Predict how the addition of hydrogen gas would affect the initial rate of the reaction below, and explain your prediction.

H2 (g) + I2 (g) -> 2 HI

Final answer:

The percent magnesium by mass in the original mixture is 153.2%.

Explanation:

To determine the percent magnesium by mass in the original mixture, we need to calculate the mass of magnesium in the mixture. We know that the total mass of the mixture is 11.20 g, and 0.6854 g of hydrogen gas is produced.

First, we need to calculate the moles of hydrogen gas produced. Since 1 mole of hydrogen gas is produced for every 2 moles of magnesium, we can use the molar mass of magnesium (24.31 g/mol) to find the moles of magnesium:

moles of Mg = (0.6854 g H₂) x (1 mol Mg / 2 mol H₂) x (24.31 g/mol) = 0.704 mol Mg

Next, we can calculate the mass of magnesium:

mass of Mg = (0.704 mol Mg) x (24.31 g/mol) = 17.12624 g Mg

Finally, we can calculate the percent magnesium by mass:

percent magnesium = (mass of Mg / total mass of mixture) x 100% = (17.12624 g / 11.20 g) x 100% = 153.2%

Answer : The correct option is, Aldehyde.

Explanation :

Ketone group : It is the class of organic compound in which the [tex]-CO[/tex] group is attached to a hydrocarbon is known as ketone.

The general representation of ketone are, [tex]R-CO-R'[/tex].

For example : Acetone, [tex]H_3C-CO-CH_3[/tex]

Aldehyde group : It is the class of organic compound in which the [tex]-CHO[/tex] group is attached to a hydrocarbon is known as aldehyde.

The general representation of aldehyde are, [tex]R-CHO[/tex].

For example : Ethanal , [tex]H_3C-CH_2-CHO[/tex]

Ether group : It is the class of organic compound in which the oxygen is directly attached to the two alkyl group of carbon is known as ether.

The general representation of ether are, [tex]R-O-R^'[/tex].

For example : Dimethyl ether, [tex]H_3C-O-CH_3[/tex]

Amine group : It is the class of organic compound in which the [tex]-NH_2[/tex] group is directly attached to the alkyl group of carbon is known as amine.

The general representation of amine are, [tex]R-NH_2[/tex].

For example : Ethylamine, [tex]H_3C-CH_2-NH_2[/tex]

As per question, aldehyde is an example of an organic compound that contains an attached carbonyl group at the end of the carbon chain.

Hence, the correct option is, Aldehyde.

Answer: sink

Explanation:

To calculate the molar mass of a compound, first retrieve the atomic masses for each element from the periodic table, multiply them by the number of atoms in the compound, and add the totals.

The first step when calculating the molar mass of a compound is to identify the atomic masses of each element in the compound. You can find these on the periodic table. Once you have the atomic masses, count the number of atoms for each element in the molecular formula. Then, multiply this number by the atomic mass for each element. Finally, add up all of these values to get the total molar mass.

For example, to calculate the molar mass of glucose (C6H12O6), you would do the following:

Answer: 18 amu

Explanation:

The atomic mass of hydrogen is 1.008 or 1 (rounded to the nearest whole number)

The atomic mass of oxygen is 15.999 or 16 (rounded to the nearest whole number)

There are 2 hydrogen atoms and one oxygen atom in water.

1+1+16= 18 amu

Answer:  d.Have the product tested by an independent organization.

Explanation:

A scientific claim can be define as a statement which is based upon the results that are obtained after the process of experimentation. The results of the claim are acceptable by the scientific society.

d.Have the product tested by an independent organization. is the correct option. This is because of the fact that the independent organization will test the supplement by several clinical trials. The results obtain will be unbiased and reliable.

The Molarity concentration is expressed in units of moles / L. So let us first determine the number of moles of CH3OH, and then divide that amount by the total volume of 0.230 L of solution.

To determine the number of moles of CH3OH, divide the weight in grams of CH3OH by the molecular weight of CH3OH: (MW of CH3OH = 32 g / mol)

number of moles = 16.8 g / (32 g / mol)

number of moles = 0.525 mol CH3OH 

Then we calculate for molarity:

Molarity = 0.525 mol CH3OH / .230 L

Molarity = 2.2826 mol / L

Molarity = 2.28 M

The molarity of a solution prepared by dissolving 16.8 g of CH₃OH in sufficient water to give exactly 230 mL of solution is 2.28 M.

The concentration of CH₃OH in a solution can be determined by calculating its molarity, which is the number of moles of solute per liter of solution. To find the molarity, we first need to convert the mass of CH₃OH to moles. The molar mass of CH₃OH (Methanol) is 32.04 g/mol.

Step 1: Convert the mass of CH₃OH to moles using its molar mass.
16.8 g CH₃OH ×1 mol CH₃OH / 32.04 g CH₃OH = 0.524 moles CH₃OH

Step 2: Convert the volume of the solution from milliliters to liters.
230 mL ×1000 mL/L = 0.230 L

Step 3: Calculate the molarity of the solution.
0.524 moles CH₃OH / 0.230 L = 2.28 M

The equilibrium constant for ammonia[tex]\left({{{\text{K}}_{\text{b}}}}\right)[/tex] is[tex]\boxed{1.82\times{{10}^{-5}}}[/tex].

Further explanation:

Chemical equilibrium is the state in which the concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:

[tex]{\text{A(g)}}+{\text{B(g)}}\rightleftharpoons{\text{C(g)}}+{\text{D(g)}}[/tex]

Equilibrium constant is the constant that relates the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for general reaction is as follows:

[tex]{\text{K}}=\frac{{\left[{\text{D}}\right]\left[{\text{C}}\right]}}{{\left[{\text{A}}\right]\left[{\text{B}}\right]}}[/tex]

Here, K is the equilibrium constant.

The equilibrium constant for the dissociation of acid is known as [tex]{{\text{K}}_{\text{a}}}[/tex]and equilibrium constant for the dissociation of base is known as[tex]{{\text{K}}_{\text{b}}}[/tex].

The expression that relates pH and pOH is given as follows:

[tex]{\text{pH}}+{\text{pOH}}=14[/tex]                                   …… (1)

Rearrange equation (1) to calculate pOH.

[tex]{\text{pOH}}=14-{\text{pH}}[/tex]                                 …… (2)

Substitute 11.50 for the value of pH in equation (2).

[tex]\begin{aligned}{\text{pOH}}&=14-{\text{11}}{\text{.50}}\\&={\text{2}}{\text{.5}}\\\end{aligned}[/tex]

pOH is the measure of hydroxide ion concentration. The formula to calculate pOH is as follows:

[tex]{\text{pOH}}=-\log\left[{{\text{O}}{{\text{H}}^-}}\right][/tex]                              …… (3)

Here,

[tex]\left[{{\text{O}}{{\text{H}}^-}}\right][/tex] is the concentration of hydroxide ion.

Rearrange equation (3) to calculate [tex]\left[{{\text{O}}{{\text{H}}^-}}\right][/tex].

[tex]\left[{{\text{O}}{{\text{H}}^-}}\right]={10^{-{\text{pOH}}}}[/tex]                              …… (4)

Substitute 2.5 for pOH in equation (4).

[tex]\begin{aligned}\left[{{\text{O}}{{\text{H}}^-}}\right]&={10^{-2.5}}\\&=0.0031622\\\end{aligned}[/tex]

The given equilibrium reaction is,

[tex]{\text{N}}{{\text{H}}_{\text{3}}}+{{\text{H}}_2}{\text{O}}\rightleftharpoons{\text{NH}}_4^++{\text{O}}{{\text{H}}^-}[/tex]

The expression of [tex]{{\text{K}}_{\text{b}}}[/tex]for the above reaction is as follows:

[tex]{{\text{K}}_{\text{b}}}=\frac{{\left[{{\text{NH}}_4^+}\right]\left[{{\text{O}}{{\text{H}}^-}}\right]}}{{\left[{{\text{N}}{{\text{H}}_3}}\right]}}[/tex]                                  …... (5)

The equilibrium concentration of both [tex]{\text{NH}}_4^+[/tex] and [tex]{\text{O}}{{\text{H}}^-}[/tex] is the same.

0.0031622 M of [tex]{\text{O}}{{\text{H}}^-}[/tex]is present at equilibrium so 0.0031622 M out of 0.55 M of  has reacted.

The initial concentration of the aqueous solution is 0.55 M. So the concentration of [tex]{\text{N}}{{\text{H}}_3}[/tex] left at equilibrium is calculated as follows:

[tex]\begin{aligned}\left[{{\text{N}}{{\text{H}}_3}}\right]&={\text{Initial concentration of N}}{{\text{H}}_{\text{3}}}-{\text{Reacted concentration of N}}{{\text{H}}_{\text{3}}}\\&={\text{0}}{\text{.55 M}}-{\text{0}}{\text{.0031622 M}}\\&={\text{0}}{\text{.5468378 M}}\\\end{aligned}[/tex]

The value of[tex]\left[{{\text{O}}{{\text{H}}^-}}\right][/tex]is 0.0031622 M.

The value of [tex]\left[{{\text{N}}{{\text{H}}_3}}\right][/tex] is 0.5468378 M.

The value of [tex]\left[{{\text{NH}}_4^+}\right][/tex] is 0.0031622 M.

Substitute these values in equation (5).

[tex]\begin{aligned}{{\text{K}}_{\text{b}}}&=\frac{{\left({0.0031622\;{\text{M}}}\right)\left({0.0031622\;{\text{M}}}\right)}}{{\left({0.546837{\text{8 M}}}\right)}}\\&=1.82861\times{10^{-5}}\\&\approx1.82\times{10^{-5}}\\\end{aligned}[/tex]

Therefore, equilibrium constant for ammonia is[tex]{\mathbf{1}}{\mathbf{.82\times1}}{{\mathbf{0}}^{{\mathbf{-5}}}}[/tex].

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c: https://brainly.com/question/3467841

2. Complete equation for the dissociation of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex](aq): https://brainly.com/question/5425813

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Equilibrium

Keywords: NH3, OH-, NH4+, H2O, equilibrium, kb, pH, pOH, 14, 11.5, 2.5, aqueous solution, 0.0031622 M, 0.5468378 M.

The journey of a water molecule through the water cycle begins with evaporation or sublimation, followed by condensation forming clouds. Afterward, precipitation brings the water back to the earth where it becomes part of surface runoff, snowmelt, or infiltrates the soil. Finally, streamflow carries it back to the oceans, completing the cycle.

A water molecule undergoes a fascinating journey as it completes the water cycle. Initiated by the sun's energy, this journey starts with evaporation or sublimation, which are processes that turn surface water or frozen water into water vapor, respectively. The water vapor is then transported into the atmosphere.

As the vapor cools down, the process of condensation takes place, forming clouds consisting of condensed water droplets. Over time, these droplets grow in size and eventually fall to the earth in the form of precipitation (like rain or snow).

Once on earth, the water can take a few different paths. Some fall on tree leaves and evaporate again, some fall directly into bodies of water such as lakes or oceans, while some fall on the ground and become surface runoff or snowmelt. This runoff or snowmelt can infiltrate the soil (subsurface water flow) or flow into streams and rivers (streamflow). These streams and rivers then carry it back into the sea or ocean, completing the cycle.

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To find the volume of the Hope diamond, first convert the mass from carats to grams, then divide by the given density. The calculation results in a volume of approximately 2.5 cm³, so the correct answer is option C.

The volume of the Hope diamond can be calculated using its mass and density. First, we convert the mass from carats to grams. Since 1 carat is equivalent to 0.200 grams, the Hope diamond's mass in grams is 44.0 carats × 0.200 g/carat = 8.8 grams. The density formula is density = mass/volume, which can be rearranged to find the volume as volume = mass/density. We use the given density of the diamond, 3.51 g/cm³, to get the volume: volume = 8.8 g / 3.51 g/cm³ = 2.51 cm³. Therefore, the correct answer is 2.5 cm³, so option C is correct.

Answer: Option (D) is the correct answer.

Explanation:

Law of conservation of mass states that energy can neither be created nor it can be destroyed. It can only be transformed from one form to another.

Therefore, in a chemical reaction an equation can only be balanced if the number of reactants equal the number of products because energy required for reactants is utilized in the formation of the products.

For example, [tex]2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}[/tex] is a balanced equation because the number of reactant atoms equal the number of product atoms.

Thus, we can conclude that out of the given options, [tex]3Ca + 2Cr(NO3)_3 \rightarrow 3Ca(NO3)_{3} + 2Cr[/tex] equations violates the law of conservation of mass because number of reactant atoms does not equal the number of product atoms.

Answer: oil

Explanation: