Balance the following equation in acidic solution using the lowest possible integers and give the coefficient of H+.
Cr2O72-(aq) + I-(aq) ? Cr3+(aq) + I2(aq)

Finding molar mass of NaOH

We know that

Molar mass of Sodium (Na) = 23 g/mol

Molar mass of Oxygen (O) = 32 g/mol

Molar mass of Hydrogen (H) = 2 g/mol

So,

Mass of 1 mole of NaOH

=> 1 × 23 + 1 × 16 + 1 × 1

=> 23 + 16 + 1

=> 40 g/mol

Now,

Mass = 40 × 0.25

Mass = 10 g

Answer:

613.0 g

Explanation:

2 L = 2000 mL

(30.65 g/100 mL)*2000 mL = 613.0 g

Given:

Solution:

Firstly,

We converted L to mL

2 L = 2 × 1000 mL = 2000 mL [ 1 L = 1000 mL ]

Now,

Solute = 30.65/100 × 2000

= 30.65 × 2000 / 100

= 61,300/100

= 613

The balanced equation to represent the picture shown is:

4 M₂N ----> 2 M₄ + 2 N₂

The mole ratio for the equation is 4 : 2 : 2

The mole ratio in a chemical reaction refers to the ratio of the number of moles of one substance involved in the reaction to the number of moles of another substance involved.

It is determined by the coefficients of the balanced chemical equation.

In a balanced chemical equation, the coefficients represent the relative amounts of the reactants and products. These coefficients can be used to establish the mole ratio between any two substances in the reaction.

For example:

4 M₂N ----> 2 M₄ + 2 N₂

The mole ratio for the equation is 4 : 2 : 2

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The intervals on which f(x) is positive and on which f(x) is negative from f(x)=x⁵ + x⁴ - 3x³ - 3x² - 9x - 9 is

f(x) > 0 if x ∈ (-1.54, ∞)
f(x) < 0 if x ∈ (-∞, -1.54)

To determine the intervals on which f(x) is positive and negative, we need to find the roots of the equation f(x) = 0 first. Then, we can use test points to determine the sign of f(x) in each interval.

Using a graphing calculator or software, we can find that f(x) has only one real root at approximately x = -1.54. This means that f(x) is always positive or always negative, except at x = -1.54 where it changes sign.

To determine the sign of f(x) on either side of the root, we can choose test points in those intervals. For example, if we choose x = -2, we have f(-2) = -5, which is negative. If we choose x = -1, we have f(-1) = -3, which is also negative. If we choose x = 0, we have f(0) = -9, which is negative. This means that f(x) is negative for x < -1.54.

Similarly, if we choose x = -1.5, we have f(-1.5) = 1.875, which is positive. If we choose x = -1.4, we have f(-1.4) = -0.573, which is negative. This means that f(x) is positive for -1.54 < x < -1.4.

Finally, if we choose x = -1.3, we have f(-1.3) = 1.1, which is positive. If we choose x = 0, we have f(0) = -9, which is negative. This means that f(x) is positive for x > -1.54.

Therefore, we can summarize the intervals on which f(x) is positive and negative as follows:

f(x) > 0 if x ∈ (-1.54, ∞)

f(x) < 0 if x ∈ (-∞, -1.54)

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Answer:

H - Cl2 +NaBr -> Br2+2NaCl

Answer: 1.2 M

Explanation:

molarity = (moles of solute)/(liters of solution) = 2.4 / 2.0 = 1.2 M

If more \(H_{2}\) and \(N_{2}\) are added to the reaction 3\(H_{2}\) + N2 → 2\(NH_{3}\) after it reaches equilibrium, two events will occur Shift in Equilibrium and Increased Yield of \(NH_{3}\)

1. Shift in Equilibrium: According to Le Chatelier's principle, when additional reactants are added, the equilibrium will shift in the forward direction to consume the added reactants and establish a new equilibrium. In this case, more \(NH_{3}\) will be produced to counteract the increase in \(H_{2}\) and \(N_{2}\).

2. Increased Yield of \(NH_{3}\): The shift in equilibrium towards the forward reaction will result in an increased yield of \(NH_{3}\). As more \(H_{2}\) and \(N_{2}\) are added, the reaction will favor the production of \(NH_{3}\) to maintain equilibrium. This will lead to an increase in the concentration of \(NH_{3}\) compared to the initial equilibrium state.

It is important to note that the equilibrium position will ultimately depend on factors such as the concentrations of \(H_{2}\), \(N_{2}\), and \(NH_{3}\), as well as the temperature and pressure of the system. By adding more reactants, the equilibrium will adjust to achieve a new balance, favoring the formation of more \(NH_{3}\).

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If Half- life of an isotope is 30 days and it was assumed that the person ate 100 Bq of isotope, there are 50 transformations in the stomach.

The radioactive decay of a sample of an isotope can be characterized by the half-life of that isotope. When a radioisotope undergoes decay, its nucleus becomes unstable, and it emits particles or energy to become more stable. The half-life of an isotope is the time it takes for half of the original sample to decay. The question states that the half-life of an isotope is 30 days, and the person ingested 100 Bq of isotope. It also says to calculate the number of transformations in the stomach using GI track model information .

Since the isotope has a half-life of 30 days, we can use the following formula to find the number of transformations in the stomach:` N = N₀ (1/2)^(t/T₁/₂)`where: N₀ = initial number of nuclei N = final number of nuclei (after time t)T₁/₂ = half-life of the isotope The isotope has a half-life of 30 days, so T₁/₂ = 30 days. The question doesn't specify how long the person has had the isotope in their stomach, so we'll assume it's been there for one half-life, or 30 days. Therefore, t = 30 days.

Substituting into the formula:` N = 100 (1/2)^(30/30)`Simplifying:` N = 100 (1/2)^1`Evaluating:`N = 50`So after 30 days in the stomach, the person would have 50 Bq of the isotope left. Therefore, the number of transformations in the stomach is the difference between the initial number of transformations (100 Bq) and the final number of transformations (50 Bq):`Number of transformations in stomach = 100 - 50 = 50 transformations. Therefore, there are 50 transformations in the stomach.

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Option (d) is correct. To test the quality of water source in the area, the important technique that Shameka used s verification of results.

To test the quality of water, the qualitative and quantitative measurements are needed from time to time to constantly monitor the quality of water from the various sources of supply. there have to  initiate remedial measures with the suppliers when water supply from outside is polluted. Testing procedures and parameters may be grouped into physical, chemical, bacteriological and microscopic categories.

· The physical tests indicate properties detectable by the senses.

· The chemical tests determine the amounts of mineral and organic   substances that affect water quality.

· Bacteriological tests show the presence of bacteria, characteristic of  pollution.

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Answer:

no it should not that place is historical and so they should make where u can visit but protect it as a historical landmark

Answer: 3.156 * 10^7

Explanation: I do not really understand your question. You answered it yourself!

Scientific notation shortens large numbers. The number right after the decimal point can only be between 1 and 9, which you did correctly. When converting to scientific notation, the exponent of ten is based on how many places you moved the decimal and the direction you moved it (left, positive; right, negative). In this case, the exponent of ten is a positive seven.

You did everything correctly :) Good job!

The option that is not a known advantage of natural gas over other fossil fuels is: C. It is far more abundant than any other fossil fuel.

Natural gas has several advantages over other fossil fuels. Firstly, it burns more completely than other fossil fuels, which means it produces fewer pollutants and greenhouse gases.

This leads to the second advantage, as natural gas burns more cleanly than other fossil fuels, resulting in reduced emissions of pollutants such as sulfur dioxide, nitrogen oxides, and particulate matter.

However, when it comes to abundance, natural gas is not necessarily more abundant than other fossil fuels.

While natural gas reserves can be substantial, the availability and reserves of other fossil fuels like coal and oil are also significant. Therefore, option C is not a known advantage of natural gas over other fossil fuels.

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Equilibrium is a state in a chemical reaction where the rate of the forward reaction is equal to the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products. It is represented by a double arrow (⇌) in chemical equations. In the given equation: H₂O (g) + CO (g) ⇌ H₂(g) + CO₂(g) + 42 KJ

To shift the equilibrium to the right, one or more of the following could occur:

Increasing the concentration of H₂ or CO₂

Decreasing the concentration of H₂O or CO

Increasing the pressure

Removing some of the products (H₂ and CO₂)

Decreasing the temperature

To shift the equilibrium to the left, one or more of the following could occur:

Decreasing the concentration of H₂ or CO₂

Decreasing the pressure

Increasing the temperature

If pressure is increased, the equilibrium will shift in the direction that produces fewer moles of gas. In this case, since there are fewer moles of gas on the right side of the equation (H₂ and CO₂), the equilibrium will shift to the right. If heat is added, the equilibrium will shift in the endothermic direction to absorb the additional heat. In this case, the forward reaction is endothermic (42 KJ on the right side), so the equilibrium will shift to the right to consume the added heat.

If CO is added, the equilibrium will shift to the right to consume the additional CO.The concentration of H₂O and CO₂ will increase, while the concentrations of H₂ and CO will decrease until a new equilibrium is reached.

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Answer:

4

Explanation:

Answer:

12.60 m has 4 significant figures.

Explanation:

On Edge 2021

Moth flakes dissolve in methanol at a rate of A, 16.7 g/100mL.

The solubility of moth flakes in methanol is the maximum amount of the solute that can dissolve in a given amount of solvent at a given temperature.

To calculate the solubility of moth flakes in methanol, divide the mass of moth flakes by the volume of methanol and multiply by 100 to express the result as grams per 100 mL of solution.

So, the solubility of moth flakes in methanol is:

Solubility = (mass of moth flakes / volume of methanol) x 100

Solubility = (12.54 g / 75.2 mL) x 100

Solubility = 16.7 g/100mL

Therefore, the solubility of moth flakes in methanol is 16.7 g/100mL.

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Option c) NH4Cl will hydrolyze when dissolved in water.

Hydrolysis is a chemical reaction that occurs when a salt reacts with water, resulting in the formation of an acidic or basic solution. In this case, when NH4Cl (ammonium chloride) is dissolved in water, it undergoes hydrolysis.

The ammonium ion (NH4+) is the conjugate acid of ammonia (NH3), which is a weak base. When NH4Cl dissociates in water, the ammonium ion reacts with water to form NH3 and H3O+ (hydronium ion). This process is called hydrolysis.

NH4+ (aq) + H2O (l) ↔ NH3 (aq) + H3O+ (aq)

The formation of NH3 leads to an increase in the concentration of hydroxide ions (OH-) in the solution, making it slightly basic. At the same time, the presence of H3O+ ions makes the solution slightly acidic. Therefore, the hydrolysis of NH4Cl results in a slightly acidic and slightly basic solution.

In contrast, salts like NaCl and KCl do not undergo hydrolysis when dissolved in water because they consist of cations (Na+ and K+) and anions (Cl-) that do not react with water to form acidic or basic species.

Na2SO4 (sodium sulfate) is also an example of a salt that does not undergo hydrolysis. The sulfate ion (SO42-) does not react with water to form acidic or basic species, so the solution remains neutral.

Therefore, among the given options, only NH4Cl undergoes hydrolysis when dissolved in water.

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The question is incomplete. Find the full content below:

Which one of the following salts when dissolved in water will hydrolyse?

a) NaCl

b) KCl

c) NH4Cl

d)Na2SO4

Specific heat capacity of a substance is the amount of heat required to raise the temperature by one degree Celsius of one gram of a substance. Therefore, -1.046KJ of heat are given off when 5.0 g of water cools from 75 °C to 25 °C.

Enthalpy term is basically used in thermodynamics to show the overall energy that a matter have. Mathematically, Enthalpy is directly proportional to specific heat capacity of a substances.

Mathematically,

Enthalpy=mass of water× specific heat capacity of water× Change in temperature

mass of water= 5.0 g

specific heat capacity of water=4.186J/g°C

Change in temperature=final temperature - initial temperature

= 25 °C- 75 °C

=-50°C

Substituting all the given values in the above equation, we get

Enthalpy=5.0× 4.186× -50°C

Enthalpy=-1,046.5J=-1.046KJ

Therefore, -1.046KJ of heat are given off when 5.0 g of water cools from 75 °C to 25 °C.

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The final pH of the solution after adding the NaOH is approximately 4.87.

To determine the final pH after adding 7 ml of 1.5 M NaOH to a 200 ml buffer solution of a weak acid with a pKa of 6.45, we need to consider the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa and the ratio of the conjugate base to the weak acid.

First, we calculate the moles of the weak acid initially present in the buffer solution:

Moles of weak acid = volume of buffer (L) × concentration of weak acid (M)

= 0.200 L × 2.0 M

= 0.400 moles

Next, we calculate the moles of the added NaOH:

Moles of NaOH = volume of NaOH (L) × concentration of NaOH (M)

= 0.007 L × 1.5 M

= 0.0105 moles

Since NaOH is a strong base, it completely reacts with the weak acid in the buffer to form the conjugate base.

Moles of conjugate base = moles of added NaOH

= 0.0105 moles

Now, we can calculate the ratio of the conjugate base to the weak acid:

Ratio of conjugate base to weak acid = moles of conjugate base / moles of weak acid

= 0.0105 moles / 0.400 moles

= 0.02625

Using the Henderson-Hasselbalch equation:

pH = pKa + log10(conjugate base/weak acid)

= 6.45 + log10(0.02625)

= 6.45 + (-1.58)

= 4.87

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Given what we know, we can confirm that the patients will have died from a "lack of oxygen" even with normal blood oxygen levels due to the fact that cyanide blocks the function of oxygen in the body.

The main function of cyanide is that it blocks the oxygen from being used in the electron transport chain. This chain creates ATP, the energy needed for the body to complete any functions. Since the oxygen cannot be used despite it being present, the patients die of a lack of oxygen even with normal levels.

Therefore, we can confirm that the patients will have died from a "lack of oxygen" even with normal blood oxygen levels due to the fact that cyanide blocks the function of oxygen in the body.

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To determine the domain and range of a relation, find the set of possible input values (domain) and the set of possible output values (range).


The domain of a relation is the set of all possible input values, whereas the range of a relation is the set of all possible output values. To determine the domain and range of a relation, follow these steps:

Step 1: Identify all the input values that are allowed for the relation.

Step 2: Determine the output values that correspond to each input value.

Step 3: Write down the set of all the input values as the domain of the relation, and the set of all the output values as the range of the relation. It's essential to note that some relations may have restrictions on their domains and ranges, so it's necessary to pay attention to any such restrictions.

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The enzyme β-ketothiolase cleaves off the acetyl-CoA molecule from the 3-ketoacyl-CoA, releasing acetyl-CoA and the remaining fatty acid chain forms acyl-CoA, which is two carbons shorter than the original fatty acid chain.

The complete reaction scheme for the catabolism of Oleoyl-CoA is given below:Oleoyl-CoA is broken down into acetyl-CoA, releasing 150 ATP molecules by the process of Beta-oxidation. The complete reaction scheme for the catabolism of Oleoyl-CoA is given below:

Step 1: Oleoyl-CoA is transported to the mitochondria matrix from the cytoplasm with the help of the carnitine shuttle system.

Step 2: The enzyme Acyl-CoA dehydrogenase catalyzes the removal of two hydrogen atoms from the alpha and beta carbons in the fatty acid chain and oxidizes it. This process forms a double bond between the alpha and beta carbon atoms, leading to the formation of trans-Δ2-enoyl-CoA.

Step 3: The enzyme enoyl-CoA hydratase adds a water molecule to the trans-Δ2-enoyl-CoA, converting it into L-3-hydroxyacyl-CoA.

Step 4: The enzyme L-3-hydroxyacyl-CoA dehydrogenase oxidizes L-3-hydroxyacyl-CoA, releasing a hydrogen ion (H+) and two electrons (2e-) and converts it into 3-ketoacyl-CoA.

Step 5: The enzyme β-ketothiolase cleaves off the acetyl-CoA molecule from the 3-ketoacyl-CoA, releasing acetyl-CoA and the remaining fatty acid chain forms acyl-CoA, which is two carbons shorter than the original fatty acid chain.

The cycle starts again, and this process is repeated until the fatty acid chain is completely degraded.

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The balanced equation of magnesium and sulfuric acid is:

The student was have seen an explosive reaction if sodium was placed in water.

The order of the reactivities of the metals from most to least reactive is: lithium >> calcium >> copper

It would be difficult to determine the order of reactivity of magnesium and zinc because they show the same reaction. This can be determined by finding out which metal displaces the other from the solution of their salt.

Reactive metals are metals that easily form positive ions by readily giving up or donating their electrons.

The reactive metals are mainly found in group 1A and group 2A of the periodic table.

Reactive metals react with water to liberate hydrogen gas.

For example: 2 Li (s) + 2 H₂O ---> 2 LiOH (aq) + H₂ (g)

Reactive metals react with dilute acids to liberate hydrogen gas and form salts.

Mg + 2 HCl (aq) ---> MgCl₂ (aq) + H₂ (g)

The rate of the reactions above decreases with a decrease in the reactivity of the reactive metals.

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Answer:

Explanation: l m n o p q r s t u v w x y z

They are extremely small infront of earth size.

The percent change in volume is 21 %.

given that :

temperature , T1 = 273 -28 = 245 K

temperature , T2 = 273 + 9 = 282 K

pressure P1 = P

pressure P2 = P - O.O5 P = 0.95 P

Volume V1 = V

volume V2 = ?

using ideal gas equation , we get :

P1 V1 / T1 = P2 V2 / T2

V2 = (P1 V1 T2 ) / P2 T1

V2 = ( P × V × 282 ) / ( 0.95 P × 245 P )

V2 = 1.211 V

change in volume is given as = V2 - V1

                                                 =  1.211 V - V

                                                 = 0.211 V

Percent change in volume is = (0.21 V)) / V ) × 100 %

                                                 = 21 %

Thus, For many purposes we can treat propane C3H8 as an ideal gas at temperatures above its boiling point of −42.°C Suppose the temperature of a sample of propane gas is raised from -28.0°C to 9.0°C, and at the same time the pressure is increased by 5.0% . the percent change in volume in 21 %.

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Acetic acid is the best option for making a buffer with pH 2.00. Option 3 is the answer.

What is the best weak acid option for making a buffer with pH 2.00?

To make a buffer with a pH of 2.00, we need to select a weak acid with a pKa close to this value. The pKa of an acid is the pH at which half of the acid molecules are dissociated.

For a buffer, we want to choose an acid where the pH is close to its pKa value because at this point, the concentration of both the acid and its conjugate base will be approximately equal.

The Henderson-Hasselbalch equation can be used to calculate the pH of a buffer solution:

\(\mathrm{pH} &= \mathrm{pKa} + \log{\left(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right)} \\\)

Where [A-] is the conjugate base concentration and [HA] is the weak acid concentration.

At pH 2.00, the [H+] concentration is 10⁻² M. Therefore, we need to choose a weak acid with a pKa close to 2.00.

One possible option is acetic acid (CH₃COOH), which has a pKa of 4.76. Using the Henderson-Hasselbalch equation, we can calculate the required ratio of [A-]/[HA]:

\(2.00 &= 4.76 + \log{\left(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right)}\)

\(\log{\left(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right)} &= -2.76 \\\\\frac{[\mathrm{A^-}]}{[\mathrm{HA}]} &= 10^{-2.76} \\\\\frac{[\mathrm{A^-}]}{[\mathrm{HA}]} &= 1.63 \times 10^{-3}\)

Thus, for a buffer with a pH of 2.00, we could mix acetic acid and its conjugate base (sodium acetate) in a 1:163 ratio, which is option 3.

The complete question is -
A student must make a buffer with a ph of 2.00. determine which weak acid is the best option at the specified pH?
1. Sodium disulfate monohydrate

2. Propionic acid

3. Acetic acid

4. Formic acid

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The IUPAC name of the compound is   c. (s)-4-methyl-2-hexyne.

The IUPAC naming system is used to systematically name organic compounds based on their structure and functional groups. To determine the IUPAC name of the given compound, we need to consider the following:

1. Numbering of the carbon chain: Start numbering from one end of the longest carbon chain, giving priority to the functional groups present. In this case, the longest carbon chain is six carbons long.

2. Identify and name the substituents: Determine the substituents attached to the main carbon chain and assign appropriate locants (numbers) to each substituent. In this case, we have a methyl group attached to the fourth carbon of the chain.

3. Identify the triple bond: Determine the position of the triple bond and include it in the name. In this case, the triple bond is between the second and third carbons of the chain.

4. Assign stereochemistry (R/S) if applicable: Determine the stereochemistry of chiral centers, if present, by assigning priority to the substituents based on the Cahn-Ingold-Prelog priority rules. In this case, there are no chiral centers present, so stereochemistry is not applicable.

Based on the above considerations, the IUPAC name of the compound is c. (s)-4-methyl-2-hexyne

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Complete Question:

What is the IUPAC name of the following compound: (R)-3-methyl-4-hexyne, (S)-3-methyl-4-hexyne, (S)-4-methyl-2-hexyne, or (R)-4-methyl-2-hexyne?

Answer:9.18 m/s

Explanation:

The average speed for the entire trip is found ......total distance/total time. Remember r*t=d, so divide both sides by t and get r = d/t.

So the cyclist went 800+500+1200 m = 2500 m for total distance.

10t =800 leads to t=80 sec

5t=500 leads to t=100 sec

13t = 1200 leads to t = 92.3 sec

total time is 272.3 sec

Average speed for the entire trip is 2500 / 272.3 = 9.18 m/s