really need help on this not understanding

Answer:

Decomposers are used to break down substances. In chemistry, this is the breaking down of organic molecules into simpilar compounds that are able to be sused

Answer:

=50g NaOH) then,no.of moles of the NaOH solution=50g/40g=1.25 moles/litre

Explanation:

answer=50g NaOH

Answer:

Explanation

a. chlorine as it has 17 electrons arranged in three quantum shells.

b. thallium

c. sodium

d. fluorine

e. zinc

f. iron

g. calcium

h. silicon

I. indium

j. selenium

k. lithium

hint: the number of valence electrons depend on the group in which an element is in.

ans energy level depend upon the period in which the element is in.

Answer:

c

Explanation:

a p e x :)

The charge an element would have if it lost or gained electrons. Hence, option C is correct.

Oxidation: The loss of electrons

Reduction: The gain of electrons

Oxidation number: the charge based on the number of electrons assigned to an atom as compared to the neutral atom.

Hence, the charge an element would have if it lost or gained electrons is the oxidation number.

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The heat of combustion per mol of the glucose is 2840 kJ.

A reaction that involves the burning of a compound in the presence of air or oxygen is called combustion. In the absence of oxygen combustion of a compound cannot take place.

It is given that the combustion of glucose produces carbon dioxide and water. The balanced chemical reaction for the given process is expressed as:

C6H12O6 (s) + 6O2 (g) ------> 6CO2 (g) + 6H2O (l)

The energy released from this reaction is -2840 kJ. The above equation clearly shows that 1 mol of glucose is used during the reaction and the energy produced is -2840 kJ.

Therefore, the heat of combustion per mol of the glucose is 2840 kJ.

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The name for the ionic compound pu(oh)3  is Plutonium(III) hydroxide.

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Answer:

YEESSS

Explanation:

Because that seems like the only logical answer.

Answer:

YES

Explanation:

YES

The balanced equation for the reaction of iron with water that results in the production of iron oxide and hydrogen gas is

3 Fe (s) + 4H₂O (g) → Fe₃O₄ (s) + 4H₂ (g)

Iron does not react directly with liquid water but react with water vapour. When the reaction happens, it results in the formation of a solid and a gas. The products of the reaction are Iron oxide and hydrogen. The equation of the reaction would be

Fe (s) + H₂O (g) → Fe₃O₄ (s) + H₂ (g)

Now, we need to balance the equation. On the right-hand side, we have 3 Fe, 4 O and 2 H. Similarly on the left-hand side there are 1 Fe, 1 O and 2 H.

To balance the equation, we add 3 to Fe, 4 to H₂0 and 4 to H₂.

As a result, the balanced chemical equation for the reaction would be

3 Fe (s) + 4H₂O (g) → Fe₃O₄ (s) + 4H₂ (g)

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The spontaneous reaction for the standard cell made using gold (Au) and gold (III) nitrate, cobalt and cobalt (II) nitrate is:
Au + Co2+ → Au3+ + Co+

The reaction in cell notation form is:

Au | Au3+ || Co2+ | Co+ | Co

Where Au represents the electrode made of gold, Au3+ represents the gold (III) nitrate solution, Co2+ represents the cobalt (II) nitrate solution, Co+ represents the cobalt electrode, and the double line represents the salt bridge.

For the standard cell made using the given components, we first need to determine the half-reactions. They are:

Au³⁺(aq) + 3e⁻ → Au(s) [Reduction]
Co(s) → Co²⁺(aq) + 2e⁻ [Oxidation]

Now we can balance the electrons and write the spontaneous reaction:

2Au³⁺(aq) + 3Co(s) → 2Au(s) + 3Co²⁺(aq)

For the cell notation, we can represent it as follows:

Co(s)|Co²⁺(aq)||Au³⁺(aq)|Au(s)

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Answer:

Explanation:Using the equation, PV=nRT, you can solve this problem.

P= 3.5 kPa V=? n= 0.55 moles of air T= 295K R= 8.314 (L kPa/mol K)

We use the constant rate of 8.314 because we are using kilopascals in this problem.

Using simple algebra we get-

V = (0.55 moles)(8.314 L kPa/mol k)(295 K) / 3.5 kPa

Using the calculator, we would get the answer of 385.4132857 L.

And whatever the significant figure is asked for, for example, 3 significant figures: the answer would be 385 liters.

If 2 moles of oxygen are consumed during cellular respiration, 36.03 grams of water are produced.

Using the cellular respiration reaction  C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O. For every 6 moles water produced, 6 mole of oxygen is consumed as well. Therefore, for 2 moles of oxygen consumed,

2/6 x 6 mol of H₂O = 2 mol of H₂O

To convert moles of water to grams, we need to use the molar mass of water,

2 mol of H₂O x 18.015 g/mol = 36.03 g of H₂O

Therefore, if 2 moles of oxygen are consumed during cellular respiration, 36.03 grams of water are produced.

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Complete question - If you have 2 moles of oxygen (O₂), how many grams of water are produced during cellular respiration? Solve and record your answer

C₆H₁₂O₆ + 602 → 6CO₂ + 6H₂O

Answer:

Same i have headaches daily, maybe from infections such as meningitis or Intracranial pressure that's either too high or too low or id.k.

Answer:

.....

Explanation:....

The vapor pressure of carbon disulfide (CS₂) at 0°C is approximately 0.178 torr.

To determine the vapor pressure of carbon disulfide (CS₂) at 0°C, we can use the Clausius-Clapeyron equation:

ln(P₂/P₁) = -(ΔHᵥᵃᵖ/R) * (1/T₂ - 1/T₁)

Where:

P₁ = vapor pressure at temperature T₁ (known)

P₂ = vapor pressure at temperature T₂ (unknown)

ΔHᵥᵃᵖ = molar enthalpy of vaporization

R = gas constant (8.314 J/(mol·K))

T₁ = initial temperature (known)

T₂ = final temperature (unknown)

Given:

ΔHᵥᵃᵖ = 26.74 kJ/mol (convert to J/mol: 26.74 kJ/mol * 1000 J/1 kJ = 26740 J/mol)

T₁ = 46°C (convert to Kelvin: 46°C + 273.15 K = 319.15 K)

T₂ = 0°C (convert to Kelvin: 0°C + 273.15 K = 273.15 K)

R = 8.314 J/(mol·K)

Substituting the known values into the equation:

ln(P₂/P₁) = -(26740 J/mol / 8.314 J/(mol·K)) * (1/273.15 K - 1/319.15 K)

Simplifying the equation:

ln(P₂/P₁) = -3216.11 * (-0.003663 + 0.003128)

ln(P₂/P₁) = -3216.11 * 0.000535

ln(P₂/P₁) ≈ -1.723

To solve for P₂/P₁, we can take the exponential of both sides:

P₂/P₁ = e^(-1.723)

Using a calculator, e^(-1.723) ≈ 0.178

Finally, to find P₂ (vapor pressure at 0°C), we multiply P₁ by P₂/P₁:

P₂ = P₁ * (P₂/P₁)

P₂ = 1 torr * 0.178

P₂ ≈ 0.178 torr

Therefore, the vapor pressure of carbon disulfide (CS₂) at 0°C is approximately 0.178 torr.

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Answer:

different forms of genes for a single trait

Explanation:

There are three main forms  of genes  that can exist for a single trait and they are:

A dominant allele is a stronger version of a gene that shows up when at least one copy of it is present in an organism's genetic makeup. It hides the effects of the weaker gene. In a person with two different alleles, the one that is more powerful decides how the person looks or behaves.

Recessive allele: A recessive allele is only expressed when there are two of the same recessive alleles present. It is hidden by a strong gene. The recessive allele can only be seen when a person has two copies of it.

Co-dominant alleles: When co-dominant alleles are together, both are shown. In this situation, neither gene is stronger than the other, and both play a role in what an organism looks like. One example of co-dominance is the ABO blood type system. In this system, both the A and B alleles are of equal importance.

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Answer:

0.347

Explanation:

Al is the limiting reagent.

Explanation:

The stoichiometry of the reaction is such that 4 moles of Al are required for every 3 moles of diatomic oxygen. This means that if the ratio of Al to diatomic oxygen is greater than 4/3, then the oxygen is the limiting reagent. If the ratio of Al to diatomic oxygen is less than 4/3, then Al is the limiting reagent.

The problem state that the ratio of Al to diatomic oxygen is

0.32

0.26

=

16

13

=

48

39

<

52

39

=

4

3

The ration of Al to diatomic oxygen is less than 4/3, so Al is the limiting reagent.

We can also see this if we ask the question "How much Aluminum" is required to completely react 0.26 moles of diatomic oxygen??

0.26

moles

O

2

×

4

moles

A

l

3

moles

O

2

≈

0.347

moles

A

l

So it would take 0.347 moles of Al to completely react with all of the oxygen, however there is only 0.32 moles of Aluminum present, so there is not enough Al to react with all of the oxygen, and so we say that the Al is the limiting reagent.

The mass of Aluminum are needed to produce 21.6 grams of Aluminum oxide (Al₂O₃) is

Mass of any element is equal to the product of its number of moles and molar mass.

For the compound, Aluminum oxide (Al₂O₃)

The atomic masses of Aluminum is 23 and oxygen is 16.

Molecular or molar mass = ( 2 × 23 ) + ( 3 × 16 ) units

M = 46 + 48 = 94 units

The mass of Aluminum in 1 mole of aluminum oxide ,46 grams in 94 grams

Mass of Aluminum in the given 21.6 g of Aluminum oxide will be;

94 grams of Aluminum oxide contains 46 grams of Aluminum.

21.6 grams of Aluminum oxide will contain ( 46 / 94 ) × 21.6 = 10.57g of Aluminum.

Therefore, 10.57g of Aluminum are needed to produce 21.6 grams of Aluminum oxide (Al₂O₃)

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The volume of 6 M HCl required to achieve a 150 mL solution of 0.4 M HCl is 10 mL (0 decimal places).

To calculate the volume of 6 M HCl required to achieve a 150 mL solution of 0.4 M HCl, you can use the dilution formula:

M1V1 = M2V2

where M1 and V1 are the initial molarity and initial volume of the concentrated solution (6 M HCl), and M2 and V2 are the final molarity and final volume of the diluted solution (0.4 M HCl and 150 mL).

Rearrange the formula to solve for V1:

V1 = (M2V2) / M1

Plug in the given values:

V1 = (0.4 M × 150 mL) / 6 M

V1 = (60 mL) / 6 M

V1 = 10 mL

So, the volume of 6 M HCl required to achieve a 150 mL solution of 0.4 M HCl is 10 mL (0 decimal places).

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Answer:

I got B - 25cm³

Explanation:

First you need to find the no. of moles of H₂SO₄ and I used this formula.

no. of moles = concentration × vol.

Substitute the values;

no. of moles = 0.1 × (50/1000)

Therefore, you would have 0.005mol

Then, if 1 mole of H₂SO₄ gives me 0.005mol, 2 moles of NaOH would give me 0.01mol. You can use ratio for this part!

Now that we have the no. of mol for NaOH, we can finally calculate the volume of NaOH used in this reaction. I used this formula;

vol. of NaOH = no. of mol/concentration

I just simply moved the previous formula around and made vol. the subject this time!

Then, substitute the values;

Vol. of NaOH = 0.01/0.4

You should get 0.025dm³!

Now, we just simply have to convert 0.025dm³ into cm³ like so;

1 dm³ --> 1000cm³

0.025dm ³ --> 0.025 × 1000

And that would give you 25cm³!

I hope this helps! Please let me know if I have any misconceptions or miscalculations! :)

Answer:

B y is a complete collection for your time and energy for the spread of Islam in the Horn of the following to make sure

The percentage of carbon-14 that will remain in a sample after 1.99 x 10⁴ years is 9.01%.

Half-life is defined as the time it will took a substance to reduce to half of its initial amount. The half-life of a substance is determined with respect to the rate of the reaction. It can be calculated using the formula below.

N(t) = N₀(1/2)^(t/T)

where N(t) = amount of substance remaining after time t

N₀ = initial amount of substance

t = time = 1.99 x 10⁴ years

T = half-life = 5730 years

Plug in the given values to the equation.

N(t) = N₀(1/2)^(t/T)

N(t) = N₀(1/2)^(1.99 x 10⁴ years/5730 years)

N(t) = N₀(0.09006)

Hence, the percentage that will remain is 9.01%.

N(t) / N₀ = 0.09006

N(t) / N₀ = 9.01%

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If x = − 4,  number line shows the value of |x| is 4 tick mark to the left of zero.

given that :

x = -4

The absolute value is the distance from zero. The absolute value is always positive.

| x | = | -4 | = 4

The absolute value = distance from the zero.

a number line contains negative and positive number . number to the right side of zero is positive and to  the left side of zero is negative. Absolute value means how far the number is from zero.

Thus, If x = − 4,  number line shows the value of |x| is 4 tick mark to the left of zero.

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The presence of water molecules in the aqueous solution shifts the reduction reaction from lead ions to water molecules, resulting in the production of hydrogen gas instead of lead metal.

The difference in the products formed during the electrolysis of molten lead bromide and aqueous lead bromide can be explained by the different conditions and species present in each case.

When molten lead bromide is used, the compound is in a liquid state without water molecules present. During electrolysis, the positive lead ions (Pb²⁺) are attracted to the negative electrode (cathode) where reduction takes place.

At the cathode, the lead ions gain electrons and are reduced to lead metal (Pb). This is because the reduction potential of lead ions is higher than that of water molecules, making the reduction of lead ions more favorable in this case.

At the same time, bromide ions (Br⁻) are attracted to the positive electrode (anode), where oxidation occurs, and bromine gas (Br₂) is produced.

On the other hand, when aqueous lead bromide is used, water molecules are present along with the lead bromide compound. During electrolysis, the water molecules can be reduced at the cathode instead of lead ions.

Reduction of water molecules produces hydrogen gas (H₂) because the reduction potential of water is lower than that of lead ions. The hydrogen gas is released at the cathode, while the lead ions (Pb²⁺) remain in the solution. At the anode, the bromide ions (Br⁻) are oxidized to form bromine gas (Br₂) as before.

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Answer:

In N-O, the delta + is on N and delta - on O while in O-F the delta + is on O and delta - is on F.

Explanation:

Hope this helped! :)

Happy Early Valentine's Day!

Brainliest, Please!

The fundamental nature of strategic management requires the awareness & understanding of outside forces & encourages strategic managers to adopt new ideas is known as Adaptation.

Three definitions of adaptability are connected. First, natural selection, a dynamic evolutionary process, adapts organisms to their environments, improving their evolutionary fitness. Second, it is a state that the populace has attained along that process. Thirdly, it is a phenotypic characteristic or adaptive trait that has been preserved and has evolved via natural selection and has a functional purpose in each individual organism.

History has recorded descriptions of adaptation going back to the time of the ancient Greek philosophers Empedocles and Aristotle. Natural theology of the 18th and 19th centuries saw adaptation as proof of the presence of a deity.

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PI2 ≈ 7.4 torrPI2 ≈ 7.4 torrs because it is directly proportional to the initial pressure of HIHI.

At equilibrium, the forward and reverse reactions occur at the same rate. Once equilibrium is reached, the amounts of each reactant and product remain constant. The equilibrium constant equation is a mathematical relationship that describes how the concentrations of products vary with the concentrations of reactants.

The state of a system whose properties are fixed under unvarying external conditions is called the equilibrium state. Chemical equilibria are dynamic in nature because reactants turn into products and products turn into reactants even after equilibrium is reached. However, the speed of forward and backward reactions is the same. Examples of balance include A book placed on the table. A car that moves at a constant speed. A chemical reaction in which the forward and reverse reaction rates are equal.

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Answer:

Explanation:

Observation 1 is correct because the crystals are just turning into water. They can be frozen back.

Answer: Lanthanoids

Explanation:

Activation energy is the minimum required energy that a system must absorb before the reaction can begin.

In chemical reaction, activation energy is the minimal amount of energy necessary or required to get atoms or molecules to a state where they undergo chemical transformation.

In this chemical reaction process;

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Rapid heating of a sample can have a significant effect on the observed melting point. When a sample is heated quickly, it may not have enough time to distribute the heat evenly, resulting in uneven melting.

This means that different parts of the sample may melt at different temperatures, leading to a broad melting point range or even inaccurate results. Additionally, rapid heating can cause thermal decomposition of the sample, leading to the release of gases, which can affect the melting point.

It is important to note that the rate of heating can vary depending on the type of sample being tested. Some samples may require a slower rate of heating to ensure accurate results, while others may be able to withstand rapid heating without any adverse effects.

In conclusion, rapid heating of a sample can have a significant effect on the observed melting point. It is important to carefully control the heating rate to ensure accurate and reliable results.

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