Diatomic n2 can react with diatomic h2 to form ammonia (nh3). the balanced chemical equation is: n2 + 3 h2 ? 2 nh3 if 6 moles of h2 totally reacted with more than enough n2, how many moles of ammonia would be expected to form?

Final answer:

2.2 moles of water are needed to react with 2.2 moles of Li2O according to the 1:1 molar ratio in the chemical equation Li2O + H2O → 2 LiOH.

Explanation:

To answer how many moles of water are needed to react with 2.2 moles of Li2O, we need to analyze the given chemical equation: Li2O + H2O → 2 LiOH. According to the equation, 1 mole of Li2O reacts with 1 mole of H2O to produce 2 moles of LiOH. Since the ratio of Li2O to H2O is 1:1, it means that to react with 2.2 moles of Li2O, we will also need 2.2 moles of water.

Answer: b

Explanation:

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In the reaction of magnesium metal with hydrochloric acid to produce hydrogen gas and aqueous magnesium chloride, increasing the pressure would not significantly affect the reaction rate, as solids and liquids are not compressible under moderate pressure changes.

The effect of an increase in pressure on the reaction of magnesium metal with a solution of hydrochloric acid (Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq)) depends on the reaction conditions. If this system is at chemical equilibrium—where the rate of the forward reaction is equal to the rate of the reverse reaction—and involves gases, increasing the pressure will favor the direction in which there are fewer moles of gases. According to Le Chatelier's Principle, because the reactant side has solid magnesium and aqueous HCl, while the product side has gaseous hydrogen and aqueous magnesium chloride, increasing the pressure would shift the equilibrium to the left, potentially reducing the rate of the forward reaction.

However, this reaction is not typically presented as an equilibrium in introductory chemistry, and the scenario doesn't imply it is reversible under normal conditions. Therefore, in practice, the reaction rate of a solid and liquid reactant generating gas is usually unaffected by external pressure changes because solids and liquids are not compressible under moderate pressure changes. Thus, the reaction rate is mostly influenced by the surface area of the magnesium, the molarity of the hydrochloric acid, and the temperature of the solution.

Final answer:

The patient requires 10 ml of potassium phosphate to obtain 30 mmol of phosphate, and 18 ml of potassium chloride to cover the remaining 36 meq of potassium out of the 80 meq required.

Explanation:

To determine how many milliliters of potassium phosphate and potassium chloride will be required, we must use the information provided on the concentrations of phosphate and potassium in each solution. The patient requires 30 mmol of phosphate and 80 meq of potassium.

K-Phos contains 3 mmol of phosphate and 4.4 meq of potassium per ml. To get the required 30 mmol of phosphate, we calculate:
30 mmol ÷ 3 mmol/ml = 10 ml of K-Phos
This will also provide:
10 ml x 4.4 meq/ml = 44 meq of potassium

We still need 80 meq - 44 meq = 36 meq of potassium, which can be supplied by KCl, containing 2 meq of potassium per ml.
36 meq ÷ 2 meq/ml = 18 ml of KCl

Therefore, 10 ml of potassium phosphate and 18 ml of potassium chloride are required to meet the patient's needs.

Each oxygen atom in the carbonate ion (CO3^2−) carries a charge of approximately -2/3.

In the carbonate ion (CO3^2−), there are three oxygen atoms bonded to a central carbon atom. The overall charge on the carbonate ion is -2. Since the charge is distributed among three oxygen atoms, we can consider the charge per oxygen atom.

If we distribute the charge evenly among the three oxygen atoms, each oxygen atom would carry a charge of −2/3. This distribution accounts for the fact that the charge is shared equally among the three oxygen atoms in the carbonate ion. Therefore, each oxygen atom in the carbonate ion is considered to have a partial negative charge of approximately −2/3.

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The Certificate of Documentation must stay onboard a Vessel Registration Number while it is in US waters, in accordance with Title 46 for the Code of Federal Regulations (CFR), Part 67.

Make sure it complies with the following specifications. Your boat's or vessel's bow must have their Vessel Registration Number permanently painted or fastened to either side be shown in larger than three-inch-high figures and simple, vertical block letters. The Certificate of Documentation must stay onboard a vessel while it is in US waters, in accordance with Title 46 for the Code of Federal Regulations (CFR), Part 67.

To know more about vessel’s registration paper, here:

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Hydrochloric acid (HCl) is a strong acid that fully dissociates in an aqueous solution, meaning it releases all of its hydrogen ions, making the solution highly acidic. Unlike weak acids, it does not create a buffered solution when reacted with a strong base. Adding HCl to a solution typically lowers the pH level due to the increase in hydrogen ion concentration.

A strong acid like HCl (hydrochloric acid) fully dissociates in an aqueous solution. This essentially means that virtually every HCl molecule that dissolves in water will undergo a reaction; donating H* (hydrogen ions) and becoming highly acidic. Acids that react completely in this way are known as strong acids.

Conversely, a weak acid does not fully dissociate in solution and some of its hydrogen ions remain bonded within the compound. For instance, acetic acid, the main component in vinegar is a weak acid.

When HCl reacts with a strong base, it can produce a salt and water, but it does not create a buffered solution. It's also important to note that adding a strong acid to an aqueous solution usually lowers (not raises) the pH due to the increase in hydrogen ion concentration.

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Baking soda reacts with vinegar to produce sodium acetate, water, and carbon dioxide gas, causing bubbling and fizzing. The pH of the solution changes from acidic to closer to neutral as the reaction proceeds.

The reaction between baking soda (sodium bicarbonate, NaHCO3) and vinegar (acetic acid, CH3COOH) is a classic example of an acid-base reaction. When these two substances mix, they react to form sodium acetate (NaCH3COO), water (H2O), and carbon dioxide gas (CO2). The formation of carbon dioxide gas is responsible for the bubbling and fizzing observed during the reaction, as the gas escapes from the liquid mixture.

As the reaction proceeds, the pH of the solution initially becomes less acidic due to the neutralizing effect of the baking soda, a base, on the vinegar, which is an acid. Eventually, though, as the reactants are converted to products, the solution will stabilize at a pH determined by the resulting mixture of sodium acetate and water, which is typically closer to neutral.

Answer: A

Explanation:

The pressure of CO in the given mixture is 0.536 atm, calculated using the concept of partial pressure in ideal gas law.

In this question, we are using the concept of partial pressure which is a part of ideal gas law. The formula we use is P(K) = X(K) * P(total), where P(K) is the pressure of component K, X(K) is the mole fraction of the component K, and P(total) is the total pressure. The mole fraction is given by the ratio of the number of moles of a component to the total number of moles.

First, we calculate the total number of moles which is 0.220 + 0.350 + 0.640 = 1.210 moles. The mole fraction of CO is 0.220/1.210 = 0.1818. Now we can find the pressure of CO by multiplying the mole fraction with the total pressure. Therefore, P(CO) = 0.1818 * 2.95 atm = 0.536 atm.

So, the pressure of CO in this mixture is 0.536 atm.

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After three half-lives, 12.5% of the original radioactive parent isotope will remain.

To determine the remaining amount of a radioactive parent isotope after a given number of half-lives, you need to understand the concept of radioactive decay.

A radioactive isotope undergoes decay, reducing its quantity over a set period known as a half-life.

Each half-life represents the time it takes for half of the parent isotope to decay into a daughter isotope.

Explanation and Calculation:

1. Concept of Half-Life:

- A half-life is the time required for half of the radioactive parent isotope to decay.

2. Formula for Remaining Fraction:

- The fraction of the parent isotope remaining after n half-lives can be calculated using the formula:

[tex]\[ \left(\frac{1}{2}\right)^n \][/tex]

- Here, n is the number of half-lives.

3. Calculation for Three Half-Lives:

[tex]\text{After 1 half-life:} \\\\\left(\frac{1}{2}\right)^1 = \frac{1}{2} = 0.5 \text{ (50\% remains)}\\\text{After 2 half-lives:} \\\\\left(\frac{1}{2}\right)^2 = \frac{1}{4} = 0.25 \text{ (25\% remains)}\\\text{After 3 half-lives:} \\\\\left(\frac{1}{2}\right)^3 = \frac{1}{8} = 0.125 \text{ (12.5\% remains)}[/tex]

Conclusion:

Thus, after three half-lives, only 12.5% of the original parent isotope remains. This is calculated using the formula [tex]\(\left(\frac{1}{2}\right)^n\)[/tex], where n is the number of half-lives.

The element with an atomic number lower than aluminum and one less valence electron than the Group 16 elements is magnesium, which has an atomic number of 12 and two valence electrons.

The element you're referring to that possesses an atomic number lower than that of aluminum and one less valence electron than the Group 16 elements is magnesium. Magnesium has an atomic number of 12, which is less than that of aluminum (13), and it falls under group 2 (the alkaline earth metals) in the periodic table. Group 16 elements, also known as chalcogens, typically have six valence electrons. Therefore, with magnesium possessing two valence electrons, it has one less than the Group 16 elements if we consider the full shell configuration.

The atomic number signifies the number of protons in an atom's nucleus, providing a key characteristic of an element's identity. In the case of magnesium, this number is 12.

Concerning valence electrons, they are the electrons present in an atom's outer shell and primarily engage in bonding with other atoms. For magnesium, there are two valence electrons.

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The number of molecules resulting from a chemical reaction depends on the balanced chemical equation, which is not provided for the ethanol combustion. We would need this to determine the exact quantities of CO2, H2O, and O2 produced from the reaction involving ethanol.

To determine the number of molecules present after a chemical reaction has gone to completion, we use the balanced chemical equations. However, the initial question does not provide a complete balanced equation for the reactions involving CO2, H2O, C2H5OH (ethanol), and O2 (oxygen). We are provided information about the combustion of methane and the reaction of propane but not the specific reaction for ethanol, which is necessary to determine the exact number of molecules produced in its combustion.

Given the example of the propane combustion, which yields 3 moles of CO2 and 4 moles of H2O from 0.75 mol of propane, we can extend this stoichiometric ratio to determine the number of molecules. The number of molecules produced from the combustion of a known quantity of another substance, like ethanol, would depend on its stoichiometry. Without the balanced equation for ethanol combustion, we cannot calculate the exact quantities of CO2, H2O, and O2 produced. Therefore, the student should provide the complete balanced chemical equation or clarify if the question is about a different reaction than the ones provided.

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Answer: The molar mass of the gas is 2.02 grams.

Explanation:

We are given:

Density of gas = 0.0902 g/L

At STP conditions:

1 mole of gas occupies 22.4 L of volume

And 1 mole of a substance is known as the molecular mass or molar mass of that substance.

To calculate the mass of gas, we use the equation:

[tex]\text{Density of gas}=\frac{\text{Mass of gas}}{\text{Volume of gas}}[/tex]

Putting values in above equation, we get:

[tex]0.0902g/L=\frac{\text{Mass of gas}}{22.4L}\\\\\text{Mass of gas}=2.02g[/tex]

Hence, the molar mass of the gas is 2.02 grams.

The molar mass of an ideal gas can be calculated by multiplying its density (given as 0.0902 g/L) by the molar volume of gas at STP (which is 22.4 L/mol). This gives us a molar mass of approximately 2.02 g/mol.

For an ideal gas at Standard Temperature and Pressure (STP), which is 0°C (273.15K) and 1 atm, one mole of the gas occupies a volume of 22.4 L. This is a well-known fact. Now, molar mass is the mass of one mole of a substance. You're given the density of the gas (0.0902 g/L).

We can use this known volume of a mole of gas at STP and the given density to find the molar mass. The formula to do that would be: Molar mass = Density x Molar volume.

Substitute the values into this equation:

Molar mass = 0.0902 g/L x 22.4 L/mol

Molar mass = 2.02 g/mol

Therefore, the molar mass of the gas is approximately 2.02 g/mol.

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Answer:

Buoyant force has to be greater than the weight of the object

Explanation:

According to Archimedes principle, the upward buoyant force is equal to the weight of the liquid displaced if an object is fully or partially immersed in a liquid. When an object is immersed in water, two forces act on it :

1. Buoyant force or upthrust, an upward force, which opposes the weight of an object inside water.          

2. Gravitational force, which act in downward direction always.

The floating or sinking of an object depend on the amount of buoyant force. If the buoyant force is greater than the weight of the object, it will float and if the buoyant force is less than the weight of the object, it will sink in liquid.

Hence, the correct option is (c) " buoyant force has to be greater than the weight of the object".

To achieve a 60-gallon mixture with an octane rating of 120, the student should use 20 gallons from the tank with an octane rating of 80.

The student's question involves calculating the amount of gasoline to use from each tank to create a mixture with a desired octane rating. This is a problem that can be solved using the concept of weighted averages. Here, we need to find the number of gallons of gas with an octane rating of 80 to mix with gas with an octane rating of 140 to achieve a final mixture of 60 gallons with an octane rating of 120.

Let the amount of gasoline used from the tank with octane rating 80 be x gallons, and hence the amount used from the tank with octane rating of 140 will be (60 - x) gallons. We can write the following equation based on the weighted average of octane ratings:

80x + 140(60 - x) = 120(60)

Solving the equation:

Multiply out the terms: 80x + 8400 - 140x = 7200.

Combine like terms: -60x + 8400 = 7200.

Subtract 8400 from both sides: -60x = -1200.

Divide both sides by -60: x = 20.

So, the student should use 20 gallons from the tank with an octane rating of 80 to obtain a 60-gallon mixture with an octane rating of 120.

There are [tex]6.12 \times 10^{23}[/tex] atoms of oxygen in 30.5 grams of glucose. There is a little over one mole of oxygen in the given mass of glucose, therefore the number of O atoms should be a little over Avogadro's number.

FURTHER EXPLANATION

To get the number of atoms present in 30.5 g of glucose the following steps must be done:

STEP 1: Convert 30.5 g glucose to moles by dividing the given mass by the formula mass (molar mass).

The molar mass of glucose is 180.156 g/mol obtained as follows:

[tex]formula \ mass \ C_6H_{12}O_6 \ = (6)(12.011) \ + \ (12)(1.008) \ + \ (6)(15.999)\\\boxed {formula \ mass \ C_6H_{12}O_6 \ = 180.156}[/tex]

The mass of 1 mol of glucose is equal t its formula mass in g.

Next, set up the equation below to convert the mass of glucose to moles of glucose.

[tex]moles \ C_6H_{12}O_6 \ = 30.5 \ g (\frac{1 \ mol}{180.156 \ g})\\\\\boxed { moles \ C_6H_{12}O_6 \ = 0.1693 \ mol}[/tex]

STEP 2: Determine how many  moles of oxygen are in the given moles of glucose. The subscripts of the elements in the chemical formula will give the mole ratio: 1 mole of glucose has 6 moles of oxygen atoms.

[tex]moles \ of \ O \ = 0.1693 \ mol \ C_6H_{12}O_6 \ (\frac{6 \ mol \ O}{1 \ mol \ C_6H_{12}O_6})\\\boxed {moles \ of \ O \ = 1.0158 \ mol}[/tex]

STEP 3: Use Avogadro's number to calculate the number of atoms of O

1 mol of O has [tex]6.022 \times 10^{23}[/tex] O atoms

[tex]no. \ of \ O \ atoms \ = 1.0158 \ mol \ O \ \times \frac{6.022 \times 10^{23} \ atoms \ O}{1 \ mol \ O} \\\boxed {no. \ of \ O \ \ atoms \ = 6.117 \times 10^{23} \ atoms}[/tex]

Since the given, 30.5 g has 3 significant figures, the final answer must also have 3 significant figures. Therefore,

[tex]\boxed {atoms \ of \ O \ = 6.12 \times 10^{23} \ atoms}}[/tex]

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Keywords: moles, Avogadro's number