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Answer:

Gypsum

Explanation:

Calcium sulphate, is a naturally occurring calcium salt. It is commonly known in its dihydrate form, a white or colourless powder called gypsum.

Answer:457

Explanation:

The probability that the glucose concentration in this sample is less than 5.6 mg/dl.

0.32

Generally, Probability is the measure of the likelihood of an event occurring. It is expressed as a number between 0 and 1, where 0 indicates that the event is impossible and a 1 indicates that the event is certain.

The probability that the glucose concentration in this sample is less than 5.6 mg/dl

= (0.4*4 + 0.6*8)/(0.4*4 + 1*8 + 0.2*15 + 1*7 + 0.4*1)

= 6.4/20

= 0.32

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Answer:

one parent

Explanation:

As one parent cell divides it into two daughter cells and so on.

The concentration of iron used in the titration : 0.009 M

Given

Reaction

Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ ⇒ 2Cr³⁺ + 6Fe³⁺ + 7H₂O

0.0093 mol/L potassium  dichromate

200 cm³ of dilute acid, 25cm³   was used in the titration.

Required

the concentration of iron

Solution

Titration formula

C₁V₁n₁=C₂V₂n₂⇒ From equation : n₁=6n₂(1=Cr₂O₇, 2=Fe)

titration average : 33+32.05+32.15+32.1 / 4 = 32.325 cm³(ml)

25 cm³ of iron solution used in titration :

\(\tt V_1=32.325~ml\\\\V_2=25~ml\\\\C_1=0.0093~M\\\\\\\\C_1.V_1.n_2=C_2.V_2.n_1\\\\0.0093\times 32.325\times 6=C_2\times 25\times 1\\\\C_2=0.07215~M\)

Dilution(25 ml from 200 ml iron solution)

\(\tt C_2.V_2=C_1.V_1\\\\0.007215\times 25=C_1\times 200\\\\C_1=0.009~M\)

Potash alum, represented by the chemical formula KAl(SO4)2·12H2O, is a double salt because it consists of two different salts, potassium sulfate (K2SO4) and aluminum sulfate (Al2(SO4)3), combined together. It forms a unique crystalline structure with 12 water molecules incorporated in its composition.

Potash alum is a type of alum that is commonly used in dyeing and water purification applications. The chemical formula for alum is KAl(SO4)2·12H2O.

The term "double salt" refers to the fact that the compound is made up of two different types of ions that are combined together.
In the case of potash alum, the two different types of ions are potassium ions (K+) and aluminum ions (Al3+). These two ions are combined together in a 1:1 ratio to form a complex ion, KAl(SO4)2, which is then combined with 12 water molecules to form the final compound, KAl(SO4)2·12H2O.
The term "double salt" is used to describe this type of compound because it contains two different types of cations (positive ions) that are combined together to form a single crystal lattice structure.

The two different cations are held together by ionic bonds, which are relatively strong and help to give the compound its characteristic properties.
Potash alum is a particularly useful compound because it has a number of important properties that make it useful in a variety of applications.

For example, it is highly soluble in water, which makes it a good choice for use in water purification applications.

It is also relatively stable and non-toxic, which makes it safe to use in a variety of different settings.

Finally, it has a number of useful physical properties, such as its ability to form crystals that are highly reflective and have a characteristic octahedral shape.

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Answer:

The water cycle is often taught as a simple circular cycle of evaporation, condensation, and precipitation. Although this can be a useful ...

Explanation:

Answer: Your answer will be D)

Explanation: Hope this helped

In terms of oxidation number, oxidation can be defined as the chemical change in which there occurs an increase in the oxidation number of an atom. Reduction may be defined as a chemical change in which there occurs a decrease in the oxidation number of an atom. The correct option is D.

An oxidizing agent is a substance which suffers a decrease in the oxidation number of one or more of its elements and a reducing agent is a substance which suffers an increase in the oxidation number of one or more of its elements.

In the reaction, oxidation state of 'Ca' increases from 0 to +2, so calcium is oxidized and acts as a reducing agent.

Thus the correct option is D.

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The volume of the gas will be 560 mL when the pressure is reduced to 125.0 kPa, assuming the temperature remains constant.

Boyle's law simply states that "the volume of any given quantity of gas is inversely proportional to its pressure as long as temperature remains constant.

Boyle's law is expressed as;

P₁V₁ = P₂V₂

Where P₁ is Initial Pressure, V₁ is Initial volume, P₂ is Final Pressure and V₂ is Final volume.

Substituting the given values, we get:

P₁ = 350.0 kPa (initial pressure)

V₁ = 200 mL (initial volume)

P₂ = 125.0 kPa (final pressure)

V₂ = ?

Solving for V₂, we get:

V₂ = ( P₁ × V₁ ) / P₂

V₂ = (350.0 kPa × 200 mL) / 125.0 kPa

V₂ = 560 mL

Therefore, the final volume of the gas is  560 mL.

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This experiment shows that the elements of pure solid substance in a system closed off from the surroundings has increased temperature very fast.

Pure substances are substances that are built up of only one kind of particle and have a fixed or abiding structure. Pure substances are further confidential as elements and compounds. An element is a substance that comprises only one type or kind of atom. a pure substance has a constant chemical composition.

No affair where you sample a substance, it is the same. For chemistry, the safest examples of pure substances are elements and compounds. So, examples cover gold, silver, helium, sodium chloride, and pure water.

So we can conclude that In chemistry, a pure substance is an element or compound made up of one type of particle.

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Answer: true

Explanation:

Thermal efficiency of process heating equipment

A chemist determines that a substance is composed of 30.4% nitrogen by mass and 69.6% oxygen by mass. The molar mass of the compound is 230.5 g/mol, the molecular formula is NO₂.

We must compute the empirical formula in order to ascertain the compound's chemical composition.

If we have 100 grams of the compound.

This suggests we have:

30.4 g of nitrogen

69.6 g of oxygen

Now, we have to convert the mass of each element to moles.

The molar mass of nitrogen (N) = 14.01 g/mol

the molar mass of oxygen (O)  =  16.00 g/mol.

Number of moles of nitrogen (N):

2.17 mol

Number of moles of oxygen (O):

4.35 mol

The simplest whole-number ratio between the moles of nitrogen and oxygen must now be determined. To calculate the ratio, we divide both numbers by the smaller value.

Moles N / moles O = 2.17 mol / 2.17 mol = 1.00

Moles O / moles O = 4.35 mol / 2.17 mol = 2.00

The ratio is approximately N₁O₂.

We divide the subscripts by their greatest common divisor to obtain the simplest ratio, since we are looking for the empirical formula. The empirical formula is NO₂ since the greatest common divisor in this situation is 1.

The molecular formula of the compound is NO₂.

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Answer:

41826.05 Calories

Explanation:

1 J = 0.239006 Calories

175 KJ

= 175 x 1000 J

= 175000 J

175000 J to Calories

= 175000 x 0.239006

= 41826.05 Calories

Answer: 41.8

Explanation:

Acellus verified ✅

Answer:

ρ = 2.07 g/mL

Explanation:

Step 1: Given data

Step 2: Calculate the density (ρ) of the block G

The density is an intrinsic property, equal to the mass of the block divided by its volume.

ρ = m/V

ρ = 49.231 g/23.8 mL

ρ = 2.07 g/mL

The density of the block G is 2.07 g/mL.

Answer:

Explanation:

1.1 we have to find mass of Na3PO4;

for that we have to Calculate the moles of Na3PO4 needed:

volume is 100mL = 0.1L

Molar concentration = Moles of solute / Volume of solution in L

0.1 mol/L = Moles of Na3PO4 / 0.1 L

Moles of Na3PO4 = 0.1 mol/L * 0.1 L

Moles of Na3PO4 = 0.01 mol

Now, Calculate the mass of Na3PO4 needed:

so, Mass = Moles of Na3PO4 * Molar mass of Na3PO4

Mass = 0.01 mol * 164 g/mol

Mass = 1.64 g of Na3PO4.

1.2 materials and glassware needed:

1.64 g Sodium phosphate (Na3PO4)

100 mL volumetric flask

weighing balance

Distilled water

Glass rod

Pipette and burette

Answer:

Pressure goes up

Explanation:

The pressure is inversely proportional to the volume.  So when volume is decreased the pressure rises.

The molecules are getting pushed closer together and more collisions occur which will also increase the temperature.

Answer:

Explanation:

The amount of heat that must be added to heat 50 grams of water from 42.0° C to 100.0° C can be calculated using the specific heat capacity formula: Q = m*C*ΔT, where Q is the heat required, m is the mass of the water, C is the specific heat capacity of water, and ΔT is the temperature change.

Using the values given, we have:

Q = 50 g * 1 cal/g °C * (100.0°C - 42.0°C)

Q = 50 g * 1 cal/g °C * 58.0°C

Q = 2,900 calories

Therefore, the amount of heat that must be added to heat 50 grams of water from 42.0° C to 100.0° C is 2,900 calories. The answer is option (C).

The structural features which contributes to a high thermal conductivity are the direction of travel of heat and the nature of materials.

This is referred to as a measure of its ability to conduct heat or current and it is dependent on some factors such as the nature of the materials.

For example , metals conduct heat and electricity easily because the materials allow free electrons to move between the atoms and also the direction of travel of heat which is towards them is also a factor in this regard.

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Answer:

Explanation:

The plant cell has a cell wall, chloroplasts, plastids, and a central vacuole—structures not found in animal cells. Plant cells do not have lysosomes or centrosomes.

Approximately 194.0 grams of glucose (C6H12O6) would be required to prepare a 5000 mL solution with a concentration of 0.215 M.

To determine the mass of glucose (C6H12O6) required to prepare a 0.215 M solution in 5000 mL, we need to use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

First, let's convert the volume of the solution from milliliters (mL) to liters (L):

5000 mL = 5000/1000 = 5 L

Now, we can rearrange the formula to solve for moles of solute:

moles of solute = Molarity (M) x volume of solution (L)

moles of solute = 0.215 M x 5 Lmoles of solute = 1.075 mol

Since glucose (C6H12O6) has a molar mass of approximately 180.16 g/mol, we can calculate the mass of glucose using the equation:

mass of solute = moles of solute x molar mass of solute

mass of glucose = 1.075 mol x 180.16 g/mol

mass of glucose = 194.0 g (rounded to three significant figures)

Therefore, approximately 194.0 grams of glucose (C6H12O6) would be required to prepare a 5000 mL solution with a concentration of 0.215 M. It's important to note that the molar mass of glucose used in this calculation may vary slightly depending on the level of precision required.

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Answer:

32.8g/mole

Explanation:

Given parameters:

Mass of sample of gas = 32.8g

Volume  = 22.4L

Unknown:

Molecular weight  = ?

Solution:

To solve this problem we must understand that at rtp;

             1 mole of gas occupies a volume of 22.4L

Number of mole of the gas  = 1 mole

 Now;

   Mass  = number of moles x molecular weight

    molecular weight  = \(\frac{mass}{molecular weight}\)    = \(\frac{32.8g}{1mole }\)      = 32.8g/mole

The wavelength of light that should be employed for nickel(II) ion is roughly 400–500 nm, according to the absorbance spectra displayed. This is due to the graph's observation of an absorbance peak with a maximum of roughly 0.8 in this region.

The plot of a substance's absorbance as a function of light wavelength is called the absorbance spectrum. The absorbance spectrum in this instance is for the nickel(II) ion.

The peak in absorbance shows that the nickel(II) ion's electrons are most easily moved from a lower energy state to a higher energy one by this particular wavelength of light. As a result, more light is absorbed, increasing the measured absorbance.

Depending on the particular experiment and the tools being used, a specific wavelength of light should be employed. However, judging by the absorbance spectrum displayed, measuring nickel(II) ion absorbance should be possible at a wavelength of roughly 400–500 nm.

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The "Rate=k" rate law constitutes a first-order reaction.

Half-life is the time required for the concentration of a reactant or product in a chemical reaction to decrease to half of its initial value. The concept of half-life is commonly used in nuclear chemistry and radioactivity, but it is also applicable to other types of reactions. The half-life of a reaction depends on the order of the reaction, which describes the relationship between the rate of the reaction and the concentration of the reactants. In a first-order reaction, the half-life is constant and is independent of the initial concentration of the reactant.

For a general nth order reaction "A → products", the rate law is given by:

Rate = k[A]^n

The half-life of an nth order reaction is written as follows:

t1/2 = [ 2^(n-1) / k[A]^(n-1) ]

We can derive this expression using the integrated rate law for an nth order reaction:

ln([A]t/[A]0) = -kt^(1-n)/(1-n)

where [A]t represents the concentration of A at time t and [A]0 represents the initial concentration of A.

For a half-life, we know that [A]t = 0.5[A]0 and t = t1/2. When these values are interchanged into the integrated rate law, we get:

ln(0.5) = -kt1/ 2ⁿ⁻¹ /(1-n)

Rearranging this equation and solving for t1/2 gives:

t1/2 = [ 2ⁿ⁻¹  / k[A]ⁿ⁻¹ ]

Therefore, the half-life for an nth order reaction can be written as t1/2 = [ 2ⁿ⁻¹ / k[A]ⁿ⁻¹ ] where k is the rate constant and [A] is the concentration of the reactant.

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Taking into account the animals and the landscape being mentioned, we can confirm that the combination of these forms part of an ecosystem.

An ecosystem is described as an area consisting of:

and many other factors, which all work together to form a small pocket of life.

Ecosystems are described as containing both living and non-living elements, meaning biotic and non-biotic parts. A community or population, on the other hand, are used to consider only the biotic or living factors of an ecosystem. Therefore, since the question includes the rock bottom (a non-biotic factor) as a part of the makeup, the only correct grouping is that of an ecosystem.

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In the given reaction, the limiting reactant is C₆H₆ (benzene).

To determine the limiting reactant as well as calculate the mass of HCl produced, compare the moles of each reactant.

The number of moles for each reactant:

Molar mass of Cl₂ = 35.5 g/mol + 35.5 g/mol = 71 g/mol

Moles of Cl₂ = mass of Cl₂ / molar mass of Cl₂

                     = 45 g / 71 g/mol

                     = 0.634 moles of Cl₂

Molar mass of C₆H₆ (benzene) = 12 g/mol + 6(1 g/mol) = 78 g/mol

Moles of C₆H₆ = mass of C₆H₆ / molar mass of C₆H₆

                        = 45 g / 78 g/mol = 0.577 moles of C₆H₆

Determine the stoichiometry between Cl₂ and HCl:

Cl₂ + C₆H₆ → C₆H₅Cl + HCl

Here, we can see that 1 mole of Cl₂ produces 1 mole of HCl.

Thus, the limiting reactant is C₆H₆ (benzene).

Calculate the mass of HCl produced:

Molar mass of HCl = 1 g/mol + 35.5 g/mol = 36.5 g/mol

Moles of HCl produced = moles of C₆H₆ = 0.577 moles

Mass of HCl produced = moles of HCl produced × molar mass of HCl

Mass of HCl produced = 0.577 moles × 36.5 g/mol

                                     ≈ 21.04 g

Therefore, approximately 21.04 grams of HCl will be produced.

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Answer:

The molar mass of Ca(OH)2 is 74.092 g/mol

Explanation:

First your going to find the atomic mass of each element on the periodic table.

Ca(Calcium) = 40.078

O(Oxygen) = 15.999

H(Hydrogen) = 1.008

Then we can create the equation:

40.078 + (15.999 + 1.008) × 2

40.078 + (17.007) × 2

40.078 + 34.014

= 74.092 g/mol

Answer:

74 gmol-1

Explanation:

Molar masses

Ca= 40

O = 16

H = 1

Molar mass of Ca(OH)2 = 40+16*2+1*2

= 40+32+2

=74