Os etapas que devem sem compridas para obter separadamente os componentes (a as substâncias ) de uma mistura de água ,sal e areia ,são respectivamente : A) filtração e sifonação B) decantação e sifonação C) filtração e destilação simples D) catação e decantação

The answer is evaluating the reliability of your experiment.

(By the way, I love your profile picture)

After you have finished a research experiment and come to a conclusion, Create a graph should your next step Option C is correct.

Experimental research of any physical or chemical thing  is research conducted with a scientific theories using two sets which are variables.

The first set works as a constant and cannot be changed  use to measure the differences of the second set which is a variable. Quantitative research like if  you don’t have enough data to make results  you must first learn about the facts.

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Answer:

See explanation

Explanation:

Now , we have the equation of the reaction as;

2H2S(g) + 302(g)------->2SO2(g) + 2H2O(g)

This equation shows that SO2 gas is produced in the process. Let us recall that this same SO2 gas is the anhydride of H2SO4. This means that it can dissolve in water to form H2SO4

So, when SO2 dissolve in rain droplets, then H2SO4 is formed thereby lowering the pH of rain water. This is acid rain.

Answer:

protons, neutrons and electrons are the sub atomic particles

Answer:

30000

Explanation:

6/100×500000 =30000

If the temperature increases, then the concentration of sodium hydroxide (NaOH) will decrease because the equilibrium will shift in the backward direction.

Sodium hydroxide is an ionic compound consisting of Na⁺ and OH⁻. It is also called as caustic soda. It is an inorganic compound with chemical formula NaOH.

When NaOH dissolve in water it is exothermic in nature. As the temperature increases, the equilibrium will shift in the backward direction.Therefore,as to nullify the effect of increased temperature. It will decrease the solubility of NaOH in water.

Thus, If the temperature increases, then the concentration of sodium hydroxide (NaOH) will decrease.

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Explanation:

K = °C + 273

1. -234 + 273 = 39K

2. 84 + 273 = 357K

3. 70 + 273 = 343K

4. -24 + 273 = 249K

5. 134 + 273 = 407K

6. 120 + 273 = 393K

-234 °C = 39K

84 °C = 357K

70 °C = 343K

-24 °C = 249K

134 °C = 407K

120 °C = 393K

Answer:

-234°C = 39K

84°C = 357K

70°C = 343K

-24°C = 249K

134°C = 407K

120°C = 393K

Explanation:

Hope it helps:)

Answer:

yea hold them in your hand for a while and your hand will look like a rainbow

Explanation:

Answer:

I could be wrong but I believe it's 99.9

An acid HX is 15 % dissociated in the water. If the equilibrium concentration of the HX is 0.39 M,  the ka value for HX is 0.0116.

The percentage of the dissociation = 15 %

The amount dissociated = a × 15 / 100

The amount dissociated = 0.15a

The chemical equation is as :

HX    ⇄   H⁺   +   A⁻

a              0          0

a - 0.15a    0.15a    0.15a

The equilibrium concentration HX = 0.39 M

a - 0.15a = 0.85 a = 0.39 M

a = 0.45 M

The expression for the ka :

ka = ( 0.15 a)² / (a - 0.15a)

ka = ( 0.15 × 0.45)² / 0.39

ka = 0.0116

The value for the ka is 0.0116.

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The sprinter's speed after 2 seconds is 9 m/s.

Use the equation of motion to solve this problem:

v = u + at

where

v = final velocity (what we need to find)

u = initial velocity (0 m/s for a stationary object)

a = acceleration (4.5 m/s²)

t = time (2 s)

Plugging in the values:

v = 0 + (4.5 m/s²) × (2 s)

v = 9 m/s

Therefore, the sprinter's speed 2 seconds later is 9 m/s.

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Complete question:

A sprinter leaves the starting blocks with an acceleration of 4.5m/s^2. What is the sprinter's speed 2 s later?

Pitt inorganic chemistry refers to the study of inorganic chemistry at the University of Pittsburgh.

Inorganic chemistry is a branch of chemistry that deals with the study of inorganic compounds and their reactions. It is concerned with the properties and behavior of inorganic compounds, which are compounds that do not contain carbon-hydrogen bonds, such as metals, salts, and minerals.

The Department of Chemistry at the University of Pittsburgh offers courses and research opportunities in the field of inorganic chemistry, including the synthesis and characterization of inorganic compounds, the study of their properties and reactivity, and the development of new materials and catalysts.

Inorganic chemistry is an important field of study because it helps us to understand the properties and behavior of a wide range of compounds that are essential for many industrial processes, such as the production of metals, ceramics, and semiconductors. It is also important for the development of new materials and technologies, such as batteries, fuel cells, and solar cells.

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Consider the following reaction:4 X(s) + 3 Y2(g) ⇌ 2 X2Y3(s)The equilibrium constant expression for the given reaction is:Kc = [X2Y3] / [X]4 [Y2]3Here, the initial moles are: [X] = 4.62 mol, [Y2] = 0.612 mol, [X2Y3] = 6.18 mol and volume, V = 3.9 L.

Let's calculate the concentration of each substance:[X] = 4.62 mol / 3.9 L = 1.185 M[Y2] = 0.612 mol / 3.9 L = 0.157 M[X2Y3] = 6.18 mol / 3.9 L = 1.585 M.

Now substitute the given values in the expression of Kc:Kc = [X2Y3] / [X]4 [Y2]3Kc = 1.585 / (1.185)4 (0.157)3Kc = 0.684 M-1Therefore, the value of Kc for this reaction is 0.684 M-1 if 4.62 mol X, 0.612 mol Y2, and 6.18 mol X2Y are present in a 3.9-L container at equilibrium.

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Answer:

Explanation:

Divide each weight by their respective element weights.

Na\(_2\)S\(_2\)O\(_3\) is the empirical formula for the compound that has 46 grams of sodium, 64 grams of sulfur, and 48 grams of oxygen.

In chemistry, the equation of maybe a chemical compound is the simplest complete number percentage of atoms that make up a molecule. The empirical formulas for sulfur monoxide, SO, and disulfur dioxide, S2O2, are two simple instances of this concept.

They do not, nonetheless, share comparable molecular formulas, which indicate how many elements are contained in each chemical compound's molecules. An empirical formula does not specify the arrangement or number of atoms. Several polymers, including such silicon dioxide, nor ionic compounds, including such calcium chloride (CaCl2), have this property (SiO2).

moles of sodium = 46/23=2

moles of sulfur = 64 / 32=2

moles of oxygen =48 / 16 =3

Na\(_2\)S\(_2\)O\(_3\)  is the empirical formula

Therefore, Na\(_2\)S\(_2\)O\(_3\) is the empirical formula.

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Answer:

4.70 grams of NH4Cl is needed to prepare 350 mL of a 0.25 M ammonium chloride solution.

We need approximately 4.68 grams of NH4Cl to prepare a 0.25 M ammonium chloride solution with a volume of 350 mL.

To determine the mass of NH4Cl needed to prepare the solution, we us use the formula:

m=M x V x MM ... (i)

where,

m= mass in grams

M=molarity of solution

MM= molar mass of compound

V= volume in litres

The number of moles of NH4Cl needed can be calculated using:

  Moles = Molarity x Volume ...(ii)

  Moles = 0.25 mol/L x 0.350 L

  Moles = 0.0875 mol

Hence we can replace M x V with number of moles in equation i.

The molar mass of NH4Cl is :

  Molar mass of NH4Cl = (1 x 14.01 g/mol) + (4 x 1.01 g/mol) + (1 x 35.45 g/mol)

  Molar mass of NH4Cl = 53.49 g/mol

We have all the variables

Putting them in equation i.

Hence,

  Mass (g) = Moles x Molar mass

  Mass (g) = 0.0875 mol x 53.49 g/mol

  Mass (g) = 4.68 g

Therefore, you would need approximately 4.68 grams of NH4Cl to prepare a 0.25 M ammonium chloride solution with a volume of 350 mL.

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Both calcium and magnesium reacts with water form their corresponding hydroxides along with the evolution of hydrogen gas. Calcium forms calcium hydroxide whereas magnesium forms magnesium hydroxide with water. But calcium reacts more violently with water.

The reactivity of metals depends on their capacity to lose electrons. The nucleus of calcium is farther away from the valence electrons and thus exert a smaller electrostatic pull on those electrons than the nucleus of magnesium.

Reactivity depends on ability to lose electrons. So calcium can lose electrons more easily and hence it is more reactive.

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Aluminum is used to make the metal cylinder.

To determine the substance of the metal block, we need to calculate its specific heat capacity, which is the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius.

The equation for calculating the heat absorbed by a substance is:

q = m * c * ΔT

where q is the heat absorbed, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

In this case, we have:

m = 500.0 g (given)

ΔT = 50.0 K (given)

q = unknown

To find c, we can rearrange the equation to:

c = q / (m * ΔT)

We know that the metal block absorbed some amount of heat, which we can calculate by multiplying the mass of the block by the specific heat capacity of water (which is 4.184 J/g*K) and the temperature change:

q = (500.0 g) * (4.184 J/g*K) * (50.0 K) = 104,600 J

Now we can calculate the specific heat capacity of the metal block:

c = q / (m * ΔT) = (104,600 J) / (500.0 g * 50.0 K) = 4.184 J/g*K

This specific heat capacity is very close to that of aluminum (which is 0.903 J/g*K) and therefore, we can conclude that the metal block is made of aluminum.

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Answer: Cyclohexylamine is a stronger base than aniline because in aniline electron pair is involved in conjugation, which makes the electron pair unavailable, where as in cyclohexyl amine, the −NH

2

​

 group is out side the ring and there is no resonance in the ring also, thereby the lone pair present on the nitrogen is freely available, resulting in a stronger base.

Explanation: I dont have one...

Answer:

Two or more!

Explanation:

Answer: Answer to #1 is 3A and ns2np1

Answer to #2 is 4A and ns2np2

Explanation:Group number is the number of valance electrons and generic outer electron configuration has to add up to equal the number of valance electrons

The element in the first diagram is in group 13 while the element in the second diagram is in group 14.

The Lewis structure of an atom of an element is a representation of the number of valence electrons on the outermost shell of the atom of that element. Every group of elements have a generic electron configuration which represents the number of valence electrons contained by atoms of elements in that group.

In the first diagram, the element X belongs to group 3A(13) and has a generic outer electron configuration, ns2np1. In the second diagram, the element X is in group 4A(14) and has outer electron configuration ns2np2.

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Answer:

It is a Binary Compound

Explanation:

Answer:

[binary compound]

Explanation:

A(n) A binary compound is one that contains two, and only two, elements.

Answer:

Explanation:

Whole formula equation:

Cl₂(g) + 2NaBr(aq) → 2NaCl(aq) + Br₂(g)

We can expand this to get the full ionic equation:

Cl₂(g) + 2Na⁺(aq) + 2Br⁻(aq) → 2Na⁺(aq) + 2Cl⁻(aq) + Br₂(g)

Cancelling out spectator ions (ions that don't change) on both sides:

Cl₂(g) + 2Na⁺(aq) + 2Br⁻(aq) → 2Na⁺(aq) + 2Cl⁻(aq) + Br₂(g)

And our final result, is the net ionic equation:

Cl₂(g) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(g)

The weight percent of copper in bornite is approximately 63.316%.

The weight percent of copper (Cu) in bornite (Cu5FeS4) can be calculated by considering the atomic masses of copper (Cu), iron (Fe), and sulfur (S) and using the formula:

\(\[\text{{Weight percent of Cu}} = \frac{{\text{{Atomic mass of Cu}} \times \text{{Number of Cu atoms}}}}{{\text{{Formula mass of Cu5FeS4}}}} \times 100\%\]\)

Given that the atomic mass of Cu is 63.546 g/mol, the atomic mass of Fe is 55.845 g/mol, the atomic mass of S is 32.065 g/mol, and the formula mass of Cu5FeS4 is 501.849 g/mol, we can substitute these values into the formula:

\(\[\text{{Weight percent of Cu}} = \frac{{5 \times 63.546}}{{501.849}} \times 100\%\]\)

Simplifying the calculation gives:

\(\[\text{{Weight percent of Cu}} = 63.316\%\]\)

Therefore, the weight percent of copper in bornite is approximately 63.316%.

To calculate the gross value of the mining operation, we multiply the weight of the ore body (20,000,000 tons) by the cost per ton ($85):

\(\[\text{{Gross value}} = \text{{Weight of ore body}} \times \text{{Cost per ton}}\]\)

\(\[\text{{Gross value}} = 20,000,000 \times 85 = \$1,700,000,000\]\)

The expenses for the mining operation can be calculated by multiplying the weight of the ore body (20,000,000 tons) by the cost per ton ($85):

\(\[\text{{Expenses}} = \text{{Weight of ore body}} \times \text{{Cost per ton}}\]\)

\(\[\text{{Expenses}} = 20,000,000 \times 85 = \$1,700,000,000\]\)

The net value (profit or loss) of the mining operation can be obtained by subtracting the expenses from the gross value:

\(\[\text{{Net value}} = \text{{Gross value}} - \text{{Expenses}}\]\)\(\[\text{{Net value}} = \$1,700,000,000 - \$1,700,000,000 = \$0\]\)

Therefore, the net value of this mining operation is zero, indicating that there is neither profit nor loss.

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Answer:

New growth of trees is an exception

Explanation:

when volcanoes erupt, they release hot magma that is destructive to the environment causing personal damage, lack of breathable air and death.

heat produce cannot in any way help in growth of trees

The correct name for HBrO is hypobromous acid. The reasoning behind this answer is based on the naming rules for acids.

When an acid has a hydrogen ion (H+) combined with a polyatomic ion that ends in -ite, the acid is named with the prefix "hypo-" and the suffix "-ous" added to the root of the element name, followed by the word "acid." In the case of HBrO, the polyatomic ion is BrO-, which is the bromite ion. Therefore, the correct name for this acid is hypobromous acid.

In general, the naming rules for acids are as follows:

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A balanced chemical equation is where the number of reactants is equal to several products.

A balanced chemical equation is one in which the number if reactants is equal to number of products.

1) Zn + 2AgCl → ZnCl₂ + 2Ag

2) FeCl₂ + 2HbS →  FeCl₂ + 2HCl + 2S

3) 2Ag₂O   →   4 Ag + 2O₂

4)3AlO₂     →  3Al + 3 O₂

5) K₂   + 2HbO → 2KOH + 2Hb

6) 3CaCO₂  →  3CaO + 3CO₂

7) 2Mg + O₂     → 2MgO

8) 2CO + 2FeO₂  →  2Fe + 2CO ₂

Therefore, A balanced chemical equation is where the number of reactants is equal to several products.

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If a solution originally 0.532 m in acid ha is found to have a hydronium concentration of 0.112 m at equilibrium, what is the percent ionization of the acid is 21.1%.

When an atom, molecule, or other material loses or acquires one or more electrons, the process is known as ionization, which produces charged particles known as ions. Chemical reactions, exposure to ionizing radiation, or other physical processes can all result in this process. Ionization is the process by which an acid molecule contributes a proton (H+) to a water molecule, resulting in the formation of a hydronium ion (H3O+).

The percent ionization of an acid can be calculated using the following formula:

% ionization = [H3O+]eq / [HA]initial × 100%

where [H3O+]eq is the hydronium ion concentration at equilibrium and [HA]initial is the initial concentration of the acid.

In this case, the initial concentration of the acid is 0.532 M, and the hydronium ion concentration at equilibrium is 0.112 M.

Therefore,

% ionization = 0.112  / 0.532 × 100%

                    = 21.1%

Therefore, the percent ionization of the acid is 21.1% (rounded to the nearest tenth).

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Answer:

K₅Mn₅O₁₆

Explanation:

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

Potassium (K) = 27.5 %

Manganese (Mn) = 37.5 %

Oxygen (O) = 35 %

Empirical formula =?

The empirical formula for the compound can be obtained as follow:

K = 27.5 %

Mn = 37.5 %

O = 35 %

Divide by their molar mass

K = 27.5 / 39 = 0.705

Mn = 37.5 / 55 = 0.682

O = 35 / 16 = 2.188

Divide by the smallest

K = 0.705 / 0.682 = 1

Mn = 0.682 / 0.682 = 1

O = 2.188 / 0.682 = 3.2 = 32/10 = 16/5

K = 1

Mn = 1

O = 16/5

Multiply by 5 to express in whole number

K = 1 × 5 = 5

Mn = 1 × 5 = 5

O = 16/5 × 5 =16

Thus, the empirical formula for the compound is K₅Mn₅O₁₆

The answer is Two molecules of carbon dioxide are released during one turn of the citric acid cycle.

Two carbons - from acetyl CoA-enter the citric acid cycle in each turn, and two carbon dioxide molecules are released. However, the carbon dioxide molecules don't actually contain carbon atoms from the acetyl CoA that just entered the cycle.

Instead, the carbons from acetyl CoA are initially incorporated into the intermediates of the cycle and are released as carbon dioxide only during later turns. After enough turns, all the carbon atoms from the acetyl group of acetyl CoA will be released as carbon dioxide.

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Among the given reactions of alkenes, the reaction of addition of HBr (treatment with HBr) to an alkene is stereo-specific.

What is stereo-specific?

Stereo-specificity is a chemical concept that describes a reaction's capability to differentiate between stereo-isomers. A stereo-specific reaction is one in which only one stereo-isomer is generated from a given stereo-isomeric reactant. The stereo-chemistry of the reactant, the product, or both is generally associated with stereo-specificity. Stereo-specific reactions are also referred to as stereo-selective reactions.

When an alkene reacts with HBr in the presence of a peroxide initiator, the hydrogen (H) and the bromine (Br) atoms add to the carbon atoms of the double bond in a syn addition. This means that the hydrogen and the bromine atoms are added to the same side of the double bond, resulting in the formation of a single product with a specific stereo-chemistry.

Therefore, the reaction of addition of HBr to an alkene is stereo-specific, because the stereo-chemistry of the product is determined by the syn addition of the hydrogen and the bromine atoms to the double bond. The other reactions listed are not stereo-specific.

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