What type of organic compound is a reactant in all substitution reactions?
1.alkyne
2.alkane
3.alkene
​

Answer:

A chemistry student must write down in her lab notebook the concentration of a solution of sodium thiosulfate. The concentration of a solution equals the mass of what's dissolved divided by the total volume of the solution.

Explanation:

The concentration of a solution can be measured in terms of molarity.

The molarity of a solution can be defined as the number of moles of solute present in the total volume of the solution.

The number of moles of solute is the ratio of mass of solute to molar mass of solute.

Hence,

[tex]Molarity=\frac{mass of solute}{molar mass of solute} * \frac{1}{volume of solution in L.}[/tex]

Answer:

Work

Explanation:

Formula for work is given by;

Work = force × distance.

It is clear from this formula that Work depends on force and distance.

This means that force and distance are independent of the workdone and so they are classified as independent variables while work will be the dependent variable.

Answer:

The choose (d)

d. polysaccharides

5.64 I think, I'm sorry if I'm wrong

Answer:

its d

Explanation:

Answer: The empirical and molecular formula for the given organic compound is [tex]C_9H_{13}O_3N[/tex]

Explanation:

Let the mass of the compound be 100 g

Given values:

% of C = 59.0%

% of H = 7.15%

% of O = 26.20%

% of N = 7.65%

Mass of C = 59.0 g

Mass of H = 7.15 g

Mass of O = 26.20 g

Mass of N = 7.65 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

To formulate the empirical formula, we need to follow some steps:

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Molar mass of N = 14 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of C}=\frac{59.0g}{12g/mol}=4.917 mol[/tex]

[tex]\text{Moles of H}=\frac{7.15g}{1g/mol}=7.15 mol[/tex]

[tex]\text{Moles of O}=\frac{26.20g}{16g/mol}=1.6375 mol[/tex]

[tex]\text{Moles of N}=\frac{7.65g}{14g/mol}=0.546 mol[/tex]

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.546 moles

[tex]\text{Mole fraction of C}=\frac{4.917}{0.546 }=9[/tex]

[tex]\text{Mole fraction of H}=\frac{7.15}{0.546 }=13[/tex]

[tex]\text{Mole fraction of O}=\frac{1.6375}{0.546 }=2.99\approx 3[/tex]

[tex]\text{Mole fraction of N}=\frac{0.546}{0.546 }=1[/tex]

The ratio of C : H : O : N = 9 : 13 : 3 : 1

The empirical formula of the compound becomes [tex]C_9H_{13}O_3N_1=C_9H_{13}O_3N[/tex]

To calculate the molecular formula, the number of atoms of the empirical formula is multiplied by a factor known as valency that is represented by the symbol, 'n'.

[tex]n =\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex] .....(2)

We are given:

Mass of molecular formula = 183 g/mol

Mass of empirical formula = 183 g/mol

Putting values in equation 2, we get:

[tex]n=\frac{183g/mol}{183g/mol}=1[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{1\times 9}H_{1\times 13}O_{1\times 3}N_{1\times 1}=C_9H_{13}O_3N[/tex]

Hence, the empirical and molecular formula for the given organic compound is [tex]C_9H_{13}O_3N[/tex]

Answer:

531.6g

Explanation:

Total moles of glucose in this case is: 886/180= 4.922 (mole)

For every 1 mole glucose we get 6 mole water

-> Mole of water is: 4.922 * 6= 29.533 (mole)

weight of water is 18. Therefore, total weight of water that we will have from 886g of glucose are: 25.933*18= 531.6g

The actions taken during quality assurance by the part of quality assurance in which they should be taken is to document procedures and keep suitable records. The correct option is a.

Quality assurance is checking the quality of objects and services. They are assured in the companies and factories and other places to check the quality of the products.

The different type of quality assurance is: There are different types of quality assurance.

They work in the set quality and set requirements. They maintain the quality and develop those sets. Furthermore, they manage waste and quality.

Thus, the correct option is a. Document procedures and keep suitable records.

To learn more about quality assurance, refer to the link:

https://brainly.com/question/13164793

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Explanation:

For preparing lyophobic sol, the substance in bulk is broken down into particles of colloidal dimensions (Dispersion) or aggregating smaller particles into particles of colloidal dimensions (condensation).

Answer:

2-isopropyl-4-methylphenol.

Explanation:

Hey there!

In this case, according to the given information, it turns out possible for us to assign the appropriate IUPAC name of the given compound, by considering that the phenol stands for the parent chain and we have isopropyl methyl radicals which the former is called first due to the alphabet consideration.

In such a way, the name would be 2-isopropyl-4-methylphenol.

Regards!

Answer:

The tasks which form the preliminaries to a descriptive investigation are: Making careful objective observations. Asking the relevant scientific questions.

Answer:

INCREASE in the difference between the melting point measured and the true melting temperature.

Explanation:

Melting point of a compound is defined as the temperature at which the soils compound changes into liquid at the atmospheric pressure. There are different circumstances that can lead to inaccurate melting point. These include:

--> presence of impurities in the compound,

--> Molecular composition,

--> Force of attraction, and

--> Rapid heating of the compound.

Under the circumstances of rapid heating of the compound, there would be an increase in the melting point range when compared with the true melting point range of the compound.

The higher the heating rate, the more rapid the rise in oven temperature, increasing the difference between the melting point measured and the true melting temperature.

Answer:

The answer is C.

Explanation:

In the chemical equation in C, there are 15 Carbon atoms and 1 Phosphorus atom in PC15, and 3 Hydrogen atoms and 6 Oxygen atoms in 3H2O on the left hand side. On the right hand side, there are 3 Hydrogen atoms, 1 Phosphorus atom and 46 Oxygen atoms in H3PO46, and 5 Hydrogen atoms and 5 Chlorine atoms in 5HCl.

Therefore, tallying up, there are 6 Oxygen atoms on the left hand side and 46 Oxygen atoms on the right hand side. This means the chemical equation in C is unbalanced.

Hope this helped!

With the help of hess's law:

ΔHf(MnO)=−248.9 kJ−12(272.0) kJ=−384.9 kJ(per mole) Δ H f ( M n O ) = − 248.9 k J − 1 2 ( 272.0 ) k J = − 384.9 k J ( p e r m o l e )

Hess's law of constant heat summation, also known simply as Hess' law, is a relationship in physical chemistry named after Germain Hess, a Swiss-born Russian chemist and physician who published it in 1840.

Moreover, hess's Law of Constant Heat Summation (or just Hess's Law) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.

Therefore, hess' law is based on the state function character of enthalpy and the first law of thermodynamics. Energy (enthalpy) of a system (molecule) is a state function.

Learn more about hess's law:

https://brainly.com/question/14561923

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Answer:

Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.

2 S + 3 O₂ → 2 SO₃

The stoichiometric calculations is as follows:

6 g S * 1 mol/32.06 g S = 0.187 mol S

Moles O₂ needed = 0.187 mol S * 3 mol O₂/2 mol S = 0.2805 mol O₂

Since the molar mas of O₂ is 32 g/mol,

Mass of O₂ needed = 0.2805 mol O₂ * 32 g/mol = 8.976 g O₂

1 NaCl + 1 HCl ➡️ 1 NaCl + Water (H2O) .

Answer:

1.75L

Explanation:

Reaction of decomposition is:

2KClO₃(s) →  2KCl(s) + 3O₂(g)

We determine moles of salt:

4.35 g . 1 mol /122.55 g = 0.0355 moles

Ratio is 2:3. 2 moles of salt can produce 3 moles of oxygen

Then, our 0.0355 moles of chlorate may produce (0.0355 . 3)/ 2 = 0.0532 moles.

We have determined, moles of gas and we have data of pressure and temperature. To find out the volume, we apply the Ideal Gases Law:

We convert T° from °C to K → 27°C + 273 = 300K

P . V = n . R . T

0.75 atm . V = 0.0532 mol . 0.0821 L.atm/mol.K . 300K

V = (0.0532 mol  . 0.0821 L.atm/mol.K . 300K) / 0.75 atm

V  = 1.75 Liters

Answer:

Decane> 2-octanone> acetaldehyde> acetic acid> 1-butanol

Explanation:

One of the most important factors that affect the retention factor (Rf) in chromatography is the polarity of the mobile phase.

We have to recall that effective separation in chromatography depends on the movement of the mobile phase over a stationary phase.

The more polar the mobile phase, the more it interacts with the solute and the more the distance it travels from the baseline. Thus high solvent polarity leads to higher Rf value.

Therefore, the order of increasing Rf values for the solvents in the question is; Decane> 2-octanone> acetaldehyde> acetic acid> 1-butanol

Answer:

Boiling T° of solution = 100.6

Explanation:

Formula for elevation of boiling point is:

ΔT = Kb . m . i

where ΔT means Boiling T° of solution - Boiling T° of pure solvent

Our solute is a non ionizing compound.

i = 1, because it is a non ionizing compound. i, indicates the ions dissolved in solution.

m = molality (moles of solute dissolved in 1 kg of solvent)

90 g of solvent = 0.09 kg of solvent

We convert mass of solute to moles (by the molar mass):

10 g . 1 mol /92.09 g = 0.108 moles

m = 0.108 mol /0.09 kg = 1.21 m

Let's replace data: Boiling T° of solution - 100°C = 0.51 °C/m . 1.21 m . 1

Boiling T° of solution = 0.51 °C/m . 1.21 m . 1 + 100°C

Boiling T° of solution = 100.6

Answer:

see explanation

Explanation:

SO₃(g) + 395.77 Kj/mole => S°(s) + 3/2O₂(g)

The standard heat of formation for SO₃(g) is given by the following rxn:

S°(s) + 3/2O₂(g) => SO₃(g) + 395.77 kJ/mole. Reversing this reaction is the decomposition of SO₃(g) into its basic elements in their standard state (25°C, 1atm) and is endothermic with +295.77Kj/mole.

Explanation:

ethoxypropane

Ch3-ch2-o-ch2-ch2-och3

Explanation:

[tex]2H_2O_2 \rightarrow 2H_2O + O_2[/tex]

First convert the amount of water into moles:

360 g H2O × [tex]\left(\dfrac{1\:\text{mol}H_2O}{18.015\:\text{g}H_2O}\right)[/tex]

[tex] = 20. \:\text{mol}H_2O[/tex]

Now let's calculate the number of moles of O2 gas produced.

20 mol H2O × [tex]\left(\dfrac{1\:\text{mol}O_2}{2\:\text{mol}H_2O}\right)=10\:\text{mol}O_2[/tex]

The volume of gas at 10°C and 5 atm can be found using the ideal gas law:

[tex]PV=nRT[/tex]

[tex]V= \dfrac{nRT}{P}[/tex]

[tex]= \dfrac{(10)(0.082)(283)}{(5)}=46.4\:L[/tex]

Answer:

Lo siento, no sé la respuesta, pero espero que lo hagas bien

Answer:

Because of its weak intermolecular forces.

Explanation:

Hello there!

In this case, according to the given description, it turns out possible for us to recall the chemical structures of both ethanol and dimethyl ether as follows:

[tex]CH_3CH_2OH\\\\CH_3COCH_3[/tex]

Thus, we can see that ethanol have London dispersion forces (C-C bonds), dipole-dipole forces (C-O bonds) and also hydrogen bonds (O-H bonds) which make ethanol a liquid due to the strong hydrogen bonds. On the other hand, we can see that dimethyl ether has just London and dipole forces, which are by far weaker than hydrogen bonding, that makes it unstable when liquid and therefore it tends to vaporize quite readily.

Regards!

Answer:

Empirical formula has a molar mass of 30.01g/mol and molecular formula is C₆H₁₂O₆

Explanation:

Molar mass of a molecule is the sum of the molar mass of each atom. In CH2O we have:

1C = 1*12.01g/mol = 12.01g/mol

2H = 2*1g/mol = 2g/mol

1O = 1*16g/mol = 16g/mol

Empirical formula of CH2O is:

12.01g/mol + 2g/mol + 16g/mol = 30.01g/mol

As the molecular compound has a molar mass of 180.0g/mol the molecular formula is:

180.0g/mol / 30.01g/mol = 6 times the empirical formula. That is:

Answer:

48

Explanation:

Answer:

6.02 × 10²⁷ km

Explanation:

Step 1: Calculate the height of the stack of pennies

A penny has a thickness of approximately 1.0 mm. If you stacked Avogadro's number of pennies (6.02 × 10²³ pennies) one on top of the other on Earth's surface, the height of the stack of pennies would be:

6.02 × 10²³ pennie × 1.0 mm/1 pennie = 6.02 × 10²³ mm

Step 2: Convert 6.02 × 10²³ mm to kilometers

We will use the following conversion factors.

6.02 × 10²³ mm × 1 m/10³ mm × 1 km/10³ m = 6.02 × 10²⁷ km