importance of hematology​

Answer:

a) Linear polymerization

b) cyclic polymerization

Answer:

Please find the complete solution in the attached file.

Explanation:

Answer:

10.96

Explanation:

A solution is prepared at 25 °C that is initially 0.14 M in diethylamine, a weak base with Kb = 1.3 × 10⁻³, and 0.20 M in diethylammonium chloride. Calculate the pH of the solution. Round your answer to 2 decimal places.

Step 1: Calculate the pOH of the solution

Diethylamine is a weak base and diethylammonium (from diethylammonium chloride) its conjugate acid. Thus, they form a buffer system. We can calculate the pOH of this buffer system using the Henderson-Hasselbach's equation.

pOH = pKb + log [acid]/[base]

pOH = -log 1.3 × 10⁻³ + log 0.20 M/0.14 M

pOH = 3.04

Step 2: Calculate the pH of the solution

We will use the following expression.

pH + pOH = 14

pH = 14 - pOH = 14 -3.04 = 10.96

Answer:

30%

Explanation:

This is the chemical formula for zinc bromate: Zn(BrO₃)₂. Calculate the mass percent of oxygen in zinc bromate. Round your answer to the nearest percentage.

Step 1: Determine the mass of 1 mole of Zn(BrO₃)₂

M(Zn(BrO₃)₂) = 1 × M(Zn) + 2 × M(Br) + 6 × M(O)

M(Zn(BrO₃)₂) = 1 × 65.38 g/mol + 2 × 79.90 g/mol + 6 × 16.00 g/mol

M(Zn(BrO₃)₂) = 321.18 g/mol

Step 2: Determine the mass of oxygen in 1 mole of Zn(BrO₃)₂

There are 6 moles of atoms of oxygen in 1 mole of Zn(BrO₃)₂.

6 × m(O) = 6 × 16.00 g = 96.00 g

Step 3: Calculate the mass percent of oxygen in Zn(BrO₃)₂

%O = mO/mZn(BrO₃)₂ × 100%

%O = 96.00 g/321.18 g × 100% ≈ 30%

Answer: The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is 0.0709 M.

Explanation:

Given: Concentration of hydrogen fluoride = 0.126 M

Concentration of fluoride ions = 0.1 M

Volume of HCl = 9.0 mL

Concentration of HCl = 0.01 M

Volume of HCl = 25.0 mL

Moles of [tex]F^{-}[/tex] ions are calculated as follows.

[tex]Moles of F^{-} = molarity \times volume\\= 0.1 M \times 0.025 L\\= 0.0025 mol[/tex]

Moles of HF are as follows.

[tex]Moles of HF = Molarity \times Volume\\= 0.126 M \times 0.025 L\\= 0.00315 mol[/tex]

Moles of HCl are as follows.

[tex]Moles of HCl = Molarity \times volume\\= 0.01 M \times 0.009 L\\= 0.00009 mol[/tex]

Now, reaction equation with initial and final moles will be as follows.

                        [tex]H^{+} + F^{-} \rightarrow HF[/tex]

Initial:      0.00009  0.0025    0.00315

Equilibrium:      (0.0025 - 0.00009)    (0.00315 + 0.00009)

                                = 0.00241                    = 0.00324

Total volume = (9.00 mL + 25.0 mL) = 34.0 mL = 0.034 L

Hence, concentration of fluoride ions is calculated as follows.

[tex]Concentration = \frac{moles}{volume}\\= \frac{0.00241 mol}{0.034 L}\\= 0.0709 M[/tex]

Thus, we can conclude that concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is 0.0709 M.

Answer:

reductants or redox reactions

Answer:

1) 2.69 * 10²³ PBr₅

2) 6.02 * 10²⁴ C₁₂H₂₂O₁₁

Explanation:

Question 1)

We want to convert 192 grams of phosphorus pentabromide to molecules. Note that 192 is three significant figures.

Phosphorus pentabromide is given by PBr₅.

To convert from grams to molecules, we can convert from grams to moles first, and then from moles to molecules.

To convert from grams to moles, we will find the molar mass of PBr₅.

Since the molar mass of P is 30.974 g/mol and the molar mass of Br is 79.904 g/mol, the molar mass of PBr₅ is:

[tex](30.974)+5(79.904) = 430.494\text{ g/mol}[/tex]

And since we want to convert from grams to moles, we can write the following ratio:

[tex]\displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}[/tex]

Where grams is in the denominator, which allows us to cancel them out, leaving us with only moles.

To convert from moles to molecules, we can use the definition of the mole: a mole of one substance has 6.022 * 10²³ amount of that substance.

So, a mole of PBr₅ has 6.022 * 10²³ molecules of PBr₅. Since we want to cancel out the moles, we can write the ratio:

[tex]\displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}[/tex]

In combination, starting with 192 grams of PBr₅, we will acquire:

[tex]\displaystyle 192\text{ g PBr$_5$} \cdot \displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}[/tex]

Cancel like units:

[tex]\displaystyle = 192 \cdot \displaystyle \frac{1 }{430.494}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1}[/tex]

Multiply. Hence:

[tex]=2.6858...\times 10^{23}\text{ PBr$_5$}[/tex]

Since the final answer should have three significant digits, our final answer is:

[tex]= 2.69\times 10^{23} \text{ PBr$_5$}[/tex]

So, there are about 2.69 * 10²³ molecules of PBr₅ in 192 grams of the substance.

Question 2)

We want to convert 3.42 kilograms of table sugar (C₁₂H₂₂O₁₁) to molecules. Note that this is three significant figures.

3.42 kilograms is equivalent to 3420 grams of table sugar.

Again, we can convert from grams to moles, and then from moles to molecules.

First, we will find the molar mass of table sugar. The molar mass of carbon is 12.011 g/mol, hydrogen 1.008 g/mol, and oxygen 15.999 g/mol. Thus, the molar mass of table sugar will be:

[tex]12(12.011)+22(1.008)+11(15.999) = 342.297\text{ g/mol}[/tex]

To cancel units, we can write our ratio as:

[tex]\displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}[/tex]

With grams in the denominator.

And by definition:

[tex]\displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}[/tex]

Combining the two ratios and the starting value, we acquire:

[tex]3420 \text{ g C$_{12}$H$_{22}$O$_{11}$}\cdot \displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}[/tex]

Cancel like units:

[tex]=3420 \cdot \displaystyle \frac{1}{342.297}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1}[/tex]

Multiply:

[tex]\displaystyle = 60.1677... \times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}[/tex]

Rewrite:

[tex]\displaystyle = 6.01677... \times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}[/tex]

The resulting answer should have three significant digits. Hence:

[tex]=6.02\times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}}[/tex]

So, there are about 6.02 * 10²⁴ molecules of table sugar in 3.42 kilograms of the substance.

Answer:

2.69×10²³ molecules of PBr₅

6.02×10²⁴ molecules of C₁₂H₂₂O₁₁

Explanation:

To solve the first problem, we want to first find formula for phosphorus pentabromide, which is PBr₅. Now, we need to know the molar mass of PBr₅, which is about 430.49 g/mol. To get to molecules, we need to use Avogadro's number, which is 6.022×10²³ molecules/mol.

[tex]192g*\frac{1mol}{430.49g} *\frac{6.022*10^{23}molecules}{1mol} =2.69*10^{23} molecules[/tex]

Now, we know that there are about 2.69×10²³ molecules of PBr₅.

To solve the second problem, we need to use Avogadro's number, along with finding the molar mass of C₁₂H₂₂O₁₁, and converting kilograms to grams.

[tex]3.42 kg*\frac{1000g}{1kg} *\frac{1mol}{342.3g} *\frac{6.022*10^{23} molecules}{1mol} =6.02*10^{24} molecules[/tex]

Now, we know that there are about 6.02×10²⁴ molecules of C₁₂H₂₂O₁₁.

Answer:

[tex]\boxed {\boxed {\sf 63.0 \%}}[/tex]

Explanation:

The percent yield is the ratio of the actual yield to the theoretical yield.

[tex]percent \ yield = \frac{actual \ yield}{theoretical \ yield} * 100[/tex]

We can substitute the known values into the formula.

[tex]percent \ yield= \frac{ 14.8 \ g}{23.5 \ g}*100[/tex]

Divide.

[tex]percent \ yield = 0.629787234043*100[/tex]

Multiply.

[tex]percent \ yield = 62.9787234043[/tex]

The original measurements for the theoretical and actual yields have 3 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place.

The 7 to the right, in the hundredths place, tells us to round the 9 up to a 0. Since we rounded up to 0, we have to move to the next place to the left and round the 2 up to a 3.

[tex]percent \ yield \approx 63.0[/tex]

The percent yield is approximately 63.0 percent.

Explanation:

For a food with a relative humidity of 81, what is the water activity in the food?

Water activity in a food = relative humidity / 100

Substitute the given values in this formula to get water activity:

Water activity in a food = 81 / 100

Hence, water activity =0.81

Answer:

CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)

Explanation:

Chromium (II) chloride is a strong electrolyte, that is, when dissolved in water, it completely dissociates into the ions. The cation is chromium (II) and the anion is chloride. The balanced equation for the solution of chromium (II) chloride is:

CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)

Answer:

810 pm

Explanation:

Step 1: Given and required data

Step 2: Calculate the de Broglie wavelength of the hydrogen atom

We will use de Broglie's equation.

λ = h / m × v

λ = 6.63 × 10⁻³⁴ J.s / 1.67 × 10⁻²⁷ kg × 490 m/s = 8.10 × 10⁻¹⁰ m

Step 3: Convert 8.10 × 10⁻¹⁰ m to picometers

We will use the conversion factor 1 m = 10¹² pm.

8.10 × 10⁻¹⁰ m × 10¹² pm/1 m = 810 pm

Answer:

See explanation

Explanation:

a) The balanced chemical reaction equation is;

2 LiOH + K2S ------> Li2S + 2 KOH

b) If we include the states of matter then we have;

2 LiOH(aq) + K2S(aq) -----> Li2S(s) + 2 KOH(aq)

c) Complete ionic equation;

2Li^+(aq) + 2OH^-(aq) + 2K^+(aq) + S^+(aq) ----> Li2S(s) + 2K^+(aq) + 2OH^-(aq)

Net ionic equation;

2Li^+(aq) + S^2-(aq) -----> Li2S(s)

Answer:

matter

1. Candle. not matter

1. light

2. Unburned Particles

2. heat

3. Ash

3. energy

Explanation:

1.The candle is experiencing a solid phase into a gas phase because the heat given off causes smoke.

2. Chemical Potential Energy to Heat EnergyThe candle has Chemical Potential Energy then when it gets lit by the flame heat energy is released.

Answer:

The chlorine element belongs to group 17 because it has 7 valence electron . Its valency is 1 . It can gain one electron from any other atom to become stable. This means that it can never form a double or triple bond.

Answer:

1). 1-pentanol - Primary

2). 3-ethyl-3-pentanol - Tertiary

3). 2-hexanol - Secondary

4). Alcohol with two other carbons attached to the carbon with the hydroxyl group - Secondary

5). Alcohol with one other carbon attached to the carbon with the hydroxyl group - Primary

6). Alcohol with three other carbons attached to the carbon with the hydroxyl group - Tertiary

Explanation:

The distinct types of alcohol have been matched with the categories above as per their descriptions provided. In chemistry, alcohols have been categorized into three different categories namely primary, secondary, and tertiary.

In the primary type, those alcohols are involved in which there is an association of hydroxyl group to a primary atom of carbon along with a minimum of two atoms of hydrogen. Example; ethanol.

In the secondary type, the alcohols have an association of carbon atoms to hydroxyl with a single atom of hydrogen and has a formula of '-CHROH.' Example: 2 - propanol.

In the tertiary alcohols, here the association is between the hydroxyl group with the carbon atom that is saturated and possessing 3 atoms of carbon associated with it. It has a formula of '-CR2OH.' Example:  3-ethyl-3-pentanol, -tert -butyl alcohol, etc.

Answer:

absorption, fluorescence = phosphorescence

Explanation:

Given a particular fluorophore, the wavelength of absorption of the fluorophore is always shorter. Both fluorescence and phosphorescence are kinds of photoluminescence.

Recall that both fluorescence and phosphorescence occur at a longer wavelength. The difference between the two is only in the time taken during the process. While fluorescence takes a shorter time to occur, phosphorescence takes a longer time to occur.

The major difference between fluorescence and phosphorescence is that change of spin occurs during phosphorescence but not fluorescence.

Answer:

97.0%m/m es la concentración de la solución

Explanation:

El porcentaje masa/masa (%m/m) se define como 100 veces el radio entre la masa de soluto (300g de HCl) y la masa de la solución. Para hallar la masa de la solución debemos hallar la masa del agua (Solvente) haciendo uso de la ecuación del volumen de un cono. Con el volumen del cono podemos hallar la masa del agua haciendo uso de la densidad, así:

Volumen:

Volumen Cono = π*r²*h / 3

Donde r es el radio = 0.300m

h la altura = 5m

Volumen = π*(5m)²*0.300m / 3

7.85m³

Masa Agua:

7.85m³ * (1.2g / m³) = 9.42g Agua

Masa solución:

300g HCl + 9.42g Agua = 309.42g Solución

%m/m:

300g HCl  / 309.42g * 100 =

Answer:

0.000731 grams aluminium sulfate

46.0 mols ammonia

Explanation:

ALS = shorthand for aluminium sulfate which has a molar mass of 342.15 g/mol

[tex]ALS: \frac{0.250mols}{1} *\frac{1g}{342.15mols} = \frac{0.250g}{342.15}=0.0007307 g[/tex]

NH3 has a molar mass of 17.031 g/mol

[tex]NH3: \frac{2.70g}{1} *\frac{17.031mols}{1g} = \frac{0.250g}{342.15}=45.9837 mols[/tex]

Aluminium sulphate (AlS) whose molar mass is= [tex]\sf{ 342.15\dfrac{g}{mol} }[/tex]

we have to find the 0.250 moles of aluminum sulphate.

[tex]\implies AlS=\dfrac{1g}{342.15~mole}×0.250~mole \\\\\implies AlS=\dfrac{0.250}{342.15}\\\\\implies \dfrac{\frac{250}{1000}}{\frac{34215}{100}}\\\\\implies \dfrac{250}{1000}×\dfrac{100}{34215}\\\\=0.00073067\approx{0.0007307~g}[/tex]

[tex]\\\\\\[/tex]

Ammonia(NH3) whose molar mass is =[tex]\sf{17.031\dfrac{mol}{g} }[/tex]

We have to find 2.70 grams of ammonia

[tex]\implies NH_{3}=\dfrac{17.031~mol}{1g}×2.70g\\\\ 17.031×2.70\\\\\dfrac{17031}{1000}×\dfrac{270}{100}\\\\ \dfrac{4598370}{100000}\\\\=45.9837\approx{46~mole}[/tex]

Answer:

1360kJ are evolved

Explanation:

When 1mole of H2 reacts with 1/2 moles O2 producing 1 mole of water and 241.8kJ.

To solve this question we need to find the limiting reactant knowing were added 90g of H2 and 90g of O2 as follows:

Moles H2 -Molar mass: 2g/mol-

90g H2 * (1mol / 2g) = 45 moles

Moles O2 -Molar mass: 32g/mol-

90g * (1mol / 32g) = 2.81moles

For a complete reaction of 2.81 moles of O2 are needed:

2.81 moles O2 * (1mol H2 / 1/2 mol O2) = 5.62 moles H2

As there are 45 moles, H2 is the excess reactant and O2 the limiting reactant.

As 1/2 moles O2 produce 241.8kJ, 2.81 moles will produce:

2.81 moles O2 * (241.8kJ / 1/2moles O2) =

The quantity of heat evolved when 180 g mixture containing equal parts of H₂ and O₂ burned is

The equation for the combustion of hydrogen is given as:

[tex]\mathbf{H_2 + \dfrac{1}{2}O_2 \to H_2O \ \ \ \ \Delta H_r^0 = -241.8\ kJ/mol}[/tex]

Recall that:

Since the mass of 180 g is equally shared by H₂ and O₂, then:

The number of moles of the reactant can be determined as follows:

For H₂:

[tex]\mathbf{no \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]

[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{2.016 \ g/mol}}[/tex]

no of moles = 44.6 mol

For O₂:

[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{32 \ g/mol}}[/tex]

no of moles = 2.8 mol

Here, since O₂ has a lesser amount of mole, then O₂ is regarded as the limiting reagent here:

If 1/2 moles of O₂ produces -241.8 kJ/mol of water;

Then, the quantity of heat that will evolve when 180 g mixture containing equal parts of H₂ and O₂ burned is:

[tex]\mathbf{= \Big (\dfrac{2.81 \ mol}{\dfrac{1}{2 } \ mol }\Big) \times (-241.8 \ kJ) }[/tex]

[tex]\mathbf{= \Big (5.62\Big) \times (-241.8 \ kJ) }[/tex]

= - 1358.91 kJ

≅ - 1360 kJ

Therefore, we can conclude that the quantity of heat evolved is - 1360 kJ

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Answer:

2AgNO, K Cro-> 2KNO; 1Cro,

The Environmental Protection Agency (EPA) refrigerant management regulations would exempt any refrigerant from venting problem when it determines that the refrigerant in an appliance do not pose a threat to the environment (surrounding) if released.

A refrigerant can be defined as any chemical substance that undergoes a phase change (liquid and gas) so as to enable the cooling and freezing of materials. They are typically used in air conditioners, refrigerators, water dispensers, etc.

In the United States of America, the Environmental Protection Agency (EPA) is a governmental agency which was established by U.S Congress and it is saddled with the responsibility of overseeing all aspects of pollution, environmental clean up, degradation, pesticide use, contamination, and hazardous waste spills. Also, EPA research solutions, policy development, and enforcement of regulations through the resource Conservation and Recovery Act.

Simply stated, the Environmental Protection Agency (EPA) is the governmental agency set up to ensure that various industries, factories and people comply with laws and regulations concerning the environment.

In conclusion, appliances such as a refrigerator, air conditioner (AC), etc., whose refrigerants do not pose a threat (potentially cause damage) to the environment if released are typically exempted by the EPA's refrigerant management regulations.

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Answer:

R.F.M of Iron (II) oxide :

[tex]{ \tt{ = (56 \times 2) + (16 \times 3)}} \\ = 160 \: g[/tex]

Moles :

[tex]{ \tt{ \frac{35.2 \times {10}^{ - 23} }{160} }} \\ = 2.2 \times {10}^{ - 24} \: moles[/tex]

Molecules :

[tex]{ \tt{ = 2.2 \times {10}^{ - 24} \times 6.02 \times {10}^{23} }} \\ = 1.3244 \: molecules[/tex]

The number of molecules of Iron(II) oxide present in 35.2 ×10⁻²³ g of Iron(II) oxide is equal to 2.95.

Avogadro’s number can be described as the proportionality constant that is used to represent the number of entities or particles in one mole of any substance. Generally, it is used to count atoms, molecules, ions, electrons, or protons, depending upon the chemical reaction or reactant and product.

The value of Avogadro’s constant can be represented as numerically approximately equal to 6.022 × 10²³ mol⁻¹.

Given, the mass of the iron oxide = 35.2 ×10⁻²³ g

The molar mass of the Iron(II) oxide, FeO = 71.84 g/mol

71.84 g of Iron (II) oxide have molecules = 6.022 × 10²³

35.2 ×10⁻²³ g of FeO have molecules = 6.022 × 10²³ × (35.2 ×10⁻²³ /71.84)

The number of molecules of FeO in a given mass = 2.95 molecules

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Answer:

[tex]NCl_3[/tex]

Explanation:

An octet rule is a thumb rule in the chemical sciences in which there is a natural tendency for an atom to prefer eight electrons in the valence shell of the atom. When there are less than eight electrons in the atom, they react with other atoms and form more stable compounds.

In the context, nitrogen trichloride, [tex]NCl_3[/tex], is an example of molecular compound which obeys the octet rule having a zero formal charges on each atom of the compound.

Answer:

the constant temperature will be 12 .F bec in places it is cold

Answer:

0.030 mole

Explanation:

Mass of crucible + lid = 23.422 g

Mass of crucible + lid + compound = 24.746 g

Mass of crucible + lid + compound - water = 24.213

Mass of water = Mass of crucible + lid + compound + heat

       = 24.746 - 24.213

                = 0.533 g

Mole of water in the hydrated compound = mass of water in the compound/molar mass of water

    = 0.533/18

         = 0.0296 mole = 0.030 mole

Answer:

A. 3-chloro-1-methylcyclobutane.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to infer that the name of this compound is A. 3-chloro-1-methylcyclobutane because of the fact that the parent chain is a cyclobutane which starts by the methyl radical as it has the priority over the chlorine radical which is actually named first at the third carbon (clockwise).

Therefore the name is given in A, accordingly to the IUPAC rules of nomenclature.

Regards!

Explanation:

here's the answer to your question

Explanation:

example is copper iron...........