What is the molecular formula of the compound CH2 with molar mass = 42.0 g/mol? A. C2H18 B. C2H4 C. CAHg O D. C3H6​

Answer:

273.15

Explanation:

A buffer is made from an acid - base conjugate pair as shown by option A

Buffers are often composed of weak acids and their conjugate bases (or weak bases and their conjugate acids). The weak acid can contribute a proton to balance any new base, whereas the conjugate base can absorb a proton to do so.

This balance between the acid and its conjugate base allows the buffer to survive pH changes. Buffers are essential in biological systems because many biochemical processes are particularly sensitive to pH changes.

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Answer:

1. Empirical formula => CH2O2

2. Molecular formula => C2H4O4

Explanation:

From the question given above, the following data were obtained:

Carbon (C) = 26.4 %

Hydrogen (H) = 3.3 %

Oxygen (O) = 70.3 %

Molar mass of compound = 91.0 g/mol

Empirical formula =..?

Molecular formula =..?

1. Determination of the empirical formula of the compound.

C = 26.4 %

H = 3.3 %

O = 70.3 %

Divide by their molar mass

C = 26.4 /12 = 2.2

H = 3.3 /1 = 3.3

O = 70.3 /16 = 4.39

Divide by the smallest

C = 2.2 /2.2 = 1

H = 3.3 /2.2 = 2

O = 4.39 /2.2 = 2

Empirical formula => CH2O2

2. Determination of the molecular formula of the compound.

Molar mass of compound = 91.0 g/mol

Empirical formula => CH2O2

Molecular formula => [CH2O2]n

We shall determine the value of n as follow:

[CH2O2]n = 91

[12 + (2×1) + (2×16)]n = 91

[12 + 2 + 32]n = 91

46n = 91

Divide both side by 46

n = 91/46

n = 2

Molecular formula => [CH2O2]n

Molecular formula => C2H4O4

In the given case where 26.4 % Carbon , 3.3 % Hydrogen , 70.3 % Oxygen , and Molar Mass: 91.0 g/mol the:

Given:

Carbon (C) = 26.4 %  

Hydrogen (H) = 3.3 %  

Oxygen (O) = 70.3 %  

Molar mass of compound = 91.0 g/mol  

Determination of the empirical formula of the compound:-

The ratio is CHO = 1 : 1.5 : 2

multiply with 2 to find correct and complete number ratio

C = 1 × 2 = 2

H = 1.5 × 2 = 3

O = 2 × 2 = 4

Thus,  the Empirical formula => \(C_2H_3O_4\)

Mass × n = molar mass

12 × 2 + 1 × 3 + 16 × 4 = 91

24 + 3 + 64 = 91

91 = 91

Thus moles are 1 which means

molecular formula =  (\(C_2H_3O_4\))\(_1\)

Thus, here in the given data:

Molecular Formula: (\(C_2H_3O_4\))\(_1\)

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The process through which a reactant loses electrons during a chemical reaction is known as oxidation. Reduction refers to the process through which a reactant acquires electrons during a reaction.

Such a thing typically happens if an acid reacts with a metal. Oxidation describes the process through which a reactant loses electrons during a chemical reaction.

Adding oxygen or any electronegative element, or taking away hydrogen or any electropositive one, is what is meant by the term "oxidation" in Classical and earlier concepts. Adding hydrogen or another electropositive element, or getting rid of oxygen or another electronegative one, is what reduction is all about, according to the classical or older idea.

An example of oxidation is C + O₂→ CO₂

An example of reduction is N₂ + 3H₂ → 2NH3

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Answer:

it's-B.....................

Answer:

66.4g

Explanation:

CO:CH4

2:1

CH4 moles= 8.3 ÷ 2

= 4.15 moles

CH4 = 12+ 4(1)= 16g

1 mole of CH4-------> 16g

4.15moles

(4.15x16)÷1

=66.4g

According to the given statement is Standard C2H2 formation enthalpy is +224 kJ/mole.

As a result, an exothermic reaction takes place, and if it releases more energy than it absorbs, its enthalpy will be negative. Think of this as a portion of the reaction's heat evaporating (or being eliminated). Because endothermic reactions take or utilize more energy than they generate, they have positive enthalpies.

Enthalpy is a characteristic or state variable which mirrors energy; it shares energy's dimensions and is quantified in joules and ergs as a result. Enthalpy's value is purely influenced by a system's composition, environment, and physical properties—not by its past.

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The IUPAC name for a binary covalent compound will lack roman numerals.

Binary covalent compound comprises of two elements which forms a

compound through the sharing of electrons.

The sharing of electrons is referred to as covalent bonding and naming such

compounds require prefixes such as mono-, di- , tri- etc.

The standard method of naming these compounds should be with the

prefixes and not with the roman numerals.

An example is carbon(IV)oxide which is wrong in  IUPAC naming.

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Answer:

B

Explanation:

Just did the test on edge 2021

When observing an organic compound's melting point with the Mel-Temp device, you should note the following temperature for your lab.

report: b. the temperature at which you see the final amount of solid turn into a liquid.

A substance's melting point is the temperature at which it transitions from a solid to a liquid condition. It's crucial to wait until the entire solid sample has melted when using the Mel-Temp equipment before taking the temperature. In contrast to a partially melted sample that can have a variable temperature due to residual heat, this assures that the temperature measured properly represents the substance's true melting point. The amount of solid that eventually turns into a liquid can be determined by watching it.

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Both solid and liquid wastes need to be transported in different ways because they both have different properties and characteristics.

solid and liquid wastes need to be disposed of in different ways as the liquid wastes are in a liquid nature and have to be transported into closed containers where as the solid are of solid nature and they require open trucks to transport them. It is important to do solid waste management so that there can be hygiene around the environment and fresh air is available.

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Answer:

Plant Reproduction

When plants reproduce asexually, they use mitosis to produce offspring that are genetically identical to the parent plant. The advantage of asexual reproduction is that it allows successful organisms to reproduce quickly. ...

When plants reproduce sexually, they use meiosis to produce haploid cells that have half the genetic information of the parent (one of every chromosome)

Explanation:

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The product composition at equilibrium of cracking pure n-butane to produce olefins is determined by the equilibrium constants.

At equilibrium, the product composition of cracking pure n-butane to produce olefins is determined by the equilibrium constants, K1 and KII.

Using these constants, we can calculate the mole fractions of the products.

The mole fraction of \(C_2H_4\) is calculated by using the formula, (1 + K1/\(KII)^{(-1/2)\), which gives a value of 0.526.

The mole fraction of \(C_2H_6\) is calculated by using the formula, K1/(1 + K1/KII), which gives a value of 0.297.

The mole fraction of \(C_3H_6\) is calculated by using the formula, KII/(1 + K1/KII), which gives a value of 0.146.

The mole fraction of \(CH_4\) is calculated by using the formula, (1 + K1/\(KII)^{(-1/2),\) which gives a value of 0.031.

Therefore, at equilibrium, the product composition is 52.6% \(C_2H_4\), 29.7% \(C_2H_6\), 14.6% \(C_3H_6\), and 3.1% \(CH_4\).

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At equilibrium, the product composition will be 50.1% C2H4, 49.9% C3H6, and negligible amounts of C2H6 and CH4. This is due to the high equilibrium constant for reaction II.

The equilibrium constant expression for each reaction is given by K1 = [C2H4][C2H6] and KII = [C3H6][CH4]/[C4H10]. Assuming x is the extent of reaction for both reactions, the equilibrium concentrations are [C4H10] = P - x, [C2H4] = [C2H6] = x/2, and [C3H6] = [CH4] = xII. Substituting these into the equilibrium constant expressions and solving for x and xII gives x = 0.459 and xII = 0.000171. Therefore, the product composition is 50.1% C2H4, 49.9% C3H6, and negligible amounts of C2H6 and CH4. This is because the equilibrium constant for reaction II is much higher than that for reaction I, meaning that more C3H6 and CH4 are formed than C2H6, making C3H6 the dominant product.

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(i) The vehicle outputs approximately 11.58 kg of CO₂ per week. (ii) The Biofuel Obligation Scheme in Ireland is implemented by requiring fuel suppliers.

To calculate the amount of CO₂ output per week, we need to determine the amount of fuel used and then use the given combustion equation to find the ratio of CO₂ produced per unit of fuel.

Given;

Average fuel efficiency: 7.5 km per litre

Distance traveled per week: 300 km

Mass of fuel per litre: 0.75 kg

First, we calculate the total fuel used per week;

Fuel used = Distance traveled / Fuel efficiency

= 300 km / 7.5 km per litre

= 40 litres

Next, we find the mass of fuel used per week:

Mass of fuel used = Fuel used × Mass of fuel per litre

= 40 litres × 0.75 kg per litre

= 30 kg

Using the combustion equation, we know that 2 moles of C₈H₁₈ produce 16 moles of CO₂. Therefore, we can calculate the moles of CO₂ produced from the given mass of fuel;

Moles of CO₂ produced = Moles of C8H18 × (16 moles of CO₂ / 2 moles of C₈H₁₈)

= (30 kg / (114 g/mole)) × (16 moles of CO₂ / 2 moles of C₈H₁₈)

= (30,000 g / 114 g/mole) × (16 moles of CO₂ / 2 moles of C₈H₁₈)

= 263.16 moles of CO₂

Finally, we convert the moles of CO₂ to kilograms;

Mass of CO₂ produced = Moles of CO₂ produced × Molar mass of CO₂

= 263.16 moles × (44 g/mole)

= 11,579.04 g

= 11.58 kg (rounded to two decimal places)

Therefore, the vehicle outputs approximately 11.58 kg of CO₂ per week.

The Biofuel Obligation Scheme in Ireland is implemented by requiring fuel suppliers to include a certain percentage of biofuels in their overall fuel sales, thereby reducing the carbon footprint by promoting the use of renewable and lower-carbon-emitting fuels.

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Answer:

832.5 calories

Explanation:

One ounce bag of Hot Cheetos contains 90 calories from fat and 150 total calories. Also a party size bag of hot Cheetos weigh 18.5 ounces.

Therefore the weight of half of a bag of party-size chips = Size of a party size bag of hot Cheetos / 2 = 18.5 ounces / 2 = 9.25 ounces.

Since one ounce bag of Hot Cheetos contains 90 calories from fat, the amount of fat contained in half a bag of party-size chips (9.25 ounces) is given as:

Calories of fat in half a bag of party-size = 90 calories per ounce × 9.25 ounces = 832.5 calories

Answer:

832.5 calories

Explanation:

i) Metals: A and D

Non-metals: B and C

ii) The valency of A is 1. This is because A has one valence electron, and elements in Group 1 (such as A) typically have a valency of 1 as they tend to lose that single valence electron to achieve a stable electron configuration.

iii)

A: Sodium (Na) - Symbol: Na

B: Chlorine (Cl) - Symbol: Cl

C: Oxygen (O) - Symbol: O

D: Calcium (Ca) - Symbol: Ca

iv) The compound formed by the combination of A and B: Sodium Chloride (NaCl) The compound formed by the combination of C and B: Oxygen Chloride (OCl2)

v) Element D (Calcium) is more active than element A (Sodium). This is because elements in Group 2 (such as D) tend to lose two valence electrons, which requires less energy compared to losing only one valence electron (as in the case of A).

vi) The elements that can lose the valence electron to become stable are A (Sodium) and D (Calcium).

Metals: A, B, C, D; Non-metals: None. Elementsii) The valency of A is 1 because it has only one valence electron.iii) The name and symbol of all the elements are:A - Sodium (Na)B - Chlorine (Cl)C - Argon (Ar)D - Calcium (Ca)iv) The molecular formula of the compound formed by the combination of A and B is NaCl. The molecular formula of the compound formed by the combination of C and B is BCl2.

v) A is more active than D because A is a metal and metals are more active than non-metals. A tends to lose electrons more easily than vi) The elements which can lose the valence electron to become stable are metals.

i) Metals: A and D Non-metals: B and C ii) The valency of A is 1. This is because A has one valence electron, and elements in Group 1 (such as A) typically have a valency of 1 as they tend to lose that single valence electron to achieve a stable electron configuration. iii) A: Sodium (Na) - Symbol: Na B: Chlorine (Cl) - Symbol: Cl C: Oxygen (O) - Symbol: O D: Calcium (Ca) - Symbol: Ca

iv) The compound formed by the combination of A and B: Sodium Chloride (NaCl) The compound formed by the combination of C and B: Oxygen Chloride (OCl2) v) Element D (Calcium) is more active than element A (Sodium). This is because elements in Group 2 (such as D) tend to lose two valence electrons, which requires less energy compared to losing only one valence electron (as in the case of A). vi) The elements that can lose the valence electron to become stable are A (Sodium) and D (Calcium).

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it doesn't repopulate as fast because ecosystem 2 could have more trouble reproducing then ecosystem 1.

The forces of attraction that exists between positive metal ions and their surrounding valence electrons is called: metallic bonds.

A chemical bond can be defined as the forces of attraction that exists between ions, crystals, atoms or molecules and they are mainly responsible for the formation of all chemical compounds.

In Chemistry, there are different types of chemical bonds and these are:

A metallic bond can be defined as the forces of attraction that exists between positive metal ions and their surrounding valence electrons.

This ultimately implies that, metallic bond are peculiar to positively charged metals but not negatively charged metals or non-metals.

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Suppose the radius of an atom in a simple cubic unit cell is 0.17 nm. 0.34 nm is the edge length of the unit cell.

In a simple cubic unit cell, the atoms are arranged in a cube with one atom at each corner. The distance from the center of an atom to the edge of the cube is half of the edge length. Therefore, the edge length of the unit cell can be calculated as twice the radius of the atom:

A body-centered cubic unit cell has one atom in the centre and atoms at each of the cube's four corners. We must know the separation between atoms situated at the opposite corners of the cube in order to calculate the edge length of the unit cell.

Think of a diagonal line joining the cube's two opposed corners while traversing its centre. The hypotenuse of a right triangle with two sides equal to the edge length of the unit cell might be imagined as being this diagonal line.
\(Edge length = 2radius\)
Edge length = 2 x 0.17 nm
Edge length = 0.34 nm
So the edge length of the simple cubic unit cell is 0.34 nm.

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Answer:

satellite

Explanation:

found it on the internet

The correct answer is Radar

Answer:

The pH scale measures acidity of a substance. known as potential of hydrogen, it varies from 0 to 14 with 7 being the pH value of a neutral solution. Below 7 shows the substance is acidic in nature and above 7 is alkaline in nature. pH 0-3 are considered strong acids while pH 4-6 are weak acids. pH 8-10 are weak alkalines and pH 11-14 are strong alkalines. This is a general trend and there may be exeptions especially if the substance has a negative pH. However, it would not be covered likely unless you are doing university chemistry.

When a cream includes 20 g of an active ingredient and 60 g of a cream base, the strength is 20:60.

A ratio with the form 1 in r serves as an expression for ratio strength. The comparable fraction's numerator would be 1, making it. According to accepted practice, the weight and volume are stated in grams and milliliters, respectively, when ratio strength denotes a solid in a liquid using units of weight and volume.

Ingredients are the materials required to create anything, particularly all the many foods you use to prepare a certain dish. Add the remaining components and combine. Countable noun One of a situation's key components is an ingredient.  Any component that is used in a food to produce a certain result is considered an ingredient. Food additives are considered food ingredients.

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The complete question is:

A cream contains 20 g of the active ingredient in 60 g of cream base. What is the ratio strength?

Select one:

2: 60

20: 60

200: 60

200: 600

Explanation:

The union of vinegar and bicarbonate produces carbon dioxide

The carbonic acid, which is weaker, in turn breaks down into water and carbon dioxide, which being volatile separates

will observe a yellow color, confirming that BASIC HYDROLYSIS has taken place. To the touch the bottle cools and in the end a white deposit may remain on the bottom.

7.133 is the rate constant increase if the temperature is raised to 37c.

factor=rate(T2)/rate(T1) .

The proportionality constant (k) that links the pace of the reaction to the concentrations of reactants is known as the specific rate constant. For any chemical reaction, the rate law and the particular rate constant must be ascertained through experimentation. Temperature affects the rate constant's value.a chemical reaction in which two or more substances interact and change into new ones, or where one substance transforms into another.

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Answer:

0.0875M of Acetic Acid

Explanation:

Based on the reaction of acetic acid (HX) with sodium hydroxide (NaOH):

HX + NaOH → NaX + H₂O

1 mole of acetic acid reacts with 1 mole of NaOH

That means the moles of sodium hydroxide added = Moles the acetic acid at the equivalence point.

To solve this question we have to find the moles of NaOH = Moles HX. With the moles and the volume (25.0mL = 0.0250L) we can find the molarity of the acetic acid:

Moles NaOH = Moles Acetic acid:

12.5mL = 0.0125L * (0.175mol / L) = 0.00219 moles Acetic Acid

Molarity acetic acid:

0.00219 moles Acetic Acid / 0.0250L =

Answer: To determine the most abundant isotopic form of an element, compare given isotopes to the weighted average on the periodic table.

Answer:

c

Explanation:

The substances initially involved in a chemical reaction are called reactants.

They are the substances that are present at the start of the reaction and are transformed into different substances, known as products, during the reaction. The reactants are written on the left side of a chemical equation and the products are written on the right side.

In a chemical reaction,

Example:

In the reaction between hydrogen gas (H2) and oxygen gas (O2) to form water (H2O), the reactants are H2 and O2, and the product is H2O. This reaction can be represented by the equation: H2 + O2 -> H2O

In this equation, the reactants H2 and O2 are listed on the left side of the arrow and the product H2O is listed on the right side.

It is important to note that the number of atoms of each element on the reactants side and products side must be equal. This principle is called the law of conservation of mass.

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Answer:

1. Ga

2. O

3. Cl

4. Br

Answer:

1. carboxylic acid

2. ether

3. aldehyde

4. ketone

5. amine

6. alcohol

Explanation:

Know the functional groups. If you would like me to teach you them let me know.

- Hope that helps! Please let me know if you need further explanation.

The amount of calcium ingested by drinking eight 8-oz glasses of local water is 0.0041655 mol. By drinking 1.0L of local water, approximately 166.85 mg of calcium is ingested, which is 14.5% of the daily requirement of calcium.

1. To calculate the amount of calcium ingested by drinking eight 8-oz glasses of local water, we need to know the volume of water and the concentration of calcium ions present in it.1 oz = 29.57mL

So, eight 8-oz glasses of water = 8 x 8 oz = 64 oz= 64 x 29.57 mL = 1892.48 mL1 L = 1000 mL

Therefore, the volume of water in liters = 1892.48/1000 = 1.8925 L

Concentration of Ca ions in water = 0.0022 mol/L

So, the amount of calcium ingested = concentration of calcium x volume of water= 0.0022 mol/L x 1.8925 L= 0.0041655 mol of calcium

2. To calculate the percentage of the daily requirement of calcium met by drinking 1.0L of local water, we need to convert the amount of calcium in moles to milligrams (mg). The molar mass of calcium is 40.08 g/mol.0.0041655 mol of calcium weighs = 0.0041655 mol x 40.08 g/mol = 0.16685464 g or 166.85 mg

The percentage of the daily requirement of calcium met by drinking 1.0L of local water is given by the formula:

Percentage of daily requirement met = (amount of calcium in water/daily requirement of calcium) x 100%Daily requirement of calcium = 1150 mg

Percentage of daily requirement met = (166.85 mg/1150 mg) x 100%≈ 14.5 %

Therefore, by drinking 1.0L of local water, approximately 166.85 mg of calcium is ingested, which is 14.5% of the daily requirement of calcium.

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