / mass of one mole) × (6.02 × 10²³ molecules / 1 mole)

Therefore,Molecules of O₂F₂ = mass of O₂F₂ × (1 mole O₂F₂ / 70 g O₂F₂) × (6.02 × 10²³ molecules /one mole of O₂F₂)

Calculate the moles of H3PO4 that reacted (8). Consult the coefficients in the balance chemical reaction to obtain the mole ratio. Show your calculation here.

Answers

Answer:

The number of moles of H₃PO₄ that reacted is 0.000343 moles

Note: Some data is missing. Data from the attachment is used for the calculationsinnthe explanation below.

Explanation:

The reaction is a neutralization reaction between NaOH and H₃PO₄. The equation of the reaction is given as follows:

3 NaOH + H₃PO₄ ---> Na₃PO₄ + 3 H₂O

The molarity of the NaOH solution is 0.238 mol/L.

Average volume of NaOH used during the titration to arrive to endpoint = (4.6 + 3.9 + 4.5) mL / 3 = 4.33 mL

Molarity is defined ratio of the number of moles of solute to the volume of solution. Mathematically, molarity = number of moles/volume in Litres

Number of moles of NaOH reacted = 0.238 mol/L × (4.33mL × 1 L/1000 mL)

Number of moles of NaOH = 0.00103 moles

From the equation of the reaction, 3 moles of NaOH reacts with 1 mole of H₃PO₄

0.00103 moles of NaOH will react with 0.00103 x 1/3 moles of H₃PO₄ = 0.000343 moles of H₃PO₄.

Therefore, number of moles of H₃PO₄ that reacted is 0.000343 moles

what does LPG stand for? mention one important source of LPG give sort answer​

Answers

Answer:

liquefied petroleum gas

LPG is prepared by refining natural gas. it is made by refining crude oil or from extracted natural gas streams as they emerge from the ground.

What is the law of multiple proportions?
A. The proportion of elements to compounds is constant.
B. All elements are found in equal proportions in nature.
C. Different compounds may contain the same elements but may have different ratios of those elements.
D. All compounds contain the same elements in the same proportions.

Answers

A I think I may be wrong

determine the budget for tge fitness event.consider the attendance (should tou ask for registration fee.give shirts etc)

Answers

Notch gfi.hu,ivjmgfcmhtxumgfhkvHbyrdyu,

A 46.6-mgmg sample of boron reacts with oxygen to form 150 mgmg of the compound boron oxide. Part A What is the empirical formula of boron oxide

Answers

Answer:

B₂O₃

Explanation:

Step 1: Calculate the mass of oxygen in 150 mg of boron oxide

Of 150 mg of boron oxide, 46.6 mg belong to boron. The mass of oxygen is:

150 mg - 46.6 mg = 103.4 mg

Step 2: Calculate the percent by mass of each element

We will use the following expression.

%Element = mElement/mCompound × 100%

%B = 46.6 mg/150 mg × 100% = 31.1%

%O = 103.4 mg/150 mg × 100% = 68.9%

Step 3: Divide each percentage by the atomic mass of the element

B: 31.1/10.81 = 2.88

O: 68.9/16.00 = 4.31

Step 4: Divide both numbers by the smallest one (2.88)

B: 2.88/2.88 = 1

O: 4.31/2.88 ≈ 1.5

Step 5: Multiply both numbers by 2 so that they are integers

B: 1 × 2 = 2

O: 1.5 × 2 = 3

The empirical formula is B₂O₃.

Underneath the ocean floor, methane can exist as a solid known as methane hydrate. Methane hydrate forms when large amounts of methane become trapped in the crystal structure of water, which develops into a solid similar to ice. A scientist working for a deep water drilling company is studying the properties of the three different phases of methane: methane hydrate, liquified natural gas, and natural gas. Which of the properties is true

Answers

Answer:

Methane is present in solid, liquid and gaseous form.

Explanation:

Methane hydrate is present in solid state when the hydrogen-bonded water and methane gas come into contact at high pressures and low temperatures in the deep oceans while on the other hand, methane which is present inside the earth surface in gaseous form due to non-availability of water that can combine with methane. Methane is also found in liquid form when it is cool with high pressure and low temperature.

pls help ive been stuck on this question for a while im not good with chemistry lol.

Answers

Answer:

The answer would be B, putting thermal energy into something means you're adding heat into it.

Energy Changes in Reactions
Independent Practice
In a synthesis reaction, one reactant contains 256 J of chemical energy, and one reactant contains 328 J of
chemical energy. The product contains 407 J of chemical energy.
Because energy is conserved, what energy change occurs during the reaction?
A. 177 J of energy is released.
B. 177 J of energy is absorbed.
C. 335 J of energy is released,
D. 335 J of energy is absorbed.
SAVE
SAVE AND CLOSE

Answers

It would be 335 J and of energy is released

What would be the specific mathematical effect on the reaction rate if you carried out the sodium iodide-in-acetone reactions on the alkyl halides using an iodide solution half as concentrated? ("Slower" or "faster" is not specific enough.)

Answers

Answer:

Slower

Explanation:

The reaction between alkyl halides and sodium iodide-in-acetone is an SN2 reaction. The rate of reaction depends on the concentration of the alkyl halide as well as the concentration of the sodium iodide. It is a bimolecular reaction.

This means that if the concentration of any of the reactants is halved, the rate of reaction decreases accordingly.

Therefore, if the iodide solution is half as concentrated, the reaction is observed to be slower in accordance with the rate law;

Rate = k[alkyl halide] [iodide]

If 6.50 L of water vapor at 50.2 °C and 0.121 atm reacts with excess iron, how many grams of iron(III) oxide will be produced?

2Fe(s)+3H2O(g)⟶Fe2O3(s)+3H2(g)

Answers

Answer:

1.60 g of Fe₂O₃

Explanation:

We'll begin by calculating the number of mole water that reacted. This can be obtained as follow:

Volume (V) = 6.50 L

Temperature (T) = 50.2 °C = 50.2 + 273 = 323.2 K

Pressure (P) = 0.121 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =?

PV = nRT

0.121 × 6.5 = n × 0.0821 × 323.2

0.7865 = n × 26.53472

Divide both side by 26.53472

n = 0.7865 / 26.53472

n = 0.03 mole

Thus, 0.03 mole of water reacted.

Next, we shall determine the number of mole of Fe₂O₃ produced from the reaction. This can be obtained as follow:

2Fe + 3H₂O —> Fe₂O₃ + 3H₂

From the balanced equation above,

3 moles of H₂O reacted to produce 1 mole Fe₂O₃.

Therefore, 0.03 mole of H₂O will react to produce = (0.03 × 1)/3 = 0.01 mole of Fe₂O₃.

Thus, 0.01 mole of Fe₂O₃ was produced from the reaction.

Finally, we shall determine the mass of 0.01 mole of Fe₂O₃. This can be obtained as follow:

Mole of Fe₂O₃ = 0.01 mole

Molar mass of Fe₂O₃ = (56×2) + (16×3)

= 112 + 48

= 160 g/mol

Mass of Fe₂O₃ =?

Mass = mole × molar mass

Mass of Fe₂O₃ = 0.01 × 160

Mass of Fe₂O₃ = 1.60 g

Therefore, 1.60 g of Fe₂O₃ were produced.

Ethylene glycol flows at 0.01 kg/s through a 3-mm diameter, thin-walled tube. The tube is coiled and submerged in well-stirred water bath maintained at 25°C. If the fluid enters the tube at 85°C, what heat rate and tube length are required for the fluid to leave at 35°C?

Answers

Answer:

heat rate= 1281W

length = 15.8m

Explanation:

we have this data to answer this question with

Tmi = 85 degrees

Tmo = 35 degrees

Ts = 25 dgrees

flow rate = 25 degrees

using engine oil property from table a-5

Tm = Tmo - TMi/2 = 333k

u =0.522x10⁻²

k = 0.26

pr = 51.3

cp = 2562 J/kg.k

mcp(Tmo-Tmi) =

0.01 x 2562(35-85)

= 1281 W

we find the change in Tim

= [(35-25)-(85-25)]/ln[(35-25)/(85-25)]

= -50/ln0.167

= -50/-1.78976

= 27.9°c

we finf the required reynold number

4x0.01/πx0.003x0.522x10⁻²

= 0.04/0.00004921

= 812.8

= 813

we find approximate correlation

NuD = hd/k

NuD = 3.66

3.66 = 0.003D/0.26

cross multiply

0.003D = 3.66x0.26

D = 3.66x0.26/0.003

= 317.2

As = 1281/317x27.9

= 0.145

As = πDL

L = As/πD

= 0.145/π0.003

= 0.145/0.009429

L = 15.378

Pl hep help help me

Answers

It is true

I’ve done this before

Silver has two naturally occurring isotopes with the following isotopic masses: 10747Ag 10947Ag 106.90509 108.9047 The average atomic mass of silver is 107.8682 amu. The fractional abundance of the lighter of the two isotopes is ________.

Answers

Answer: The fractional abundance of lighter isotope is 0.518

Explanation:

Average atomic weight is the sum of the masses of the individual isotopes each multiplied by its fractional abundance. The equation used is:[tex]\text{Average atomic weight}=\sum_{i=1}^{n}\text{(Atomic mass of isotope)}_i\times \text{(Fractional abundance)}_i[/tex]           ......(1)

Let the fractional abundance of Ag-107 isotope be 'x'

For Ag-107 isotope:

Atomic mass = 106.90509 amu

Fractional abundance = x

For Ag-109 isotope:

Atomic mass = 108.9047 amu

Fractional abundance = (1 - x)

Average atomic mass of silver = 107.8682 amu

Plugging values in equation 1:

[tex]107.8682=(106.90509 \times x) + (108.9047 \times (1-x))\\\\107.8682=106.90509x+108.9047-108.9047x\\\\1.99961x=1.0365\\\\x=0.518[/tex]

Fractional abundance of Ag-107 isotope (lighter) = x = 0.518

Hence, the fractional abundance of lighter isotope is 0.518

When you hold an object with a positive electric charge near an object with a
negative electric charge, it causes the negatively charged object to move. How
could you model this interaction with magnets? Describe the types of forces
involved and the energy transformations that take place.

Answers

Answer:

A positively charged object will exert a repulsive force upon a second positively charged. This repulsive force will push the two objects apart while a negatively charged object will exert a repulsive force upon a second negatively charged object. Objects with like charge repel each other

The interaction between objects with positive and negative electric charges can be analogously modeled using magnets. The Types of Forces Involved are; Attractive Magnetic Force, Repulsive Magnetic Force and the Energy Transformations are; Potential Energy Transformation, and Kinetic Energy Transformation.

In this analogy, magnets can represent the charges, and magnetic forces can represent the electric forces.

Interaction Between Magnets

Imagine we have two magnets: one with a north pole (N) and the other with a south pole (S). When you bring the north pole of one magnet close to the south pole of the other magnet, they are attracted to each other. Conversely, if you bring the north pole of one magnet near the north pole of the other magnet, they repel each other.

Types of Forces Involved:

Attractive Magnetic Force (Analogous to Electric Attraction):

When the north pole of one magnet is brought close to the south pole of another magnet, they experience an attractive magnetic force. Similarly, when objects with opposite electric charges were brought close together, then they will experience an attractive electric force.

Repulsive Magnetic Force (Analogous to Electric Repulsion):

When two magnets with the same pole (both north or both south) are brought close to each other, they experience a repulsive magnetic force. This is analogous to the repulsion between objects with like electric charges (both positive or both negative).

Energy Transformations;

When you bring the magnets closer together or move them apart, energy transformations occur:

Potential Energy Transformation;

As the magnets are moved closer together, the potential energy of the magnetic interaction decreases. This is because the magnets' magnetic fields interact more strongly, and they tend to move toward each other due to the attractive or repulsive forces.

Kinetic Energy Transformation;

If you let the magnets go after bringing them close together, they will move towards each other (in the case of attraction) or move apart (in the case of repulsion). This movement involves a transformation of potential energy into kinetic energy. The kinetic energy increases as the magnets move, and it's at its maximum when the magnets are farthest apart (in the case of repulsion) or when they collide (in the case of attraction).

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Question 16(Multiple Choice Worth 5 points)

(04.01 LC) Which statement is true about the total mass of the reactants during a chemical change?

O It is destroyed during chemical reaction.
O It is less than the total mass of the products. O It is equal to the total mass of the products.
O It is greater than the total mass of the products.​

Answers

Answer:

It is equal to the total mass of the products.

Explanation:

Hope this helps :)

o Calculate the pH of 500 ml of buffer solution containing .2M of ammonium sulphate and 0.3M ammonium hydroxide which 5 2.1% ionized in dilute solution. (kb of ammonium hydroxide is 1.8×10^-5)

Answers

Answer is 9

pKb=−logK

b=−log10^-5=5

A mixture of ammonium hydroxide and ammonium chloride forms a basic buffer solution.

Henderson's equation for the basic buffer solution is as given below-

pOH=pKb+log[acid]*[salt]

Substitute values in the above expression.

pOH=5+log0.1*0.1=5

Hence, the pH of the solution is pH=14−pOH=14−5=9

What do you mean by pH ?potential of hydrogena measure of the acidity or alkalinity of a solution equal to the common logarithm of the reciprocal of the concentration of hydrogen ions in moles per cubic decimetre of solution. Pure water has a pH of 7, acid solutions have a pH of less than 7, and alkaline solution with a pH greater than 7.

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what is the mass of insoluble calcium phosphate produced from .555 grams of calcium chloride​

Answers

Answer:

0.518 g

Explanation:

Step 1: Write the balanced equation

3 CaCl₂ + 2 H₃PO₄ ⇒ Ca₃(PO₄)₂ + 6 HCl

Step 2: Calculate the moles corresponding to 0.555 g of CaCl₂

The molar mass of CaCl₂ is 110.98 g/mol.

0.555 g × 1 mol/110.98 g = 5.00 × 10⁻³ mol

Step 3: Calculate the moles of Ca₃(PO₄)₂ produced

5.00 × 10⁻³ mol CaCl₂ × 1 mol Ca₃(PO₄)₂/3 mol CaCl₂ = 1.67 × 10⁻³ mol Ca₃(PO₄)₂

Step 4: Calculate the mass corresponding to 1.67 × 10⁻³ moles of Ca₃(PO₄)₂

The molar mass of Ca₃(PO₄)₂ is 310.18 g/mol.

1.67 × 10⁻³ mol × 310.18 g/mol = 0.518 g

can anybody pls help me with chemistry how to balance an equation of class 7 ??? pls
really I don't like chemistry..
I need only the step(s)..

Answers

Answer:

[tex] This\:may\: help[/tex]

Answer:

(1) Write down the chemical reaction in the form of word equation,keeping reactants on left hand side and products on right hand side.

(2) Write symbol and formula of all reactants and products in word equation. (3) Balance the equation by multiplying the symbols and formula by smallest possible figures.

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