How many grams of carbon atoms are needed to make 1.50 moles of sucrose c12h22o11?

Molarity is the molar concentration of the chemical species of the solute dissolved in the solvent. The molarity of the solution when diluted is 0.0257 M.

The molar concentration or the molarity of the solution is inversely proportional to the dilution. An increased dilution will decrease the molarity of the solution.

The molarity is given by:

\(\rm Molarity = \dfrac{\text{number of moles }}{ \text{Volume of the solution in L}}\)

Also, it can be shown as:

\(\rm M_{1}V_{1}= M_{2}V_{2}\)

Given,

\(\rm M_{1}\)  = 0.513 M

\(\rm V_{1}\) = 25 mL

\(\rm M_{2}\) = ?

\(\rm V_{2}\) = 500 mL

Substitute value in the equation as:

\(\begin{aligned} \rm M_{1}V_{1}&= \rm M_{2}V_{2}\\\\0.513 \times 25 &= \rm M_{2} \times 500\\\\\rm M_{2} &= \dfrac{0.513 \times 25 }{500}\\\\&= 0.02565\;\rm M\end{aligned}\)

Therefore, 0.0257 M is the molarity after dilution.

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Answer:

● They have more than two carbons that are bonded to only one other

Explanation:

They have carbons bonded in a single row from one end of the molecule to the other end of the molecule. ... The straight-chain alkane has two more hydrogen atoms than the cycloalkane

Answer:

D. They have carbons bonded in a single row from one end of the molecule to the other end of the molecule.

Explanation:

I got it wrong when I chose C - They have more than two carbons that are bonded to only one other carbon. So yeah p.s UNUS ANNUS

Answer:

Extensive properties, such as mass and volume depend on the amount of substance present and are not useful in the identification of a substance. In this experiment, we will use three properties to identify a liquid substance: solubility, density and boiling point.

In order to deprotonate an acid, we must remove protons in order to achieve a more stable conjugate base. For this example, we can use the relationship between carboxylic acid and hydroxide.

Deprotonation is the removal of a proton from a specific type of acid in reaction to its coming into contact with a strong base. The compound formed from this reaction is known as the conjugate base of that acid. The opposite process is also possible and is when a proton is added to a special kind of base. This is a process referred to as protonation, which forms the conjugate acid of that base.

For the example we have chosen to give, the conjugate base is the carboxylate salt. This would be the compound formed by the deprotonated carboxylic acid. The base in question was strong enough to deprotonate the acid due to the greater stability offered as a conjugated base.

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Explanation:

I think

maganese is +7 and Nitrogen is +5

sorry if I am incorrect I haven't done chemistry in a year

btw for chemistry if you need help with equations I used this equation balancer called chemical aid it helps

In Mn3N4, the compound consists of three manganese (Mn) atoms and four nitrogen (N) atoms.

To determine the oxidation numbers, we can assign x as the oxidation number for manganese and y as the oxidation number for nitrogen.

Since the compound is neutral overall, the sum of the oxidation numbers must equal zero.

For the three manganese atoms:

3x = 0

x = 0

For the four nitrogen atoms:

4y = 0

y = 0

Therefore, the oxidation numbers for manganese (Mn) and nitrogen (N) in Mn3N4 are both 0.

Answer:

Catalase

Explanation:

A damaged enzyme may no longer work to catalyze a chemical reaction. Catalase is an enzyme in the liver that breaks down harmful hydrogen peroxide into oxygen and water. When this reaction occurs, oxygen gas bubbles escape and create foam.

The concentration of A → products, successive half-lives are observed to be 10.0, 20.0, and 40.0 min for an experiment in which [ A ] 0 = 0.10 M at the following times,

a. 80.0 min = 0.0107 M.

b. 30.0 min = 0.0471 M

To solve this problem, we can use the following equation for a first-order reaction:

ln([A]t/[A]0) = -kt

where [A]t is the concentration of A at time t, [A]0 is the initial concentration of A, k is the rate constant, and t is time.

From the given half-lives, we can find the rate constant k as follows:

k = (0.693/t1/2)

where t1/2 is the half-life.

For the given experiment, we have:

k1 = (0.693/10.0) = 0.0693 \(min^{-1}\)

k2 = (0.693/20.0) = 0.03465 \(min^{-1}\)

k3 = (0.693/40.0) = 0.017325 \(min^{-1}\)

a. To find the concentration of A at 80.0 min:

t = 80.0 min

[A]t = [A]0 × \(e^{(-kt)}\) = 0.10 × \(e^{(-(0.069380.0 + 0.0346580.0 + 0.017325 * 80.0))}\) = 0.0107 M

Therefore, the concentration of A at 80.0 min is 0.0107 M.

b. To find the concentration of A at 30.0 min:

t = 30.0 min

[A]t = [A]0 × \(e^{(-kt)}\) = 0.10 × \(e^{(-(0.069330.0 + 0.0346530.0 + 0.017325 * 30.0)}\)) = 0.0471 M

Therefore, the concentration of A at 30.0 min is 0.0471 M.

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The question is -

For the reaction A → products, successive half-lives are observed to be 10.0, 20.0, and 40.0 min for an experiment in which [ A ] 0 = 0.10 M . Calculate the concentration of A at the following times.

a. 80.0 min

b. 30.0 min

Answer:

Na2CO3 → Na2O + CO2  

Explanation:

CO2 is your answer

Answer:

CO2

Explanation:

I Just Took The Quiz

Have A Good Day :)

Answer:

Name                        Oxygen

Atomic Mass                 15.999 atomic mass units

Number of Protons 8

Number of Neutrons 8

Number of Electrons 8

Explanation:

Answer:

the water displaced will not help the student find the correct volume because you can not accurately calculate it. you have to get the water when it is settled then you place the wood in.

Explanation:

Answer:

Because to make us easy to understand memories and read.Scientist place all the elements in label according valency and valence electons.

The law of diminishing marginal rate of substitution indicates that the rate at which one input can be substituted for another decreases as the quantity of one input increases, leading to isoquants being bent toward the origin.

In other words, as the quantity of one good increases, the individual is willing to sacrifice fewer units of the other good to obtain an additional unit of the first good. This reflects a diminishing rate of substitution between the two goods.

The reason why the law of diminishing marginal rate of substitution suggests that isoquants must be bent toward the origin is rooted in the concept of diminishing marginal utility. As more units of a particular input (e.g., labor or capital) are added while holding other inputs constant, the additional output gained from each additional unit of the input will decrease. This diminishing marginal productivity leads to a decreasing MRS.

When isoquants (which represent different combinations of inputs that produce the same level of output) are bent toward the origin, it reflects the fact that as more of one input is used, the amount of the other input that needs to be substituted decreases. This bending signifies the diminishing MRS and captures the idea that a larger quantity of one input can be substituted for a smaller quantity of the other input to maintain the same level of output.

Overall, the law of diminishing marginal rate of substitution indicates that the rate at which one input can be substituted for another decreases as the quantity of one input increases, leading to isoquants being bent toward the origin.

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Following represents the number of Molecules of the type indicated is present as a reactant or product of the chemical reaction shown is the coefficient of the species

The number of atoms, molecules, or formula units of a reactant or product in a balanced chemical equation is the coefficient of that species then the mole ratio of two substances in a chemical reaction is the ratio of their coefficients in the balanced chemical equation and the this numbers are called as coefficient and they tell how many of each molecule are participate in the reaction

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Answer:

0

Explanation:

glycolysis requires no oxygen

Based on the information given the percentage of the new punch is juice is 62.5%.

First step is to calculate how many cups of juice will a 20-cup bowl of punch have:

Number of cups=0.75×20

Number of cups= 15 cups  

Second step is to calculate the percentage of the new punch

Percentage=Number of cups/Total new punch number of cups

Percentage=15/24×100

Percentage=62.5%

Inconclusion  the percentage of the new punch is juice is 62.5%.

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Answer:

1.66x10^-24 M

Explanation:

By definition, 1 mole = 6.02x10^23 particles, or molecules, in this case.  Use this relationship as a conversion factor:

((1 mole)/(6.02x10^23 molecules))

Use this conversion factor to calculate moles of the solute when only 1 molecule is present:

(1 molecule)/((1 mole)/(6.02x10^23 molecules)) = 1.66x10^-24 moles

The molecules cancel, leaving only moles.

Concentration, by definition, is the moles per liter of a solute.  Take the

1.66x10^-24 moles and divide by the volume, which is 1.0L here:

(1.66x10^-24 moles solute)/(1.0 Liter) = 1.66x10^-24 molar, or 1.66x10^-24 M

Answer:

Groups I & II

Explanation:

The most reactive metal groups on the periodic table are metals of group 1 and II.

Group I elements are the alkali metals

Group II elements are the alkali earth metals

Therefore, elements in groups I and II are the most reactive metallic groups.

Group I elements prefers to lose one electron to attain an octet configuration.

Group II elements prefers to lose two electrons.

Based on the volume of NaOH required to reach the equivalence point, the concentration of the acetic acid in the vinegar at is 0.492 M.

The equivalence point in a titration experiment is the point at which equal amounts of reactants have reacted.

At equivalence point, moles of vinegar = moles of NaOH

Moles of NaOH = concentration × volume

Moles of NaOH = 1.5 × 8.2 = 12.3 mmoles

Moles of vinegar = 12.3 mmoles

Concentration of vinegar = moles/volume

Concentration of vinegar = 12.3/25.0 = 0.492 M

Therefore, the concentration of the acetic acid in the vinegar is 0.492 M.

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At 0.752 atm of H2S and 1073 K, [H2] = 4.94 x 10^-12 mol/L and [S2] = 4.24 x 10^-5 mol/L. At 0.20 atm of H2 and 0.40 atm of S2, [H2] = 1.67 x 10^-12 mol/L and [S2] = 4.68 x 10^-5 mol/L.

The reaction for the decomposition of hydrogen disulfide (HS) is:

2 H2S (g) <=> 2 H2 (g) + S2 (g)

The equilibrium constant (Kc) at 1073 K is 2.0 x 10^-23.

1) At the initial conditions of 0.752 atm of H2S and the temperature of 1073 K, the equilibrium concentrations can be calculated using the equation:

Kc = [H2]^2 / [S2].

Rearranging the equation to solve for [H2], we get:

[H2]^2 = Kc * [S2]

[H2] = sqrt(Kc * [S2])

Since the initial pressure of H2S is 0.752 atm, we can convert it to concentration using the ideal gas law:

P = nRT/V, where n = number of moles, R = gas constant (8.31 J/mol*K), T = temperature (1073 K), and V = volume.

Since n = PV/RT, [H2S] = P/RT.

[H2S] = 0.752 atm / (8.31 J/mol*K * 1073 K) = 8.48 x 10^-5 mol/L.

[S2] = [H2S] / 2 = 4.24 x 10^-5 mol/L.

[H2] = sqrt(2.0 x 10^-23 * 4.24 x 10^-5) = 4.94 x 10^-12 mol/L.

2) At the new initial conditions of 0.20 atm of H2 and 0.40 atm of S2, the new equilibrium concentrations can be calculated using the Kc expression:

Kc = [H2]^2 / [S2].

[H2] = sqrt(Kc * [S2]) = sqrt(2.0 x 10^-23 * (0.40 atm / (8.31 J/molK * 1073 K))) = 1.67 x 10^-12 mol/L.

[S2] = 0.40 atm / (8.31 J/molK * 1073 K) = 4.68 x 10^-5 mol/L.

These are the new equilibrium concentrations at the new initial pressures and temperatures.

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The complete Question is:

Consider the reaction for the decomposition of hydrogen disulfide:         2H2S W=23 H2 = S2t Ko-2. Ax10 at 1073 K. A reaction vessel initially contains 0.752 atm of H2S at 1073K. Find the equilibrium concentrations of H2 and S2. The reaction in question 2 is then carried out at the same temperature but the initial pressures were 0.20 atm for H2 and 0.40 atm for S2. Calculate the equilibrium concentrations.

Answer:

HOPE THIS HELPS

Helium and Hydrogen

Explanation:

Hydrogen and helium: they are by far the most abundant elements found in the Sun, making up about 98 percent of its mass, but other, heavier elements play an important role in the physical processes that occur in the Sun.

a. The half-life of radon-222 is 3.82 days. b. It will take approximately 11.46 days for the sample to decay to 20% of its original amount.

(a) To find the half-life of radon-222, we can use the formula:

\(N = N0 * (1/2)^{(t/T)}\)

where:

\(N = amount\ remaining\ after\ time\ t\\N0 = initial\ amount\\T = half\ -life\)

We know that after 3 days, the amount remaining is 58% of the original amount, so N/N0 = 0.58 and t = 3 days. Substituting these values:

\(0.58 = (1/2)^(3/T)\)

Taking the natural logarithm of both sides:

\(ln(0.58) = ln(1/2)^{(3/T)} \\ln(0.58) = -(3/T) * ln(2)\\T = -(3/ln(2)) * ln(0.58)\\T = 3.82 days\)

(b) To find how long it will take the sample to decay to 20% of its original amount: \(N = N0 * (1/2)^{(t/T)}\)

We want to find the time t for which N/N0 = 0.20. Substituting this value and T = 3.82 days into the formula gives:

\(0.20 = (1/2)^{(t/3.82)}\)

Taking the natural logarithm of both sides:

\(ln(0.20) = (t/3.82) * ln(1/2) \\t = -(3.82/ln(1/2)) * ln(0.20)\)

\(t = 11.46 days\)

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Answer:

Low-pressure systems are associated with clouds and precipitation that minimize temperature changes throughout the day, whereas high-pressure systems normally associate with dry weather and mostly clear skies with larger diurnal temperature changes due to greater radiation at night and greater sunshine during the day.

Explanation:

Buoyancy is known as the tendency of an object to float in a fluid. Buoyancy results from the differences in pressure acting on opposite sides of an object which is immersed in a static fluid. The magnitude of the buoyant force is 2.4 N.

The upward force exerted on an object which is wholly or partly immersed in a fluid is defined as the Buoyant force. This upward force is also called the Upthrust. It is due to the buoyant force a body submerged partially or fully in a fluid lose its weight.

The buoyant force is given by the equation:

Buoyant force = weight = mg = 3 × 0.8 = 2.4 N

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