Molarity by dilution. Show steps

\( \boxed{\sf \red{answer}}\)

\(MnO + HCl \green\rightarrow \: MnCl _{2} + H_{2}O \: + Cl_{2} \)

Answer:

78.75L

Explanation:

According to Charle's Law:

V1/T1=V2/T2

35/200=V2/450

V2=(35*450)200

V2=78.75L

The volume of carbon dioxide produced is approximately (d) 11.9 liters.

To determine the amount of carbon dioxide (C\(O_2\)) produced when 37.8 grams of carbon disulfide (C\(S_2\)) reacts with excess oxygen gas (\(O_2\)), we need to use stoichiometry and the given balanced chemical equation:

C\(S_2\)(l) + 3\(O_2\)(g) → C\(O_2\)(g) + 2S\(O_2\)(g)

First, we calculate the number of moles of C\(S_2\) using its molar mass:

Molar mass of (C\(S_2\)) = 12.01 g/mol (C) + 32.07 g/mol (S) × 2 = 76.14 g/mol

Number of moles of (C\(S_2\)) = mass / molar mass = 37.8 g / 76.14 g/mol ≈ 0.496 mol

From the balanced equation, we can see that the stoichiometric ratio between (C\(S_2\)) and C\(O_2\) is 1:1. Therefore, the number of moles of C\(O_2\) produced will also be 0.496 mol.

Now we can use the ideal gas law to calculate the volume of C\(O_2\) at the given temperature and pressure. The ideal gas law equation is:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

Converting the temperature from Celsius to Kelvin:

T(K) = 28.85°C + 273.15 = 302 K

Using the ideal gas law:

V = nRT / P = (0.496 mol) × (0.0821 L·atm/mol·K) × (302 K) / (1.02 atm) ≈ 11.9 L

The correct answer is 11.9 liters.

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The standard deviation of ksp is 2.646 x 10⁻⁶ and the relative standard deviation kf ksp (%RSD) is 27.56%.

To calculate standard deviation of Ksp, first, find the deviations of each trial's Ksp value from the average Ksp value:

Trial 1 = (10 x 10⁻⁵) - (9.6 x 10⁻⁶) = 9.04 x 10⁻⁶

Trial 2 = (8.0 x 10⁻⁵) - (9.6 x 10⁻⁶) = 7.04 x 10⁻⁵

Trial 3 = (11 x 10⁻⁵) - (9.6 x 10⁻⁶) = 1.04 x 10⁻⁴

Square each deviation and find the average:

(9.04 x 10⁻⁶)² + (7.04 x 10⁻⁵)² + (1.04 x 10⁻⁴)² = 2.098 x 10⁻¹⁰

Average = 2.098 x 10⁻¹⁰ / 3 = 6.993 x 10⁻¹¹

Finally, take the square root of the average to get the standard deviation of Ksp:

Standard deviation of Ksp = √(6.993 x 10⁻¹¹) = 2.646 x 10⁻⁶

To calculate the relative standard deviation of Ksp (%RSD):

%RSD = (standard deviation / average) x 100%

%RSD = (2.646 x 10⁻⁶ / 9.6 x 10⁻⁶) x 100% = 27.56% (rounded to two decimal places)

Therefore, the standard deviation of Ksp is 2.646 x 10⁻⁶ and the relative standard deviation of Ksp (%RSD) is 27.56%.

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a) The solubility of YF3 would decrease with increasing pH of the solution. This is because the fluoride ion concentration would decrease with increasing pH, according to the equilibrium: F-(aq) + H2O(l) ⇌ HF(aq) + OH-(aq). The lower the concentration of fluoride ions in the solution, the less YF3 can dissolve, based on the solubility equilibrium.

b) The solubility of YF3 would increase with decreasing pH of the solution. This is because the fluoride ion concentration would increase with decreasing pH, according to the equilibrium: HF(aq) + H2O(l) ⇌ H3O+(aq) + F-(aq). The higher the concentration of fluoride ions in the solution, the more YF3 can dissolve, based on the solubility equilibrium.

The number of moles of the water produces by the 1.3 moles of O₂ is 1.00 mol of H₂O. The correct option is C.

The chemical equation is as :

2C₄H₁₀  +  13O₂  ---->  8CO₂ + 10H₂O

The moles of the O₂ = 1.3 moles H₂O

The molar mass of the O₂ = 32 g/mol

2 moles of the C₄H₁₀ produces the 10 moles of the water,

13 moles of O₂ produces the 10 moles of H₂O

The moles of the H₂O = (10/13) × 1.3

The moles of the H₂O = 1 mol

The number of moles of the water is 1 mol of the H₂O and the moles of the oxygen that is O₂ is the 1.3 moles.. Therefore, the correct option is C.

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Answer:

see below

Explanation:

1) here we are given that one mole of propane releases 2221kJ of energy.

now use unitary method as ,

to burn 1 mol (44g) propane 2221kJ of energy is required

to burn 1g of propane 2221kJ/44g of energy is required

therefore to burn 25g of propane 2221/44*25kJ of energy is required= 1261.93kJ ≈ 1262 kJ

Hence 1262kJ of energy is required to burn 25g propane.

2) amount of heat absorbed= 25kJ

amount of work done = 12J

change in energy= (25-12)kJ = 13kJ

Hence the value of ∆E is 13kJ .

and we are done!

\(\boxed{\sf Density=\dfrac{Mass}{Volume}}\)

\(\\ \sf\longmapsto Mass=Density\times Volume\)

\(\\ \sf\longmapsto Mass=1.5(10)\)

\(\\ \sf\longmapsto Mass=15g\)

Answer:

H3O^+ < HClO2 < H3PO4< HCOOH < H2PO4^-<CIO2^-<H2O< HCOO^-

Explanation:

The pH of a solution refers to the negative logarithm of it's hydrogen ion concentration. Generally, the hydrogen ion concentration of an acid is always greater than that of its conjugate base. Hence, the conjugate acid always has a lower pH than the conjugate base.

This can be seen in the species, H3O^+ and H2O, HClO2 and ClO2^- etc.

Answer:

Explanation:

Propionic acid  has chemical formula of  C₃H₆O₂ , and smells somewhat unpleasant (like body odour).

Answer:C

Explanation:it gets rid of waste

The mass percent of the sodium chloride in the saline solution is approximately 1.33%. The molar concentration of the saline solution is approximately 0.229 M.

To determine the mass percent and molar concentration of the saline solution, we need to analyze the mass of the sodium chloride residue and the initial mass of the sample.

Mass percent:

The mass percent is calculated by dividing the mass of the sodium chloride residue by the initial mass of the sample and then multiplying by 100%.

Mass percent = (Mass of NaCl / Initial mass of sample) × 100%

Mass of NaCl = 0.669 g

Initial mass of sample = 50.320 g

Mass percent = (0.669 g / 50.320 g) × 100% ≈ 1.33%

The mass percent of the sodium chloride in the saline solution is approximately 1.33%.

Molar concentration:

To calculate the molar concentration of the saline solution, we need to determine the number of moles of sodium chloride and the volume of the solution.

Moles of NaCl = Mass of NaCl / Molar mass of NaCl

The molar mass of NaCl is 58.44 g/mol.

Moles of NaCl = 0.669 g / 58.44 g/mol ≈ 0.01144 mol

Since the volume of the sample is given as 50.0 mL, we need to convert it to liters.

Volume of solution = 50.0 mL = 50.0 mL × (1 L / 1000 mL) = 0.0500 L

Now we can calculate the molar concentration (Molarity) using the formula:

Molarity (M) = Moles of solute / Volume of solution (in liters)

Molarity = 0.01144 mol / 0.0500 L ≈ 0.229 M

The molar concentration of the saline solution is approximately 0.229 M.

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As per the balanced equation of the reaction one mole of Mg give s one mole of H2. 1.10 g of magnesium is 0.045 moles. Thus it gives 0.045 moles of H2 which is 1.02 litres.

Magnesium chloride is an ionic compound formed by the reaction of magnesium metal with two moles of hydrochloric acid. The reaction also produce one mole  hydrogen gas.

The atomic mass of magnesium is 24 g/mol. Thus number of moles in 1.10 g is 1.10/24 = 0.045 moles.

As per the reaction equation one mole of magnesium gives one mole of hydrogen gas. Thus 0.045 moles of magnesium gives 0.045 moles of hydrogen gas.

At STP, the volume of one mole of any gas will be equal to 22.4 L. Thus volume of 0.045 moles of hydrogen at STP is:

volume = 0.045 × 22.4

             =1.02 L.

Hence, the volume of hydrogen produce here is 1.02 L.

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Answer:

Explanation:

TERTIARY ALCOHOL:

The alcohols where the carbon to which the hydroxide group is linked is 3 ' C carbon, as it has 3 alkyl groups attached to it.

TERTIARY COMPOUND:

A carbon attached to 3 other carbons is tertiary compound.

Answer:

9.33

Explanation:

balanced chemical equation:

2NH₃ + 3CuO → 3Cu + N₂ + 3H₂O

14 moles of NH3

2NH₃ → 3H₂O

14 * 2/3 = 9.33

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Ionic bonding is a type of chemical bond that occurs when electrons are transferred between atoms to form ions. The resulting ions are attracted to each other by their opposite charges and form an ionic compound. Here is a summary of the information you provided:

Element: Na (sodium)

Number of valence electrons: 1

Dot structure: Na

Transfer of electrons: Na loses 1 electron to become Na+

Ions formed: Na+ and Cl-

Compound formed: Sodium chloride (NaCl)

Name of compound: Salt

Element: F (fluorine)

Number of valence electrons: 7

Dot structure: F

Transfer of electrons: F gains 1 electron to become F-

Ions formed: F- and Mg2+

Compound formed: Magnesium fluoride (MgF2)

Element: O (oxygen)

Number of valence electrons: 6

Dot structure: O

Transfer of electrons: O gains 2 electrons to become O2-

Ions formed: Ba2+ and O2-

Compound formed: Barium oxide (BaO)

Element: Li (lithium)

Number of valence electrons: 1

Dot structure: Li

Transfer of electrons: Li loses 1 electron to become Li+

Ions formed: Li+ and Cl-

Compound formed: Lithium chloride (LiCl)

Element: P (phosphorus)

Number of valence electrons: 5

Dot structure: P

Transfer of electrons: P gains 3 electrons to become P3-

Ions formed: Al3+ and P3-

Compound formed: Aluminum phosphide (AlP)

Element: N (nitrogen)

Number of valence electrons: 5

Dot structure: N

Transfer of electrons: N gains 3 electrons to become N3-

Ions formed: Al3+ and N3-

Compound formed: Aluminum nitride (AlN)

Answer:

\(\Delta G_{rxn}=42.7\frac{kJ}{mol}\)

Explanation:

In this case, for the dissociation of hypochlorous acid, we know that the acid dissociation constant (Ka) is 2.9x10⁻⁸, which is related with the Gibbs free energy as shown below:

\(\Delta G_{rxn}=-RTln(K)\)

But in this case K is just Ka, therefore, at 296 K, it turns out:

\(\Delta G_{rxn}=-8.314\frac{J}{mol*K}*296K*ln(2.9x10^{-8})\\\\\Delta G_{rxn}=42.7\frac{kJ}{mol}\)

Such result, means that the reaction is nonspontaneous at the given temperature, it means it is not favorable (not easily occurring).

Best regards.

Answer:

Activation energy for the reaction is 39029J/mol

Explanation:

Arrhenius equation is an useful equation that relates rate of reaction at two different temperatures as follows:

\(ln\frac{K_2}{K_1} = \frac{-Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1} )\)

Where K₁ and K₂ are rate of reaction, Ea is activation energy and R is gas constant (8.314J/molK

If the reaction at 400K is 50 times more faster than at 300K:

K₂/K₁ = 50 where T₂ = 400K and T₁ = 300K:

\(ln50 = \frac{-Ea}{8.314J/molK} (\frac{1}{400K} -\frac{1}{300K} )\)

\(ln 50 = 1x10^{-4}Ea\)

Ea = 39029 J/mol

The activation energy for this chemical reaction is equal to 39,029.24 J/mol.

Given the following data:

Ideal gas constant, R = 8.314 J/molK

To determine the activation energy for this chemical reaction, we would use the Arrhenius' equation:

Mathematically, Arrhenius' equation is given by the formula:

\(ln\frac{K_2}{K_1} = \frac{-E_a}{R} (\frac{1}{T_2} - \frac{1}{T_1})\)

Where:

Substituting the given parameters into the formula, we have;

\(ln50 = \frac{-E_a}{8.314} (\frac{1}{400} - \frac{1}{300})\\\\3.9120 = \frac{-E_a}{8.314} (\frac{-1}{1200})\\\\3.9120 = \frac{E_a}{9976.8} \\\\E_a = 9976.8 \times 3.9120\\\\E_a = 39,029.24 \;J/mol\)

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When hydrogen bromide (HBr) is added to 2-methylbut-2-ene ((CH3)2CCHCH3), an electrophilic addition reaction takes place, where the π bond of the alkene is broken, and the hydrogen and bromine atoms are added to the resulting carbocation.

The reaction proceeds through a Markovnikov addition, where the hydrogen atom attaches to the carbon atom with the greater number of hydrogen atoms.

In this case, the initial addition of HBr to 2-methylbut-2-ene leads to the formation of a primary carbocation, as the positively charged carbon atom only has one alkyl group attached to it. The primary carbocation is relatively unstable, and it can undergo a rearrangement to form a more stable secondary carbocation.

The major product that is typically obtained is the 2-bromo-2-methylbutane. The hydrogen atom from HBr adds to the carbon with three hydrogen atoms (the more substituted carbon), resulting in the formation of a secondary carbocation.

On the other hand, a minor product is also formed, which is 3-bromo-2-methylbutane. This product arises from the addition of HBr to the primary carbocation, which is less stable. Although the primary carbocation is less favored, it can still be formed and lead to the formation of the minor product.

In summary, the addition of HBr to 2-methylbut-2-ene yields two products: the major product is 2-bromo-2-methylbutane, resulting from the addition of HBr to the more stable secondary carbocation, and the minor product is 3-bromo-2-methylbutane, originating from the less stable primary carbocation.

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The characteristics of the ocean ridges allow finding the results to complete the statements are:

The oceanic ridges are mountain ranges or chains of volcanic mountains at the bottom of the ocean, in these the magma that rises cools and forms young rocks, in they are pushed by the other rocks that emerge, in general they are located on the edges of the tectonic plates .

Therefore, the oceanic ridges are the point where the earth's crust is separating, therefore it has the youngest rocks and in the furthest points it has the oldest.

Let's find the correct words for each phrase to be correct:

As a consequence of the characteristics of the ocean ridges, we can find the results to complete the statements  are:  

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Answer: The rate determining step of this reaction would be step 2 and the best option to answer the question is the fourth one (letter D).

Explanation:

The question requires us to identify the rate determining step in a reaction that happens in three steps, given that steps 1 and 3 are fast, and step 2 is slow.

The rate determining step of a reaction can be compared to the neck of a funnel where water needs to flow through: the rate by which water flows will be limited by the size of the neck of the funnel (because that is the available space for water flow) . In that sense, we can say that the rate of a chemical reaction is determined by its slow step.

Therefore, considering the three steps provided by the question (step 1 - fast, step 2 - slow, step 3 - fast), the rate determining step of this reaction would be step 2 and the best option to answer the question is the fourth one (letter D).

Dissolve 60g of potassium bromide in 340g of water to produce 15% (mass/mass) aqueous solution of potassium bromide.

Here we have to prepare a total of 400 g of solution. Aqueous solution means the solvent we use here is water.

So to prepare 400 g of 15% aqueous solution of potassium bromide, we need to find out how many grams of potassium bromide need to be dissolved in water and how many grams of water must be used.

Here the weight percent is given, that is 15%

       15/100 = weight of potassium bromide/ 400 g

       0 .15      = weight of potassium bromide / 400

      weight of potassium bromide needed = 0.15 × 400

                                                                       =  60 g

So, we calculated the required amount of potassium bromide as 60 grams. The total weight of the solution to be made is 400 grams.

So amount of water required = 400 - 60

                                                =  340 g

So we need to mix 60 grams of potassium bromide in 340 grams of water to get a 15% (mass/mass) aqueous solution.

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In this ranking we would have; C > N > O > F > Ne.

The term formal charge has to do with the charge that an atom appears to have in  compound. We can obtain the formal charge from the formula;

Formal charge = [# of valence electrons] – [electrons in lone pairs + 1/2 the number of bonding electrons]

Let us now consider the atoms; F, O, N, C, Ne. We have to rank the atoms by the common number of single bonds the element makes when it bears no formal charge.

In this ranking we would have; C > N > O > F > Ne.

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Almosphere is a cloud-based platform that provides business intelligence, analytics, and machine learning capabilities to enable organizations to make data-driven decisions.

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Almosphere is a cloud-based platform that provides business intelligence, analytics, and machine learning capabilities to enable organizations to make data-driven decisions.

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Answer:

magnitude and direction