List what you are thankful for! (It's Thanksgiving! Answer if you bake! Or if you like cakes pies cookies and treats!!! )

Answer:

283.33 g of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2Mg + O2 —> 2MgO

Next, we shall determine the masses of Mg and O2 that reacted together from the balanced equation. This can be obtained as follow:

Molar mass of Mg = 24 g/mol

Mass of Mg from the balanced equation = 2 × 24 = 48 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 1 × 32 = 32 g

From the balanced equation above,

48 g of Mg reacted with 32 g of O2.

Finally, we shall determine the mass of O2 needed to react with 425 g of Mg as follow:

From the balanced equation above,

48 g of Mg reacted with 32 g of O2.

Therefore, 425 g of Mg will react with = (425 × 32)/48 = 283.33 g of O2.

Therefore, 283.33 g of O2 is needed for the reaction.

Answer:

Explanation:

2Ma+O2 ----  2MaO

mass of Mg +Mass pf O2 = Mass of MgO

(conservation of mass )

425 + X =600

so the mass of O2 = 600-425 = 175 g

The best way to measure the pH of a natural solution while out in a forest is to use a portable pH meter or pH test strips specifically designed for field use. These instruments provide accurate and reliable pH measurements and are convenient for outdoor applications.

1. Prepare the necessary equipment: Before heading out to the forest, gather the required tools. You will need a portable pH meter or pH test strips, as well as the necessary reagents or calibration solutions if using a pH meter.

2. Collect the sample: Locate the natural solution you want to measure the pH of, such as a stream, pond, or soil. Use a clean container to collect a representative sample of the solution.

3. Calibrate the pH meter (if applicable): If you are using a portable pH meter, it is essential to calibrate it before taking measurements. Follow the manufacturer's instructions to calibrate the meter using the provided calibration solutions.

4. Conduct the measurement: For pH meters, immerse the electrode into the collected sample. Allow some time for the reading to stabilize, and then record the pH value indicated on the meter's display.

5. Using pH test strips: If you are using pH test strips, dip the strip into the collected sample for the recommended amount of time. Remove the strip and compare the color change with the provided color chart. Determine the corresponding pH value from the chart.

6. Repeat for accuracy: To ensure reliability, repeat the measurement process at least once and compare the results. This step helps confirm the accuracy of your measurements.

7. Record and analyze the data: Note down the pH values obtained and any relevant observations. Analyze the data as needed for your research or monitoring purposes.

By following these steps and using the appropriate equipment, you can effectively measure the pH of a natural solution while in a forest setting.

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Answer: 12,000in = 0.18939394m

To get 0.18939394, you'll divide 12,000 by 63360, which you'll get 0.18939394. Round to the nearest thousandth and the answer will be 0.189.

Answer:

ok, here is your answer

Explanation:

AI-generated answer

To find the mass of oxygen that reacted, we need to use the Law of Conservation of Mass, which states that in a chemical reaction, the mass of the reactants equals the mass of the products.

First, we need to find the number of moles of hydrogen that reacted:

Molar mass of hydrogen (H₂) = 2.016 g/mol

Number of moles of H₂ = mass/molar mass = 46.2 g/2.016 g/mol = 22.92 mol

Next, we need to use the balanced chemical equation to find the number of moles of water produced:

hydrogen + oxygen → water

2H₂ + O₂ → 2H₂O

From the equation, we can see that for every 2 moles of H₂, 1 mole of O₂ is required to produce 2 moles of H₂O. Therefore, the number of moles of O₂ required to produce 22.92 moles of H₂O is:

Number of moles of O₂ = 1/2 x 22.92 mol = 11.46 mol

Finally, we can find the mass of oxygen that reacted by using its molar mass:

Molar mass of oxygen (O₂) = 32.00 g/mol

Mass of oxygen = number of moles x molar mass = 11.46 mol x 32.00 g/mol = 366.72 g

Therefore, the mass of oxygen that reacted is 366.72 g.

the pH of the aqueous solution is 8.51 with a hydrogen ion concentration of [H+]=3.1×\(10^-9\)M

The pH of the aqueous solution can be calculated using the formula:

pH = -log[H+]

where [H+] is the hydrogen ion concentration of the solution.

Substituting the given value, we get:

pH = -log(3.1×\(10^-9\))

pH = 8.51

An aqueous solution is one in which water serves as the solvent. It is utilised in a variety of applications, including analytical chemistry, biochemistry, and industrial chemistry. It is the most prevalent kind of solution used in chemical reactions. Water serves as both the solvent and the solute in an aqueous solution, where the solute is often a solid, liquid, or gas. Due to its high polarity and capacity to make hydrogen bonds with other molecules, water is an excellent solvent that can dissolve a variety of materials, including polar molecules and ionic compounds. Acid-base reactions, redox reactions, and precipitation reactions are just a few of the numerous chemical processes that take place in aqueous solutions. A variety of variables can have an impact on an aqueous solution's characteristics.

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Answer:

atoms contain tiny negatively charged subatomic particles or electrons

Answer:

As light is emitted from the filament, the energy in the metal is replaced as lightbulbs absorb background radiation. This prevents the filament from burning out quickly but the radiation cannot be used as power so electricity is still required for the lightbulb to work.

There are 0.84 moles of aluminum ions (Al3+) present in 0.42 mol of Al2(SO4)3.

To determine the number of moles of aluminum ions (Al3+) present in 0.42 mol of Al2(SO4)3, we need to consider the stoichiometry of the compound.

The formula of aluminum sulfate (Al2(SO4)3) indicates that for every 1 mole of the compound, there are 2 moles of aluminum ions (Al3+). This means that the mole ratio of Al3+ to Al2(SO4)3 is 2:1.

Given that we have 0.42 mol of Al2(SO4)3, we can calculate the moles of Al3+ as follows:

Moles of Al3+ = 0.42 mol Al2(SO4)3 x (2 mol Al3+ / 1 mol Al2(SO4)3)

Moles of Al3+ = 0.42 mol Al2(SO4)3 x 2

Moles of Al3+ = 0.84 mol Al3+

Therefore, there are 0.84 moles of aluminum ions (Al3+) present in 0.42 mol of Al2(SO4)3.

It's important to note that the stoichiometry of the compound determines the mole ratio between the different species involved in the chemical formula. In this case, the 2:1 ratio of Al3+ to Al2(SO4)3 allows us to determine the number of moles of Al3+ based on the given amount of Al2(SO4)3.

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Answer:

D

Explanation:

I hope this helps and have a great day

I would divided it to get the anwser and then moloeces

Explanation:

Divide

Answer:

crust

Explanation:

it is on the outside so it is exposed to physical weathering

The part of the Earth that experiences physical weathering is the crust. Therefore, option (A) is correct.

Physical weathering is when rocks and minerals break down through mechanical means, but their chemical make-up doesn't change. The topmost layer of the Earth, called the crust, is changed by things like temperature changes, friction, frost action, and the actions of wind, water, and ice.

These things can crack, break, and erode rocks, causing them to break up physically into smaller pieces without changing their chemical makeup. The Earth's mantle, inner core, and outer core, on the other hand, are not directly visible to the surface and do not go through the same physical processes that cause the crust to break down.

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Based on the mass of sodium carbonate reacting,

The equation of the reaction of sodium carbonate reacting with calcium hydroxide is given below:

Na₂CO₃ + Ca(OH)₂ ----> 2 NaOH + CaCO₃

Moles of sodium carbonate reacting = mass /molar mass

The molar mass of sodium carbonate = 106 g/mol

Moles of sodium carbonate reacting = 56/106

Moles of sodium carbonate reacting =0.528 moles

Moles of CaCO₃ produced = 0.528 moles

Mass of CaCO₃ produced = 0.528 * 100 g

Mass of CaCO₃ produced = 52.8 g

Moles of sodium hydroxide produced = 2 * 0.528

Mass of sodium hydroxide produced = 2 * 0.528 * 40 g

Mass of sodium hydroxide produced =42.24 g

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To calculate the pressure inside the canister, we can use the ideal gas law equation:

PV = nRT

Therefore, the pressure inside the canister at a campfire temperature of 1500 °C is approximately 136.39 atmospheres.

To calculate the pressure inside the canister, we can use the ideal gas law equation:

PV = nRT

Where:

P = pressure (in atmospheres)

V = volume (in liters)

n = number of moles of gas

R = ideal gas constant (0.0821 L·atm/(mol·K))

T = temperature (in Kelvin)

First, we need to convert the volume from milliliters to liters:

1500 milliliters = 1500/1000 = 1.5 liters

Next, we need to convert the temperature from Celsius to Kelvin:

1500 °C + 273.15 = 1773.15 K

Now, we can substitute the values into the ideal gas law equation:

P * 1.5 = 3 * 0.0821 * 1773.15

Simplifying the equation:

P = (3 * 0.0821 * 1773.15) / 1.5

Calculating the value:

P ≈ 136.39 atm

Therefore, the pressure inside the canister at a campfire temperature of 1500 °C is approximately 136.39 atmospheres.

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Organism – Squid

Number of chromosomes after meiosis: 26

Number of chromosomes in the body cells: 52

Number of chromosomes in gametes: 26

Number of chromosomes after mitosis: 52

Organism – Sunflower

Number of chromosomes after meiosis: 19

Number of chromosomes in the body cells: 38

Number of chromosomes in gametes: 19

Number of chromosomes after mitosis: 38

Organism – Pea plant

Number of chromosomes after meiosis: 7

Number of chromosomes in the body cells: 14

Number of chromosomes in gametes: 7

Number of chromosomes after mitosis: 14

Organism – Cow

Number of chromosomes after meiosis: 30

Number of chromosomes in the body cells: 60

Number of chromosomes in gametes: 30

Number of chromosomes after mitosis: 60

Chromosome are made up of genetic material called DNA and it looks like a thread. It is present in the inside of nucleus of a cell.

During meiosis number of chromosomes would be reduced to half whereas in mitosis number of chromosomes remains the same.

The number of chromosomes in body cells is the chromosomal number of an organism and in gametes it would be reduced to half.

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Based on the proportionality established by the given data, we predict that the force of gravity on an object of mass 1000 kg would be 50 Newtons.

To predict the force of gravity on an object of mass 1000 kg, we can use the concept of proportionality. Based on the given information, we have two data points: the mass of an object of 500 kg and the corresponding gravitational force of 25 Newtons.

Let's denote the force of gravity on the object of mass 1000 kg as F₁ and solve for it using a proportion:

mass₁ / force₁ = mass₂ / force₂

Plugging in the values we have:

1000 kg / F₁ = 500 kg / 25 N

Cross-multiplying and solving for F₁:

(1000 kg) * (25 N) = (500 kg) * F₁

25,000 kg*N = 500 kg * F₁

Dividing both sides by 500 kg:

F₁ = (25,000 kg*N) / 500 kg

F₁ = 50 N

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The equilibrium constant at this temperature is Kc= 4.17 x 10⁻⁶.

Since the equilibrium constant depends on the equilibrium concentration of both the reactants and the products of the chemical reaction.

Balanced reaction equation

2CH₄(g)⇌C₂H₂(g)+3H₂(g)

The initial concentration of the CH₄ = 0.093 M

The equilibrium concentration of the H = 0.017 M

Equilibrium constant = ?

Let's make the ice table

2CH₄(g) ⇌ C₂H₂(g) + 3H₂(g)

0.093 M 0 0

-2x +x +3x

0.093-2x x 0.017 M

3x = 0.017 M

Therefore, x =0.017 M /3 = 0.00567 M

Therefore, the equilibrium concentration of CH₄ =

0.093 M – 2x = 0.093 M – (2 x 0.00567 M) = 0.0817 M

Equilibrium concentration of the C₂H₄ = x = 0.00567 M

Let's write the equilibrium constant expression

Kc= [C₂H₄[H2]³/[CH₄]²

Let's put the values in the formula

Kc= [0.00567][0.017]³/[0.0817]²

Kc= 4.17 x 10⁻⁶

Therefore, the equilibrium constant is 4.17 x 10⁻⁶.

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Answer:

To calculate the enthalpy of vaporization for the phase change from gaseous methanol to liquid methanol:

1. Determine the standard enthalpy of formation of both states: CH3OH (g) is -201.8 kJ/mol and CH3OH (l) is -238.7 kJ/mol.

2. Subtract the standard enthalpy of formation of the liquid state from the standard enthalpy of formation of the gaseous state: -201.8 kJ/mol -(-238.7 kJ/mol) = 36.9 kJ/mol.

3. The enthalpy of vaporization for the phase change from gaseous methanol to liquid methanol is 36.9 kJ/mol.

Answer:B

Explanation:B can be reversed.

Answer:

Explanation:

Physical properties are characteristics of matter not associated with a change in chemical composition. Identity of the molecules do not change. A physical change is a change in state without changing the composition (e.g., melting wax, freezing water, grinding solids into powders).  Examples: density, color, melting and boiling points, evaporation, odor, taste, hardness, conductivity.

B. Honey dissolving in tea would be a physical change. The honey is not changing chemically but physically as it dissolves in the hot tea.

The number of moles of the gas can be determined using ideal gas equation. The number of moles of Br gas in 2.8 L at 1.38 atm and 327 K is 0.144 moles.

Ideal gas law states the relation between temperature, pressure and volume with the number of moles of a gas as written below:

PV = nRT

where, R is the universal gas constant equal to 0.082 L atm/K mol

Given that, T = 327 K

P = 1.38 atm

V = 2.8 L.

Then, n = PV/RT

Number of moles of Br gas, n = (1.38 atm ×2.8 L)/(327 K × 0.082 L atm/K mol ) = 0.144 moles.

Therefore, the number of moles of Br gas in in 2.8 L at 1.38 atm and 327 K is 0.144 moles.

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Answer:

The d-block and transition metal are not interchangeable terms because the d-block elements are a subset of the transition metal elements. The transition metals are defined as the elements that have partially filled d orbitals, which includes the d-block elements as well as other elements that have partially filled d orbitals in other blocks, such as lanthanides and actinides. Therefore, while all d-block elements are transition metals, not all transition metals are d-block elements.

Answer:

Explanations:

Beta decay is a type of radioactive decay in which a beta particle is emitted from the nucleus. A beta particle is written as:

The required balanced nuclear equation for the beta decay of Pb-212 will be expressed as:

From the reaction, you can see that the beta decay of Pb-212 produces the Bismuth element with atomic number 83 and mass number 212.

Answer:

Synthesis reaction

Explanation:

In synthesis reactions, two or more atoms/molecules combine to form a larger molecule.

These reactions follow the general structure: A + B <-----> AB.

Four properties of gases will be investigated: pressure, volume, temperature, and number of molecules. By assembling the equipment, conducting the appropriate tests, and analyzing your data and observations, you will be able to describe the gas laws, both qualitatively and mathematically.

The mass of carbon stored in the plants in grams is 7.98×10¹⁴ Kg

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

We can convert 8.80×10¹¹ tons to kilograms as illustrated below:

1 ton = 907.19 Kg

Therefore,

8.80×10¹¹ tons = (8.80×10¹¹ tons × 907.19 Kg) / 1 ton

8.80×10¹¹ tons = 7.98×10¹⁴ Kg

Thus, 8.80×10¹¹ tons is equivalent to 7.98×10¹⁴ Kg

Therefore, we can say that the mass of the carbon in the plants is 7.98×10¹⁴ Kg

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Answer:

arms

Explanation:

because arms can be metric system of our body

Answer:

nose and the tip of an outstretched middle finger.

Mass that is number of protons and  neutrons is lost in an atom  as a result of radioactive decay.

An atom is defined as the smallest unit of matter which forms an element. Every form of matter whether solid,liquid , gas consists of atoms . Each atom has a nucleus which is composed of protons and neutrons and shells in which the electrons revolve.

The protons are positively charged and neutrons are neutral and hence the nucleus is positively charged. The electrons which revolve around the nucleus are negatively charged and hence the atom as a whole is neutral and stable due to presence of oppositely charged particles.

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Explanation:

2,4-DNP is negative: aldehyde and ketone absent

Chromic acid is negative: alcohol absent.

KMnO4 test negative: alkene absent

The only positive result is for Fe(OH)2.

Nitro group-containing compounds react with ferrous hydroxide to give amine and ferric hydroxide red ppt. Hence given compound contains the nitro group (-NO2).

The Fe^3+ is reduced in the reaction while the Al is oxidized in the reaction. The species that loses electrons (is oxidized) is called the reducing agent

A redox reaction (short for reduction-oxidation reaction) is a type of chemical reaction that involves the transfer of electrons between two or more species. In redox reactions, one reactant loses electrons (is oxidized) while another gains electrons (is reduced).

The species that loses electrons (is oxidized) is called the reducing agent, since it causes another species to be reduced. The species that gains electrons (is reduced) is called the oxidizing agent, since it causes another species to be oxidized.

Reduction half equation;

2Fe^3+(aq)  + 6e -----> 2Fe(s)

Oxidation half equation;

2Al(s) →2Al3+(aq) + 6e

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385 x 42.13 x 0.079 = 1281.38395

The concentration of H₂SO₃ solution is equal to 0.123 M.

A neutralization reaction is described as a reaction in which an acid and base react to form salt and water. When a strong acid will react with a base then the salt which is formed can be neither acidic nor basic.

When H₂SO₃ (a strong acid) reacts with KOH, the resulting salt is K₂SO₃  and water.

H₂SO₃  +  2KOH   →   K₂SO₃  +   2H₂O

Given, the concentration of KOH = 0.235 M

The volume of the KOH = 65.25 ml = 0.06525 L

The number of moles of KOH, n = M × V = 0.235 × 0.06525 = 0.0153 M

The volume of the  H₂SO₃ = 62.4 ml = 0.0624 L

The number of moles of  H₂SO₃, n = 0.0153/2 = 0.00767 mol

The concentration of H₂SO₃ =0.00767/0.0624 = 0.123 M

Therefore, the molarity of  H₂SO₃ is 0.123 M.

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