how can you start preserving the gift of nature which you can apply in your day-to-day life?

Answer:  There are

Explanation:

Lead levels in plankton and algae are high, mostly as a result of environmental pollution brought on by human activity. While it is true that some brown algae species have the ability to accumulate heavy metals like lead.

Plankton and algae have high levels of lead, mostly as a result of environmental contamination brought on by human activities including mining, industrial operations, and the burning of fossil fuels.

Due to the fact that plankton and algae take up trace quantities of lead from the surrounding water, their tissues contain greater concentrations of the metal.

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Answer: Number of atoms of Fe = 5.214x10²²atoms

Explanation:

First, we'll calculate the number of moles of Fe using the given number of atoms:

Number of moles = Number of atoms / Avogadro's number-----------(i)

Number of atoms of Fe = 5.22x10²²atoms    

Avogadro's number = 6.022x10²³atoms/mol)

now putting values in equation (i)

Number of moles = 5.22x10²²atoms  ₓ 6.022x10²³atoms/mol)    

Number of moles of Fe = 0.0866 moles

To find the number of atoms of Fe, we can use Avogadro's number again:

Number of moles = Number of atoms / Avogadro's number-----------(i)

Number of moles of Fe = 0.0866 moles

Avogadro's number = (6.022x10²³atoms/mol)

Number of atoms of Fe = 5.214x10²²atoms

Answer:

Approximately 4.678 moles

Explanation:

150/32.065 (atomic weight of S)

Answer:

4.677 Moles

Explanation:

150g / 32.07g = 4.677268475 moles

Question 1 : The empirical formula for this compound is CBr₄.

Question 2: The molecular formula of the compound is CBr₄.

To determine the empirical formula of the compound, we need to find the simplest whole-number ratio between the elements present.

Empirical formula:

The compound contains 3.622% carbon and 96.38% bromine. To convert these percentages into masses, we can assume a 100 g sample of the compound.

Mass of carbon = (3.622/100) * 100 g = 3.622 g

Mass of bromine = (96.38/100) * 100 g = 96.38 g

Next, we need to find the moles of each element. We can use their atomic masses to convert the masses to moles.

Atomic mass of carbon (C) = 12.01 g/mol

Atomic mass of bromine (Br) = 79.90 g/mol

Moles of carbon = Mass of carbon / Atomic mass of carbon = 3.622 g / 12.01 g/mol ≈ 0.3017 mol

Moles of bromine = Mass of bromine / Atomic mass of bromine = 96.38 g / 79.90 g/mol ≈ 1.205 mol

To find the simplest whole-number ratio between the elements, we divide both moles by the smallest number of moles (0.3017 mol in this case):

Moles of carbon (C) = 0.3017 mol / 0.3017 mol = 1

Moles of bromine (Br) = 1.205 mol / 0.3017 mol ≈ 4

Therefore, the empirical formula of the compound is CBr₄.

Molecular formula:

The empirical formula of CBr₄ gives us the simplest whole-number ratio of the elements. To determine the molecular formula, we need the molar mass of the compound.

Given that the molecular weight (molar mass) of the compound is 331.6 amu, we can find the ratio of the molecular weight to the empirical formula weight:

Molecular weight / Empirical formula weight = 331.6 amu / (12.01 amu + (4 × 79.90 amu)) ≈ 331.6 amu / 332.64 amu ≈ 0.9965

Since the ratio is close to 1, the empirical formula is also the molecular formula. Therefore, the molecular formula of the compound is CBr₄.

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Answer: it will have a pressure of approximately 371.73 κPa.

We know that According to Boyle's Law

P1 ×V1 = P2 ×V2

therefore in order to find P2

P2 =[tex]\frac{P1 X V1}{V2}[/tex] --------------------(I)

where

P1 = initial pressure

V1 = initial volume    

P2 = final pressure

V2 = final volume

here

P1(initial pressure) = 146 kPa        

V1(initial volume) = 2.55 L

V2(final volume) = 2 L

Now putting the given values in equation (I) :

P2 = (146 kPa × 2.55 L) / 2 L

P2 = 371.73 kPa

Therefore, 2 L of gas, when it takes 2.55 L at 146 kPa, will have a pressure of 371.73 kPa.

The balance reactions are :

1) 2Ni+(aq) + 4H+(aq) → 2Ni2+(aq) + 2Ni(s) + 2H2O(l) ,

2) 3MnO42-(aq) + 4H+(aq) → 2MnO4-(aq) + MnO2(s) + 2H2O(l) ,

3) H2SO3(aq) + H2O(l) → S(s) + 2HSO4-(aq) + 2H+(aq) ,

4) Cl2(aq) + 2OH-(aq) → Cl-(aq) + ClO-(aq) + H2O(l).

1) To balance the disproportionation reaction of Ni+ in an acidic solution, we need to ensure that the number of atoms and charges are balanced on both sides of the equation. The balanced equation is as follows:

2Ni+(aq) + 4H+(aq) → 2Ni2+(aq) + 2Ni(s) + 2H2O(l)

In this reaction, Ni+ is oxidized to Ni2+ (oxidation state increases from +1 to +2) while simultaneously being reduced to Ni (oxidation state decreases from +1 to 0). The hydrogen ions (H+) act as the oxidizing agent, accepting electrons and being reduced to form water (H2O).

2) Balancing the disproportionation reaction of MnO42- in an acidic solution:

3MnO42-(aq) + 4H+(aq) → 2MnO4-(aq) + MnO2(s) + 2H2O(l)

In this reaction, MnO42- is both oxidized and reduced. The oxidation state of Mn in MnO42- changes from +7 to +6 in MnO4- (reduction) and from +7 to +4 in MnO2 (oxidation). The hydrogen ions (H+) again act as the oxidizing agent, undergoing reduction to form water.

3) Balancing the disproportionation reaction of H2SO3 in an acidic solution:

H2SO3(aq) + H2O(l) → S(s) + 2HSO4-(aq) + 2H+(aq)

In this reaction, H2SO3 is both oxidized and reduced. The sulfur (S) in H2SO3 is reduced from an oxidation state of +4 to 0 in S, while the hydrogen sulfite ion (HSO3-) is oxidized from an oxidation state of +4 to +6 in HSO4-. The water molecule (H2O) acts as a reactant and is not involved in the redox process.

4) Balancing the disproportionation reaction of Cl2 in a basic solution:

Cl2(aq) + 2OH-(aq) → Cl-(aq) + ClO-(aq) + H2O(l)

In this reaction, Cl2 is both oxidized and reduced. The chlorine (Cl) in Cl2 is reduced from an oxidation state of 0 to -1 in Cl-, while simultaneously being oxidized from an oxidation state of 0 to +1 in ClO-. The hydroxide ions (OH-) act as the reducing agent, accepting electrons and being oxidized to form water (H2O). The reaction takes place in a basic solution, hence the presence of OH- ions.

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Answer:

Empirical formula of a compound means that it provides simplest ratio of whole number.

Explanation:

Mass of boron and chlorine is 9.224% and 90.74%

The mass of 4.35 × 10^-2 mol of NaOH is 1.74 grams.

Given

Number of moles = 4.35 x 10^-2

First, we calculate the molar mass of NaOH,

Molar mass of NaOH = (1 × atomic mass of Na) + (1 × atomic mass of O) + (1 × atomic mass of H)

= (1 × 22.99 g/mol) + (1 × 16.00 g/mol) + (1 × 1.01 g/mol) = 40.00 g/mol

Molar mass of NaOH = 40.00 g/mol

Mass of NaOH = Number of moles × Molar mass

                          = 4.35 × 10^-2 mol × 40.00 g/mol  

Mass of NaOH  = 1.74 g

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Answer:

its sodium hydroxide

Explanation:

Answer:

Explanation:

To find the concentration of hydroxide ions ([OH-]) in an aqueous solution, we can use the relationship between hydrogen ion concentration ([H+]) and hydroxide ion concentration in water at 25 °C, which is given by the equation:

[H+] x [OH-] = 1.0 x 10^-14

Given that the hydrogen ion concentration ([H+]) is 5.7 x 10^-10 (derived from the H* concentration provided), we can rearrange the equation to solve for [OH-]:

[OH-] = (1.0 x 10^-14) / [H+]

[OH-] = (1.0 x 10^-14) / (5.7 x 10^-10)

[OH-] ≈ 1.754 x 10^-5

Therefore, the concentration of hydroxide ions ([OH-]) in the given aqueous solution is approximately 1.754 x 10^-5.

The column 1 has the value of Isotope, column 2 has the value of mass in atomic mass units, and column 3 has the value of abundance and the average atomic mass of strontium is 87.47 amu.

To calculate the average atomic mass of strontium using the given information, we need to multiply the mass of each isotope by its abundance and then sum up these values. Here's the calculation:

Isotope | Mass (amu) | Abundance

^84Sr | 83.913428 | 0.56%

^86Sr | 85.909273 | 9.86%

^87Sr | 86.908902 | 7.00%

^88Sr | 87.905625 | 82.58%

To find the average atomic mass, we multiply each isotope's mass by its abundance (in decimal form) and sum up the values:

Average atomic mass = ([tex]Mass of ^{84Sr}[/tex] × [tex]Abundance of^{84Sr}[/tex]) + ([tex]Mass of ^{86Sr}[/tex]× [tex]Abundance of^{86Sr}[/tex]) + ([tex]Mass of ^{87Sr}[/tex] × [tex]Abundance of^{87Sr}[/tex]) + ([tex]Mass of ^{88Sr}[/tex] × [tex]Abundance of^{88Sr}[/tex])

Average atomic mass = (83.913428 amu × 0.0056) + (85.909273 amu × 0.0986) + (86.908902 amu × 0.0700) + (87.905625 amu × 0.8258)

Calculating this expression yields:

Average atomic mass = 0.469901638 + 8.468098826 + 6.08462314 + 72.44409075

= 87.466714354 amu

Rounding the result to two decimal places, the average atomic mass of strontium is approximately 87.47 amu.

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Answer: b. Aluminum

Explanation:

Frist count all the electrons in the given model. You will get there is 13 electrons. The number of electrons in an element is equivalent to the number of protons in an element. Using a periodic table look for the element that has the equivalent amount of protons. You find that Aluminum has 13 protons, so it is the element shown in the diagram.

The two regions  that I will  compare their effects of climate change in areas are Arctic and the Amazon rainforest..

The Major Events that can be associated to Arctic region  can be described as rapid warming that affect ecosystem.

The major  that can be attributed to Amazon rainforest  can be described as increased deforestation rates.

In term of Adaptation the Arctic communities  are facing some challenges which makes some of the people to communities relocating homes away from eroding coastline.

In term of Adaptation the Amazon rainforest were seeking for the way to combat deforestation  and bring about Initiatives such as reforestation.

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Explanation: heat is transferred from the hot water to the cold water until they reach the same temperature

The concentration of hydroxide ions ([OH-]) and the concentration of hydronium ions ([H+]) are related in an aqueous solution by the equation [H+][OH-] = 1.0 x 10^-14 at 25 °C .The concentration of hydronium ions ([H+]) in the aqueous solution at 25 °C is approximately 3.03 x 10^-18.

Given that [OH-] is 3.3 x 10^3, we can substitute this value into the equation as follows:

[H+][3.3 x 10^3] = 1.0 x 10^-14

Dividing both sides of the equation by 3.3 x 10^3, we get:

[H+] = (1.0 x 10^-14) / (3.3 x 10^3)

Simplifying the expression, we have:

[H+] ≈ 3.03 x 10^-18

In summary, at 25 °C, an aqueous solution with an OH- concentration of 3.3 x 10^3 has a hydronium ion concentration of approximately 3.03 x 10^-18. The hydronium ion concentration is determined by the equilibrium constant for water dissociation and is inversely proportional to the hydroxide ion concentration. The two concentrations are related through the equation [H+][OH-] = 1.0 x 10^-14.

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The decomposition of potassium chlorate KClO₃ in the presence of manganese oxide MnO is given by the reaction equation:

KClO₃ (s) → 2KCl (s) + 3O₂ (g)

To calculate the moles of product formed from moles of reactants, the following steps are followed:

1. Balancing the equation

2. Calculating the ratio of product's stoichiometric coefficient and reactant's stoichiometric coefficient.

3. Multiplying the obtained ratio with the number of moles of reactant.

Thus, the number of moles of oxygen evolved will be calculated as:

R = [tex]\frac{coefficient of O2}{coefficient of KClO3}[/tex] = [tex]\frac{3}{1}[/tex] = 3

Number of moles of oxygen evolved = R × number of moles of KClO₃ = 3×5= 15 moles

From the ideal gas equation, 1 mole of gas is equivalent to 22400 ml or 22.4 L.

Thus, volume of oxygen evolved = 22400 × 15 = 336000 ml = 336 L

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The evidence for quantized energy states in atoms stems primarily from the photoelectric effect, bright line spectra, and diffraction phenomena. Options 1,4 and 5 are correct.

The evidence of quantized energy states in atoms comes from several experimental observations, which collectively provide a comprehensive understanding of atomic structure and the behavior of electrons within atoms.

One of the key pieces of evidence is the observation of the photoelectric effect (1). When light shines on a metal surface, electrons are ejected from the surface. The observation that electrons are only ejected if the light has a minimum frequency, regardless of its intensity, supports the idea that energy is quantized in discrete packets known as photons.

Another crucial observation is the presence of bright line or emission spectra (5). When atoms are excited, they emit light at specific wavelengths that correspond to distinct energy transitions. These discrete emission lines indicate that electrons can only exist in specific energy levels within an atom, and they transition between these levels by absorbing or emitting photons of precise energy.

The phenomenon of diffraction (4) also provides evidence for quantized energy states. Diffraction occurs when light passes through a narrow slit or encounters a periodic structure. The resulting pattern indicates that light behaves as waves with specific wavelengths. This suggests that the energy of light is quantized and can only exist in certain discrete values.

While rainbows from prisms (2) and the oil drop experiment (3) are not directly related to quantized energy states in atoms, they are important experiments in their own right. Rainbows result from the dispersion of white light into its component colors due to different wavelengths of light bending at different angles. The oil drop experiment explores the behavior of charged oil droplets in an electric field, providing insights into charge quantization.

Lastly, the gold foil experiment, also known as the Rutherford scattering experiment, is significant but not directly related to quantized energy states. It demonstrated that the atom has a small, dense, positively charged nucleus by observing the deflection of alpha particles fired at a thin gold foil.

These experimental observations support the fundamental concept that energy levels in atoms are discrete and that electrons occupy specific energy states within an atom.  Options 1,4 and 5 are correct.

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[tex]\huge\mathcal{\fcolorbox{aqua}{azure}{\red{Answer:-}}}[/tex]

Lead exists in high concentrations in plankton and algae primarily due to environmental pollution from human activities, such as industrial processes, mining, and the burning of fossil fuels. Plankton and algae accumulate trace amounts of lead from their surrounding water, resulting in higher concentrations within their tissues.

Explanation:

An oxidizer is a substance that facilitates oxidation, a chemical reaction where a substance loses electrons. It is also called an oxidizing agent or oxidant. Oxidizers are commonly used in combustion reactions, supporting the burning of fuels by providing oxygen. Examples of oxidizers include oxygen, chlorine, hydrogen peroxide, and potassium permanganate. They have applications in combustion, chemical synthesis, bleaching, rocket propellants, and cleaning. Oxidizers can be highly reactive and require proper handling and safety precautions.

Answer: Therefore, the concentration of hydroxide ions [OH-] in the given solution is 4.0 x 10⁻¹⁹M.

We know that In an aqueous solution Kw is the ionization constant of water.

Kw = [H3O⁺][OH⁻]

[OH⁻] = [tex]\frac{Kw}{[H3O^+]}[/tex]--------------------------------------(a)

Kw = ionization constant of water

[H3O⁺]= the concentration of hydronium ions

[OH⁻]  =  the concentration of hydroxide ions

Kw = 1x10⁻¹⁴M²-------------------(i)

[H3O⁺]= 2.5 x 10⁴M------------------(ii)

[OH⁻]  = ?    

NOW Putting values in  (i) and (ii) in equation (a)

[OH⁻] = [tex]\frac{1 X 10^-^1^4}{2.5 X 10^4}[/tex]

[OH⁻] =  4.0 x 10⁻¹⁹M

There are approximately 0.0112 moles of hydrogen gas in the 250 ml flask.

To determine the number of moles of hydrogen gas in the flask, we can use the ideal gas law and Dalton's law of partial pressure.

First, let's convert the given pressures to atm units:

P_total = P_hydrogen + P_water vapor

P_total = (763 mmHg - 42.2 mmHg) / (760 mmHg/atm) [Converting to atm]

P_total = 0.9524 atm

Next, let's convert the given temperature to Kelvin:

T = 35°C + 273.15 [Converting to Kelvin]

T = 308.15 K

Now we can use the ideal gas law equation: PV = nRT

R is the ideal gas constant, which has a value of 0.0821 L·atm/(mol·K)

Rearranging the equation to solve for n (moles):

n = PV / RT

Substituting the given values:

n = (0.9524 atm) * (0.250 L) / (0.0821 L·atm/(mol·K)) * (308.15 K)

Simplifying the expression:

n = 0.0112 mol

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As per the given data, 1.6 grams of oxygen reacted in this experiment.

To calculate the amount of oxygen that reacted in the experiment, we need to determine the difference in the mass of X oxide (X,O) formed and the mass of X initially used.

Given:

Mass of X = 4.6 g

Mass of X oxide (X,O) = 6.2 g

To find the amount of oxygen that reacted:

Mass of oxygen = Mass of X oxide - Mass of X

= 6.2 g - 4.6 g

= 1.6 g

Therefore, 1.6 grams of oxygen reacted in this experiment.

Calculate the mass of 1 mole of X:

Given that the mass of X is 4.6 g, we can calculate the molar mass of X by dividing the mass by the number of moles:

Molar mass of X = Mass of X / Number of moles of X

Molar mass of X = 4.6 g / 0.1 mol

Molar mass of X = 46 g/mol

Therefore, the mass of 1 mole of X is 46 grams.

Thus, the answer is 46 grams.

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The pH of the resulting solution after mixing 50 mL of 0.180 M [tex]NH_{3}[/tex] with 5 mL of 0.360 M HBr is approximately 11.56.

To calculate the pH of the resulting solution after mixing NH3 and HBr, we need to consider the reaction between NH3 (ammonia) and HBr (hydrobromic acid).

First, let's write the balanced chemical equation for the reaction:

[tex]NH_{3} + HBr - > NH_{4+} + Br-[/tex]

We can see that [tex]NH_{3}[/tex] acts as a base and HBr acts as an acid, forming the ammonium ion ([tex]NH_{4+}[/tex]) and bromide ion (Br-).

Next, we'll determine the initial moles of NH3 and HBr:

Moles of NH3 = concentration (M) × volume (L) = 0.180 M × 0.050 L = 0.009 mol

Moles of HBr = concentration (M) × volume (L) = 0.360 M × 0.005 L = 0.0018 mol

Since NH3 and HBr react in a 1:1 ratio, NH3 will be completely consumed, and we'll be left with 0.009 - 0.0018 = 0.0072 mol of NH4+ ions.

Now, let's calculate the concentration of NH4+ ions in the final solution:

Volume of the final solution = 50 mL + 5 mL = 55 mL = 0.055 L

Concentration of NH4+ ions = moles / volume = 0.0072 mol / 0.055 L = 0.131 M

Next, we need to calculate the pOH of the solution using the Kb of ammonia:

[tex]Kb = [NH_{4+}][OH-] / [NH_{3}][/tex]

Since the concentration of NH4+ is equal to the concentration of OH- in this case, we can rewrite the equation:

[tex]Kb = [OH-]^2 / [NH3]\\[OH-] = sqrt(Kb * [NH3]) = sqrt(1.77*10^-5 * 0.131) = 3.62*10^-3 M[/tex]

Now, we can calculate the pOH:

[tex]pOH = -log10([OH-]) = -log10(3.62*10^-3) = 2.44[/tex]

Finally, we can calculate the pH using the equation:

pH = 14 - pOH = 14 - 2.44 = 11.56

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Some boron compounds and the determination of boron's oxidation number. In those boron hydrides that contain one or more B-B bonds, the oxidation number of boron can be less than +3 and more than 0.

Thus, The same molecule will have various boron atom types with various oxidation values in such a complex. Therefore, the average oxidation number would be determined using the formula for such a molecule.

Tetraborane (B4 H10) and decaborane (B10 H14) are two examples of such compounds that are displayed in the table's final two entries.

These substances are less stable and have complicated structures. The majority of stable boron compounds have boron with an oxidation number of +3.

Thus, Some boron compounds and the determination of boron's oxidation number. In those boron hydrides that contain one or more B-B bonds, the oxidation number of boron can be less than +3 and more than 0.

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The species that should be present in the net-ionic equation are:

[tex]Hg_{2}, 2 CI^-[/tex]and [tex]Hg_{2}Cl_{2}[/tex]

To determine the species that should be present in the net-ionic equation for the reaction of strontium chloride (SrCl₂) and mercury(I) nitrate (Hg2(NO3)2), let's analyze the reactants and products:

Reactants:

Strontium chloride (SrCl₂): Sr2+, Cl-

Mercury(I) nitrate (Hg2(NO3)2): Hg2^2+, NO3-

Products:

Strontium nitrate (Sr(NO3)2): Sr2+, NO3-

Mercury(I) chloride (Hg2Cl2): Hg2^2+, Cl-

The species that should be present in the net-ionic equation are:

[tex]Hg_{2}, 2 CI^-[/tex]and [tex]Hg_{2}Cl_{2}[/tex]

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Correct answers are:

i. Coal, charcoal, oil, and gas - E. Fuels

ii. It supports combustion - D. Fire triangle

iii. Fire associated with electrical equipment - G. Class E fire

iv. A chemical change occurring in iron or steel - C. Rusting

v. Oxygen, heat, and fuel - D. Fire triangle

vi. Fire involving flammable liquids - G. Class B fire

vii. Coating of iron and steel with Zinc - L. Galvanizing

viii. Monoammonium phosphate with Nitrogen carrier - M. Fire extinguisher

ix. A team which put off fire when it's out of control - J. Fire squad

x. It uses oxygen when burning but produces soot - N. Non-luminous flame

Coal, charcoal, oil, and gas are commonly used as fuels. Therefore, they are matched with E. Fuels. Electrical fires are classified as Class E fires. Therefore, it is matched with G. Class E fire. Fires that involve flammable liquids are classified as Class B fires. Therefore, it is matched with G. Class B fire.

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When cyclohexene is mixed with a sulfonitric mixture (H2SO4/HNO3), it reacts to form nitrocyclohexane and sulfur dioxide.

This reaction proceeds in two steps. Firstly, cyclohexene undergoes electrophilic addition with the nitronium ion (NO2+), which is generated from the reaction between HNO3 and H2SO4. This results in the formation of nitrocyclohexane, giving the initial pale yellow color to the solution.

In the second step, nitrocyclohexane reacts with the excess sulfuric acid present in the mixture. This step is highly exothermic, releasing a significant amount of energy. The sudden release of energy causes an explosion. The exact mechanism of the explosive reaction is complex, involving the generation of reactive intermediates. It is believed that the reaction proceeds via a radical mechanism, where nitrocyclohexane decomposes into highly reactive nitrogen and carbon-centered radicals. These radicals further react with sulfur dioxide, which is produced in the reaction, to form stable compounds. As a result, the solution turns dark brown after the explosion.

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