Which statements are TRUE about fossil fuels? (Select all that apply.)

They are in a limited supply.

They do not replenish themselves.

They are expensive to extract compared to other forms of energy.

They release large amounts of carbon dioxide when burned
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The amount of volume of KOH solution that should be added to make 500mL of a 0.300M solution is 200mL.

The volume of a solution given the concentration can be calculated using the following expression;

CaVa = CbVb

Where;

According to this question, we are to calculate how many mL of a 0.75 M OH solution that should be added to a 500 mL flask to make 500 mL of a 0.300 M KOH solution.

0.75 × Va = 500 × 0.3

0.75Va = 150

Va = 150/0.75

Va = 200mL

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Answer:

9.98%

Explanation:

To find the percent of NaCl in the mixture, we need to divide the mass of NaCl by the total mass of the mixture, and then multiply by 100 to express it as a percentage.

Step 1: Find the total mass of the mixture

total mass = mass of NaCl + mass of water

total mass = 23.5 g + 212 g

total mass = 235.5 g

Step 2: Calculate the percent of NaCl

% NaCl = (mass of NaCl / total mass) x 100

% NaCl = (23.5 g / 235.5 g) x 100

% NaCl = 0.0997876857 x 100

% NaCl = 9.978768677%

% NaCl = 9.98%

Therefore, the percent of NaCl in the mixture is 9.98%.

B. To plot the data on a bar chart, draw a horizontal axis for metals and a vertical axis for time to complete the reaction. Then, draw bars for each metal that represent the amount of time required to complete the reaction. The height of the bars must match the time values ​​in the table.

C. No, Emilia was not correct in her forecast. According to the data, aluminum reacted 100 seconds faster than magnesium, which reacted in 50 seconds. Thus aluminum reacts more rapidly with hydrochloric acid than magnesium.

From most reactive to least reactive, the metals are as follows:

This order is consistent with the reactivity series, which is:

We are unable to estimate the reactivity of potassium, sodium, calcium, copper, silver, or gold from this experiment because those variables are not present in the data.

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the oxidation number of copper in copper oxide is 2...

Answer: +2

Explanation: Copper has a +2 oxidation number in CuO.

This is due to the fact that oxygen has an oxidation number of 2, and the entire chemical has a neutral charge. Consequently, the following equation can be used to determine copper's oxidation number:

(+2) + (-2) = 0

In order to counteract the -2 oxidation number of oxygen in CuO, copper must have an oxidation number of +2.

Answer:

3.8 ⋅*10^-14

Explanation:

The standard free energy change (ΔG°) for the reaction at 298 K can be calculated using the following equation:

ΔG° = ΣnΔGf°(products) - ΣnΔGf°(reactants)

where ΔGf° is the standard molar free energy of formation of the species and n is the stoichiometric coefficient.

ΔG° = [1×ΔGf°(Fe2O3)] - [1×ΔGf°(FeO) + 1×ΔGf°(Fe) + 1×ΔGf°(O2)]

ΔG° = [1×(-822.16 kJ/mol)] - [1×(-271.9 kJ/mol) + 1×(0 kJ/mol) + 1×(0 kJ/mol)]

ΔG° = -550.26 kJ/mol

The standard enthalpy change (ΔH°) and standard entropy change (ΔS°) can be used to calculate the standard free energy change (ΔG°) at any temperature using the following equation:

ΔG° = ΔH° - TΔS°

where T is the temperature in Kelvin.

ΔG° = ΔH° - TΔS° = (-550.26 kJ/mol) - (298 K)(-0.08996 kJ/K/mol) = -524.05 kJ/mol

Now, we can calculate the equilibrium constant (K) for the reaction at 298 K using the following equation:

ΔG° = -RTlnK

where R is the gas constant (8.314 J/K/mol) and T is the temperature in Kelvin.

-524.05 kJ/mol = -(8.314 J/K/mol)(298 K)lnK

lnK = -200.16

K = e^(-200.16) = 3.89×10^(-87)

Therefore, the value of K for the reaction at 25 °C is 3.89×10^(-87). Answer: 3.8 ⋅*10^-14.

Answer:

56.11 g/mol

Explanation:

To determine the molar mass of potassium hydroxide, we need to find the atomic mass of each element in the compound and add them up.

The atomic mass of potassium (K) is 39.10 g/mol, the atomic mass of oxygen (O) is 16.00 g/mol, and the atomic mass of hydrogen (H) is 1.01 g/mol.

So, the molar mass of potassium hydroxide (KOH) is:

Molar mass of K = 39.10 g/mol

Molar mass of O = 16.00 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of KOH = Molar mass of K + Molar mass of O + Molar mass of H

= 39.10 g/mol + 16.00 g/mol + 1.01 g/mol

= 56.11 g/mol

Therefore, the molar mass of potassium hydroxide (KOH) is 56.11 g/mol.

The most basic level of organization that can perform functions like converting food into energy is the cell.

The most basic level of organization that can perform functions like converting food into energy is the cell. Cells are the fundamental units of life and are capable of various functions, including metabolism, which involves converting food into energy through processes such as cellular respiration.

Cells can be found in all living organisms, from single-celled bacteria to complex multicellular organisms like plants and animals. Within a cell, various organelles such as mitochondria, which are responsible for energy production, carry out specialized functions to support the overall cellular function. Therefore, the cell is the smallest and most basic level of organization that is capable of performing functions like converting food into energy.

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Johannes Brsted and Thomas M. Lowry, two chemists, identified the Bromsted-Lowry acids and bases as a particular kind of acid-base reaction in 1923.

Acids are substances that give a base a proton (H+), whereas bases are substances that take a proton from an acid. In a Bromsted-Lowry acid-base reaction, the acid gives the base a proton in order to create the conjugate base and the conjugate acid, two new compounds.

Nitric acid (HNO3), sulfuric acid (H2SO4), and hydrochloric acid (HCl) are a few examples of acids. Sodium hydroxide (NaOH), ammonium hydroxide (NH4OH), and potassium hydroxide (KOH) are a few examples of bases.

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The method for determining the sum of Vitamin C in fruit juice ordinarily includes adding an indicator (such as DCPIP) to the juice test and titrating the test with a standard arrangement of ascorbic corrosive until the marker changes colour.

In any case, there are a few variables that seem to make this strategy troublesome to utilize for other natural product juices such as cranberry, blackcurrant, or pomegranate juice:

In this manner, whereas the strategy for deciding the sum of Vitamin C in natural product juice can be a valuable apparatus, it may not be appropriate for all sorts of natural product juices and may have to be be adjusted or adjusted to account for these variables. 

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Answer:

A position-time graph can support the statement that a positive slope results when an individual is moving away from the origin. This is because a positive slope on a position-time graph represents a positive velocity, which means that the object is moving in a positive direction (away from the origin). Conversely, a negative slope would indicate that the object is moving in a negative direction (towards the origin).

Option A and B are incorrect. A negative slope on a position-time graph indicates that the object is moving towards the origin, not away from it. A horizontal line on a position-time graph indicates that the object is not moving at all (velocity is zero), not moving at a non-zero velocity.

Option D is also incorrect. The speed of an individual can be determined from a position-time graph by calculating the slope of the graph at any point, which gives the velocity (speed and direction) of the individual at that point..

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The theoretical yield of MgO for Trial 1 is 0.348 g, and for Trial 2 is 0.307 g. The percent yield of MgO for Trial 1 is 58.0% and for Trial 2 is 159.2%. The average percent yield of MgO for the two trials is 108.6%.

To calculate the theoretical yield of MgO, we need to use the balanced chemical equation for the reaction between magnesium (Mg) and oxygen (O2) to form magnesium oxide (MgO):

2Mg + O₂ → 2MgO

According to the stoichiometry of this equation, 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO. Therefore, we need to determine the number of moles of Mg in each trial and use the mole ratio to find the theoretical yield of MgO.

For Trial 1:

The mass of Mg used is: 26.682 g - 27.012 g = 0.330 g

The molar mass of Mg is 24.31 g/mol, so the number of moles of Mg is:

0.330 g / 24.31 g/mol = 0.0136 mol Mg

According to the balanced equation, 2 moles of Mg produce 2 moles of MgO, so the theoretical yield of MgO is:

0.0136 mol Mg x (2 mol MgO / 2 mol Mg) x (40.31 g MgO/mol) = 0.348 g MgO

For Trial 2:

The mass of Mg used is: 26.987 g - 26.695 g = 0.292 g

The number of moles of Mg is:

0.292 g / 24.31 g/mol = 0.0120 mol Mg

The theoretical yield of MgO is:

0.0120 mol Mg x (2 mol MgO / 2 mol Mg) x (40.31 g MgO/mol) = 0.307 g MgO

To calculate the percent yield of MgO, we need to use the following formula:

Percent yield = (actual yield / theoretical yield) x 100%

For Trial 1:

The actual yield of MgO is: 27.214 g - 27.012 g = 0.202 g MgO

The percent yield of MgO is:

(0.202 g / 0.348 g) x 100% = 58.0%

For Trial 2:

The actual yield of MgO is: 27.183 g - 26.695 g = 0.488 g MgO

The percent yield of MgO is:

(0.488 g / 0.307 g) x 100% = 159.2%

To calculate the average percent yield of MgO for the two trials, we add the percent yields and divide by 2:

Average percent yield = (58.0% + 159.2%) / 2 = 108.6%

Therefore, the theoretical yield of MgO for Trial 1 is 0.348 g, and for Trial 2 is 0.307 g. The percent yield of MgO for Trial 1 is 58.0% and for Trial 2 is 159.2%. The average percent yield of MgO for the two trials is 108.6%.

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The maximum mass of water that can be produced by the reaction is 43.3 g, rounded to three significant figures.

The balanced chemical equation for the reaction between butane and oxygen is:

C4H10 + 13/2 O2 → 4 CO2 + 5 H2O

From the equation, we can see that 1 mole of butane reacts with 13/2 moles of oxygen to produce 5 moles of water.

moles of butane = 34. g / 58.12 g/mol = 0.585 mol

moles of oxygen = 200. g / 32.00 g/mol = 6.25 mol

Determining the limiting reactant.

butane : oxygen = 0.585 mol : 6.25 mol

= 0.0936 : 1.00

stoichiometric ratio = 1 : 13/2

= 0.7692 : 1.00

Since the actual ratio is lower than the stoichiometric ratio for oxygen, it is the limiting reactant.

The maximum amount of water that can be produced is determined by the amount of limiting reactant (oxygen).

moles of water = 5/13 * 6.25 mol

= 2.403 mol

Finally, we can convert the moles of water to grams:

mass of water = 2.403 mol * 18.015 g/mol

= 43.3 g

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1) Bromination of propylene to form 2-bromopropane using NBS and a Lewis acid catalyst.

Bromination is a chemical process in which bromine is added to a molecule. This can be done by either direct substitution or as a substitution reaction, allowing for the addition of one or more bromine atoms to the molecule. Bromination is a commonly used organic reaction, particularly in the laboratory, and can be used to alter the properties of a compound. It can also be used to produce a wide range of products, including aromatics and halogenated compounds. Bromination is particularly useful in pharmaceutical synthesis, as the products of this reaction often have desirable bioactivity.

2) Reduction of 2-bromopropane to 2-propanol using NaBH₄
3) Reaction of 2-propanol with phosphorus tribromide to form 2-bromopropanol
4) Alkylation of 2-bromopropanol with methyl iodide to form 2-bromopropyl methyl ether
5) Reduction of 2-bromopropyl methyl ether to 2-methoxypropane using NaBH₄
6) Reaction of 2-methoxypropane with phosphorus tribromide to form 2-bromo-2-methoxypropane
7) Reduction of 2-bromo-2-methoxypropane to Compound X using NaBH₄

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The value of H (in kJ) when 5.10 g of H2(g) is formed is 326 kJ (option B).

The given reaction is: CH3OH(I) —> CO(g)+2H2(g)

From the given value of H, we know that when one mole of CH3OH reacts, 128.1 kJ of heat energy is absorbed.

The molar mass of H2 is 2 g/mol. So, 5.10 g of H2 is equivalent to 5.10/2 = 2.55 moles of H2.

From the balanced equation, we can see that two moles of H2 are produced for each mole of CH3OH that reacts.

So, 2.55 moles of H2 are produced by 1.275 moles of CH3OH reacting (2.55/2).

Therefore, the amount of heat energy absorbed when 1.275 moles of CH3OH reacts can be calculated as:

Q = n x ΔH = 1.275 mol x 128.1 kJ/mol = 163.28 kJ

Since this amount of heat energy is absorbed when 1.275 moles of CH3OH reacts, to find the amount of heat energy absorbed when 2.55 moles of H2 is formed, we can simply double the value of Q:

Q = 2 x 163.28 kJ = 326.56 kJ

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the original temperature of the h ot iron bar was 327.9°C.

We can use the specific heat of iron to do this:

Q1 = m1 * C1 * (Ti - 32.8°C)

Q1 = 408 g * 0.45 J/g °C * (Ti - 32.8°C)

Q1 = 183.6 J/g °C * (Ti - 32.8°C)

where m1 is the mass of the iron bar, C1 is the specific heat of iron, and Ti is the initial temperature of the iron bar.

Next, let's calculate the heat gained by the cold water when it is heated from 20.0°C to 32.8°C:

Q2 = m2 * C2 * (32.8°C - 20.0°C)

Q2 = 1000 g * 4.184 J/g °C * (32.8°C - 20.0°C)

Q2 = 52272 J

where m2 is the mass of the water, C2 is the specific heat of water.

Since the energy lost by the iron bar is gained by the water, we can set Q1 equal to Q2:

Q1 = Q2

183.6 J/g °C * (Ti - 32.8°C) = 52272 J

Now, let's solve for Ti:

183.6 J/g °C * Ti - 60236.8 J = 0

183.6 J/g °C * Ti = 60236.8 J

Ti = 327.9°C

Therefore, the original temperature of the h ot iron bar was 327.9°C.

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According to the information, we can infer that the difference between photographs 1 and 2 originate from the translation of the Moon around the earth (option C).

To explain the difference between both images we must take into account the movement patterns of the earth and the moon. In the case of the earth, it has 2 main movements, which are rotation on its own axis and translation around the sun.

On the other hand, the moon has a translational movement around the earth, which is what causes the different lunar phases. This motion causes the moon to appear partially shadowed from the earth because the earth blocks the sunlight.

Based on the above, we can infer that the correct answer is option C because this phenomenon is caused by the translation of the moon.

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The molar enthalpy for the reaction 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) is 266.4 kJ/mol.

The molar enthalpy is determined from Hess's law as follows:

Equation 1 x2:

2 N₂ (g) + 2 O₂ (g) → 4 NO (g) ∆H° = 361.2 kJ/mol

Equation 3 x3, :

6 H₂ (g) + 3 O₂ (g) → 6 H₂O (g) ∆H° = -1451.1 kJ/mol

Equation 2 x -4:

-8 N₂ (g) - 12 H₂ (g) → -8 NH₃ (g) ∆H° = 367.2 kJ/mol

Adding the equations together:

-6 N₂ (g) - 6 H₂ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) - 8 NH₃ (g) ∆H° = 266.3 kJ/mol

Multiplying the equation above by -1/2:

3 N₂ (g) + 3 H₂ (g) - 5/2 O₂ (g) → -2 NO (g) - 3 H₂O (g) + 4 NH₃ (g) ∆H° = -133.2 kJ/mol

Multiplying the above equation by -2:

4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) ∆H° = 266.4 kJ/mol

This is the molar enthalpy of the given reaction

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Explanation:

The temperature change of a substance when it absorbs or loses energy can be calculated using the specific heat capacity of the substance. The specific heat capacity of water is approximately 4.18 J/(g°C), which means that it takes 4.18 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.

To calculate the temperature change of the water sample when 50 joules of energy is added, we need to use the following equation:

q = m * c * ΔT

where q is the amount of energy absorbed by the water, m is the mass of the water sample, c is the specific heat capacity of water, and ΔT is the resulting temperature change.

Rearranging the equation to solve for ΔT, we get:

ΔT = q / (m * c)

Plugging in the values, we get:

ΔT = 50 J / (m * 4.18 J/(g°C))

We need to know the mass of the water sample to calculate the temperature change. Let's assume a mass of 10 grams:

ΔT = 50 J / (10 g * 4.18 J/(g°C))

ΔT = 1.2°C

Therefore, if 50 joules of energy is added to a 10-gram sample of water, the resulting temperature change will be approximately 1.2 degrees Celsius.

A mass of 37.0 g of sulfur must be used to produce 25.7 L of gaseous sulfur dioxide at STP.

The molar ratio of S₈ to SO₂ is 1:8.

At STP, one mole of gas occupies 22.4 L. Therefore, 25.7 L of SO₂ gas will contain;

25.7 L / 22.4 L/mol = 1.15 mol of SO₂.

The number of moles of S₈ needed is calculated as;

= 1.15 mol SO₂ / 8 mol S₈ per 1 mol SO₂

= 0.144 mol S₈.

The mass of S₈ needed is calculated as;

0.144 mol S₈ × 256.6 g/mol = 37.0 g of S₈.

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Answer:

A solvent front is the point on a chromatography paper or plate where the solvent has reached the end of the stationary phase and has migrated as far as it can go. It is the farthest point reached by the solvent in the chromatography process.

Answer:

V ≈ 2144.66 m³

Explanation:

Volume of sphere formula is:

V = 4/3 πr³

Radius is half the diameter so we divide the given diameter, 16 by 2 to get 8, the radius. Now we can solve

V = 4/3 π (8)³

V = 4/3 (512π)

V = 2048/3 π

V ≈ 2144.66 m³

Answer:

4/3 x π

Explanation:

First, we need to calculate the heat gained by the cold water and the heat lost by the hot water:

Qcold = mcΔT = (30 g)(4.184 J/g-°C)(1.93°C) = 242.06 J

Qhot = mcΔT = (70 g)(4.184 J/g-°C)(-1.93°C) = -546.53 J

Since energy is conserved, we can assume that the heat gained by the cold water and calorimeter is equal to the heat lost by the hot water:

Qcold + Qcalorimeter = Qhot

Qcalorimeter = Qhot - Qcold

Qcalorimeter = -546.53 J - 242.06 J = -788.59 J

Therefore, the heat capacity of the calorimeter can be calculated as:

Ccalorimeter = Qcalorimeter / ΔT

Ccalorimeter = (-788.59 J) / (1.93°C)

Ccalorimeter ≈ -408.4 J/°C

Note that the negative sign indicates that the calorimeter loses heat when the system gains heat, which is expected since the calorimeter is absorbing some of the heat from the hot water.

Answer:

D

Explanation:

to keep seeds safe and make them easier to spread

Answer:

D. to keep seeds safe and make them easier to spread

Explanation:

The main reason plants grow fruit is to aid in the protection and spreading of seeds. The fruit protects the seeds and also helps to spread them. Many fruits are good to eat and attract small animals, such as birds and squirrels, who like to feed on them. The seeds pass through them unharmed and then get spread through their droppings. So, the correct answer would be D.

Answer:

3.29 grams

Explanation:

This is found by multiply 2.35 L by 1.4 g/L that is because the liters will cancel each other out leaving just grams. [tex]\frac{g}{L} * \frac{L}{1}[/tex]

Answer:

90g

Explanation:

Ans. 90 gram

we know that,

n = wt/m.wt

where, n=  moles

wt.= weight

m.wt = molecular weight

putting values we get

5 = wt./18 ( molecular weight of water is 18

wt.= 90

hence  ans.= 90 gram

When water and oxygen react in the presence of an acid, the oxygen can oxidize the acid to produce a compound and release hydrogen ions.

Reaction:

The reaction is as follows and it's diagram mentioned below.

Acid + Oxygen + Water → Compound + Hydrogen ions

if we take the acid hydrochloric acid (HCl), the reaction with oxygen and water can produce the compound chlorine dioxide ([tex]ClO_{2}[/tex]) and hydrogen ions ([tex]H^{+}[/tex]):

2 HCl + [tex]O_{2}[/tex] + [tex]H_{2}O[/tex] → 2 [tex]ClO_{2}[/tex] + 4 [tex]H^{+}[/tex]

This type of reaction is known as an oxidation-reduction reaction or a redox reaction, where one species is oxidized (loses electrons) while the other is reduced (gains electrons).

What is redox reaction?

A redox reaction, also known as an oxidation-reduction reaction, is a type of chemical reaction in which there is a transfer of electrons between two species. One species undergoes oxidation, meaning it loses electrons, while the other species undergoes reduction, meaning it gains electrons.

Redox reactions are fundamental to many processes in nature and technology, including photosynthesis, respiration, corrosion, and energy production in batteries and fuel cells. They are also important in many industrial processes, such as the production of metals, chemicals, and pharmaceuticals.

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First, we need to write the balanced chemical equation for the reaction:

2 NO2(g) + H2O(l) → HNO3(aq) + NO(g)

From the equation, we can see that 2 moles of NO2 react with 1 mole of H2O to produce 1 mole of HNO3 and 1 mole of NO. Therefore, we need to determine which reactant is limiting and calculate the amount of HNO3 that can be produced based on that.

To do this, we can use the mole ratio of NO2 to H2O:

5.0 mol NO2 × (1 mol H2O / 2 mol NO2) = 2.5 mol H2O

Since we have 11.0 mol of H2O, it is not limiting and we will use up all of the NO2.

Therefore, we can calculate the amount of HNO3 that can be produced from 5.0 mol of NO2:

5.0 mol NO2 × (1 mol HNO3 / 2 mol NO2) = 2.5 mol HNO3

Therefore, the largest amount of HNO3 that could be produced is 2.5 mol, rounded to the nearest 0.1 mol.

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The term that would best describe the solution formed if 80g KBr dissolved in 100g water is unsaturated solution.

A saturated solution is a solution with solute that dissolves until it is unable to dissolve anymore, leaving the undissolved substances at the bottom.

On the other hand, an unsaturated solution is that solution that is capable of dissolving more of a solute at the same temperature.

According to this question, 80 grams of KBr were dissolved in 100 grams of water at 35 degrees Celsius. This means that the solution is unsaturated because it can still dissolve more KBr.

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Answer: