What is the volume in milliliters of 8.00 g of acetone?

Answer:

C. the uneven heating of Earth's surface

Answer:

The correct answer is - 15 hours.

Explanation:

Half-life an element is the time required to decay the half amount of the initial amount of the element and is represented by t1/2.

The formula for the calculating initial amount

n(i)=n(r)⋅2^n

where

n(i) - the initial number of moles of element

n(r) - the number of moles remained amount after the given time

n - the number of periods,

=  time T, where T is the half-life.

n(i)=n(r)⋅2^n

n(i)/n(r) =2^n

1/16 =2^n

n= 4

The half life = 60/n

= 60/4

= 15 hours.

Answer:

There are two -NH2 groups around the C=O. bond in the urea molecule structure. There are one lone pairs on each nitrogen atoms and two lone pairs on oxygen atom.

The molecule urea that the molecular geometry of the nitrogen atoms is tetrahedral.

The term molecular geometry has to do with the arrangement of atoms around the central atom. On this case, we are looking at the arrangement of atoms around nitrogen.

We can see from the molecule urea that the molecular geometry of the nitrogen atoms is tetrahedral.

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The mass percentage of each element in baking soda is:

The calculation of mass percentage for each element is as follows:

In this way, the mass percentage should be calculated by dividing each one from the baking soda sample.

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Answer:

The mass percentages of sodium, hydrogen, carbon, and oxygen are 27.37%, 1.20%, 14.30%, and 57%, respectively.

Explanation:

The mass percentage (%mass) of an element present in a compound is calculated using the following expression:

\(\rm \%mass=\dfrac{mass\;of\;the\;element}{mass\;of\;the\;compound}\times 100\%\)

According to the given problem, the mass of \(\rm NaHCO_3\) is 168.02 g.

The mass of sodium in the given sample is 45.98 g.

So, the %mass of sodium in the given sample is calculated as shown below:

\(\rm \%mass\;of\;sodium=\dfrac{45.98\;g}{168.02\;g}\times 100\%=27.37\%\)

Similarly, the mass percentages of hydrogen, carbon, and oxygen are calculated as shown below:

\(\rm \%mass\;of\;hydrogen=\dfrac{2.02\;g}{168.02\;g}\times 100\%=1.20\%\)

\(\rm \%mass\;of\;carbon=\dfrac{24.02\;g}{168.02\;g}\times 100\%=14.30\%\)

\(\rm \%mass\;of\;oxygen=\dfrac{96\;g}{168.02\;g}\times 100\%=57\%\)

Hence, the mass percentages of sodium, hydrogen, carbon, and oxygen are 27.37%, 1.20%, 14.30%, and 57%, respectively.

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Answer:

The melting/ boiling point of selenium is closer to that of sulphur (a non-metal), compared with metals and metalloids. This shows selenium is more likely to be a non-metal, just like sulphur.

Selenium is classified as a metalloid. It has properties of both metals and non-metals, making it difficult to classify as either a metal or a non-metal.

Selenium is generally considered a non-metal, but it has some characteristics of metals as well. It is a semiconductor and an excellent conductor of electricity. Selenium has six valence electrons, which is one less than a typical non-metal, but it is also not a typical metal. Its properties place it in the intermediate category of metalloids.

A type of chemical element known as a metalloid possesses a significant number of properties that fall somewhere in between those of metals and nonmetals or are a combination of them. There is no universal definition of a metalloid, and there is no consensus regarding which elements constitute metalloids.

Metalloids are typically solid at room temperature and are brittle, somewhat shiny substances. Electronic band structures of metalloids are known to be comparable to those of semiconductors or semimetals. Amphoteric or weakly acidic oxides are known to form from metalloids.

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Answer:

water

Explanation:

i cba

Answer:

\(n=1.7 mol\)

Explanation:

Hello!

In this case, according to the ideal gas equation, it is possible to notice that:

\(PV=nRT\)

Thus, since we are asked to compute moles, we proceed as follows:

\(n=\frac{PV}{RT}\)

Now, since 1 L = 1 dm³ and 20 °C equal 293.15 K, we obtain:

\(n=\frac{4atm*10 L}{0.08206\frac{atm*L}{mol*K}*293.15K}\\\\n=1.7 mol\)

Best regards!

Approximately 90.6 g of K₂Cr₂O₇ would precipitate out of the solution when cooled from 80°C to 20°C.

To determine how much solute will precipitate out of a solution of potassium dichromate (K₂Cr₂O₇) when cooled from 80°C to 20°C, we apply the following method:

At 80°C:

Initial amount of K₂Cr₂O₇ = 100 g/L

At 20°C:

Solubility of K₂Cr₂O₇ = 9.4 g/L

Amount of K₂Cr₂O₇ that would remain in solution at 20°C = 9.4 g/L x 1 L = 9.4 g

Therefore, the amount of K₂Cr₂O₇ that would precipitate out of the solution when cooled from 80°C to 20°C would be:

Amount of K₂Cr₂O₇ that precipitates out = Initial amount - Amount that remains in solution at 20°C

Amount that precipitates out = 100 g - 9.4 g = 90.6 g

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The complete question is below:

When 100 g of k2cr2o7 in 1 liter of water is cooled from 80oc to 20oc, how much solute will precipitate out of the solution?.

Answer:

Number of protons in the nucleus of an atom

Evaporation causes the dew to vanish when the temperature of the atmosphere rises with the rising of the sun.

A kind of vaporization called evaporation takes place on the surface of a liquid as it transitions into the gas phase. When humidity impacts the rate of evaporation of water, for example, a high concentration of the evaporating material in the surrounding gas considerably slows down evaporation.

Similar to how perspiration evaporates from your body on a hot day to cool you down, dew evaporates as it cools the plant. This lessens heat exhaustion in extremely hot settings. Some species, particularly desert plants, are capable of directly absorbing water via their leaves.

Thus, Evaporation causes the dew to vanish when the temperature of the atmosphere rises with the rising of the sun.

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Answer:

A mixture of ethane and ethene occupies 40 litre at 1.00 atm and at 400 K.The mixture reacts completely with 130 g of O2 to produce CO2 and H2O . Assuming ...

Missing: 32 ‎Br2

The substance begins to boil at 2750⁰C, the substance begins to melt at 1500⁰C, the temperature at which the substance is both a liquid and a gas at 2750⁰C, temperature is the substance both a solid and a liquid​ at 1500⁰C.

Heating curves are the graphical correlations between heat added to a substance. When viewed from a cooling perspective, ie. loss of heat, it is the cooling curve.

The gradient of the cooling curve is related to the heat capacity, the thermal conductivity of the substance, and the external temperature. The more heat is required to change the temperature of the substance, the slower it cools, so the smaller the gradient of the curve. The higher the thermal conductivity, the faster heat is transferred, so the faster the substance cools.

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Answer:

uncomplete question

Explanation:

The student needs to add approximately 83.3 mL of 12 M HCl solution to the existing 1 L of 2 M HCl solution to increase the concentration to 3 M. It is important to handle concentrated acids with caution and follow proper safety procedures.

To increase the concentration of a 1 L solution of 2 M HCl to 3 M, the student needs to calculate the volume of concentrated HCl needed and add it to the existing solution. Here's how the calculation can be done:

Given:

Initial concentration of HCl solution = 2 M

Final concentration desired = 3 M

Initial volume of HCl solution = 1 L

Step 1: Calculate the moles of HCl in the initial solution.

Moles of HCl = Initial concentration × Initial volume = 2 M × 1 L = 2 moles

Step 2: Calculate the moles of HCl needed for the desired concentration.

Moles of HCl needed = Final concentration × Final volume = 3 M × 1 L = 3 moles

Step 3: Calculate the moles of HCl to be added.

Moles of HCl to be added = Moles needed - Moles present = 3 moles - 2 moles = 1 mole

Step 4: Convert the moles of HCl to the required volume of concentrated HCl.

To calculate the volume, we need to know the concentration of the concentrated HCl solution. Assuming it is 12 M, we can use the following formula:

Volume of concentrated HCl = Moles of HCl to be added / Concentration of concentrated HCl

Volume of concentrated HCl = 1 mole / 12 M = 0.0833 L or 83.3 mL

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Answer:

the l sign means the substance in the chemical equation is a liquid. (hope this helped : D )

Answer:

The correct answer is 0.0033 L (33.0 mL)

Explanation:

We uses the Charles's law which describes the changes in the volume (V) of a gas and its temperature in Kelvin (T) at constant pressure. The mathematical expression is the following:

V₁/T₁ = V₂/T₂

We have the following data:

V₁= 31.0 mL = 0.0031 L

T₁= 19.0°C = 292 K

T₂= 38.0°C = 311 K

V₂= ?

We calculate V₂ from the mathematical expression, as follows:

V₂= V₁/T₁ x T₂ = 0.0031 L/(292 K) x 311 K = 0.0033 L

Answer:

0.146 M

Explanation:

Use v1s1 = v2s2

here, v1 = 500 mL, v2 = 700 mL, s1 = 0.205 M & s2 = new concentration

Types of luminous flames:

1. Yellow Luminous Flame

2. Smoky Luminous Flame

3. Orange Luminous Flame

4. Blue Luminous Flame

Luminous flames are characterized by their visible glow, which is caused by the incomplete combustion of fuel. The presence of soot particles in the flame causes the emission of light. There are different types of luminous flames, which can be classified based on their fuel composition and burning conditions. Here are some common types of luminous flames:

1. Yellow Luminous Flame: This is the most common type of luminous flame, often seen in open fires, candles, and gas stoves. It appears yellow due to the presence of soot particles in the flame. Yellow flames indicate incomplete combustion of hydrocarbon fuels, such as methane, propane, or natural gas. The high carbon content in these fuels leads to the formation of soot, which emits visible light.

2. Smoky Luminous Flame: This type of flame is characterized by a significant amount of black smoke and soot production. It is commonly observed in poorly adjusted or malfunctioning burners or engines. The excessive presence of unburned fuel in the flame results in incomplete combustion and the emission of dark smoke particles.

3. Orange Luminous Flame: An orange flame indicates a higher combustion temperature compared to a yellow flame. It is often seen in more efficient burners or when burning fuels with a higher carbon content, such as oil or diesel. The higher temperature helps in burning more of the carbon particles, reducing the amount of soot and making the flame appear less yellow.

4. Blue Luminous Flame: A blue flame is typically associated with complete combustion. It indicates efficient burning of fuel, resulting in minimal soot formation. Blue flames are commonly observed in gas burners or Bunsen burners. The blue color is a result of the combustion of gases, such as methane, in the presence of sufficient oxygen.

It's important to note that the luminosity of a flame can vary depending on factors such as fuel-air mixture, combustion temperature, and the presence of impurities. Achieving complete combustion and minimizing the production of soot is desirable for efficient and cleaner burning processes.

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Vectors may be plasmids. An example of one of several useful features of commercially prepared cloning vectors is MULTIPLE CLONING SITES.

The pUC18 vector is a widely used standardized cloning vector for replication in Escherichia coli.

A multiple cloning site can be defined as a short DNA fragment observed in genetically engineered plasmids.

These DNA fragments (multiple cloning sites) contain twenty (20) or more sites where restriction enzymes can cut in order to generate recombinant DNA molecules.

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Answer:

75g

Explanation:

n=Cmxv

=0,15x12,5

=1,875 mol

m=nxM

=1,875x40

=75g

Answer:

Explanation:

                             HERE WE TAKE CH3,

                      C = 1 × 12 = 12g

                      H₃= 3 × 1  = + 3g

                          CH3       =15 g ans  

Answer: I think so.. (sorry if i am wrong)

Explanation:  Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity. A physical property is a characteristic of matter that is not associated with a change in its chemical composition.

The part of the product label that describes the hazardous effects of a chemical is the hazard statement.

The hazard statement describes the potential health and environmental risks associated with using or coming into contact with the chemical.

It typically includes information about the severity of the hazard, the type of route of exposure (skin, inhalation, oral, etc.), the symptoms of exposure, and the necessary protective measures to be taken.

For example, a product label may contain a hazard statement such as “Harmful if swallowed. May cause respiratory irritation. Avoid contact with eyes and skin. Wear suitable protective clothing”.

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Explanation:

Answer:krypton toh superman ko maarta haiExplanation:no

The number of moles of the water that was evaporated is  12083467.5 moles of water.

We know that the number of moles that we have can be gotten as the ratio of the mass to the number of moles of the object. Now we know that from the question, we have the volume of the hydrogen to be 90,000 gallons. We have to convert this volume to liters.

Given that;

1 gallon = 3.79 L

90,000 gallons  = 90,000 gallons  * 3.79 L/1 gallon

= 341100 L

Then we have the density of the hydrogen to be 70.85 g/L

Mass of the hydrogen =  70.85 g/L *  341100 L

= 24166935 g

Number of moles of the hydrogen reacted = 24166935 g/2 g/mol

= 12083467.5 moles

If 2 mole of hydrogen gives 2 moles of water

12083467.5 moles of hydrogen gives 12083467.5 moles of water

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Answer:

gravwfewtu

Explanation:

ushifelllf

Answer:

a. SOLUTION:

Step 1: Write the balanced chemical equation.

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

Step 2: Calculate the number of moles of CO₂ formed by each reactant.

• Using C₂H₄

Based on the balanced chemical equation, 1 mole of C₂H₄ is stoichiometrically equivalent to 2 moles of CO₂.

The molar mass of C₂H₄ is 28.054 g/mol.

• Using O₂

Based on the balanced chemical equation, 3 moles of O₂ is stoichiometrically equivalent to 2 mole of CO₂.

The molar mass of O₂ is 31.998 g/mol.

Step 3: Determine the limiting reagent.

Since O₂ produced less amount of CO₂ than C₂H₄, O₂ is the limiting reagent.

Step 4: Determine the mass of CO₂ formed.

Note that the (maximum) mass of a product that can be formed is dictated by the limiting reagent. In this case, we will start at the number of moles of CO₂ formed from the limiting reagent (O₂) which is equal to 0.11022 mol.

The molar mass of CO₂ is 44.009 g.

Hence, 4.85 g of CO₂ can be formed.

------------------------------------------------------------

b. ANSWER:

The LR is O₂ and the ER is C₂H₄.

------------------------------------------------------------

c. SOLUTION:

The theoretical yield of the reaction is 4.85 g.

Hence, the percent yield of the reaction is 87.6%.

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Explanation:

brainliest pls

The percent yield of \(Fe_2O_3\)  if 57.8g was collected is 75.1%.

To calculate the percent yield, we need to compare the actual yield (the amount of\(Fe_2O_3\) collected) to the theoretical yield (the amount of \(Fe_2O_3\)that would be obtained if the reaction went to completion).

First, we need to determine the molar mass of \(Fe_2O_3\) and \(Fe_2S_3\):

Molar mass of \(Fe_2O_3\):

2(55.85 g/mol) + 3(16.00 g/mol) = 159.69 g/mol

Molar mass of\(Fe_2S_3\):

2(55.85 g/mol) + 3(32.07 g/mol) = 207.67 g/mol

Next, we can calculate the theoretical yield of\(Fe_2O_3\) using stoichiometry:

2 moles of \(Fe_2S_3\) produce 2 moles of \(Fe_2O_3\)(according to the balanced equation).

So, the molar ratio of \(Fe_2O_3\)to Fe2S3 is 2:2.

The molar mass ratio of \(Fe_2O_3\) to Fe2S3 is:

159.69 g/mol : 207.67 g/mol

To calculate the theoretical yield, we can use the following equation:

Theoretical yield of\(Fe_2O_3\) = (mass of \(Fe_2S_3\)) * (molar mass of \(Fe_2O_3\) / molar mass of \(Fe_2S_3\))

Theoretical yield of \(Fe_2O_3\) = (100.0 g) * (159.69 g/mol / 207.67 g/mol)

Theoretical yield of \(Fe_2O_3\) = 76.46 g

Now we can calculate the percent yield using the formula:

Percent yield = (actual yield / theoretical yield) * 100

Percent yield = (57.8 g / 76.46 g) * 100

Percent yield = 75.1%

Therefore, the percent yield of \(Fe_2O_3\) is 75.1%.

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