What effect would poor circulation of the melting point bath liquid have on the observed melting point?

It would a a chemical change as you cant see the boiling points changing or being formed. Physical changes are only when you visibly and physically observe the change

Answer is: the percent ionizationof formic acid is 1,82%.
Chemical reaction: HCOOH(aq) ⇄ H⁺(aq) + HCOO⁻(aq).
pKa(HCOOH) = 3,77.

Ka(HCOOH) = 1,7·10⁻⁴.

c(HCOOH) = 0,5 M.

[H⁺] = [HCOO⁻] = x; equilibrium concentration.
[HA] = 0,1 M - x.
Ka = [H⁺] · [HCOO⁻] / [HCOOH].
0,00017 = x² / 0,5 M - x.
Solve quadratic equation: x = 0,0091 M.
α = 0,0091 M ÷ 0,5 M · 100% = 1,82%.

Answer : 1.91 %

Explanation : The steps to solve this problem are explained below;

1.  HCOOH ⇄ [tex]HCOO^{-} + H^{+} [/tex]      

Here Ka =([tex] [HCOO^{-}]_{eq} X  [H^{+}]_{eq} [/tex] )/  [tex][HCOOH]_{eq}[/tex]                                                        

As the equilibrium concentration of [tex] H^{+} [/tex] will be the pH of the solution.

∴ [tex] [H^{+}]_{eq}[/tex] = [tex]10^{(-2.02)} = 9.55 x [tex] 10^{-3} [/tex] M  

2. The initial concentration of HCOOH.  When it loses x moles from it as the acid undergoes dissociation to form [tex]HCOO^{-}[/tex] and [tex]H^{+}[/tex].  

3.  The moles present will be as

         [HCOOH] (M)                 [tex] [H^{+}] [/tex](M)           [tex][HCOO^{-}] [/tex](M)  

Initial       0.50                                0.00                                          0.00

After Change  -x                              +x                                            +x

Equilibrium      ( 0.50 -x)                     x                                              x

∴ Ka   =   (x) x (x)  / (0.50 - x)  

4.  Assuming that all of the [tex] H^{+}[/tex] comes from the acid, and none from water.  

As [tex] [H^{+}]_{eq}[/tex] = 9.55 x[tex]  10^{-3}[/tex] which is much higher than the 1.0 x[tex]  10^{-7 } [/tex] M [tex[H^{+}[/tex] from water.  

Also, the concentration of HCOOH will change very little, from 0.50  to 0.50 - 9.55 x [tex] 10^{-3}[/tex].  

The change in concentration can be ignored if it is less than 5% of the original concentration.  

∴ 0.50 M x 5% = 0.025, so the change in [HCOOH] in this problem can be ignored.    

Now,  Ka = (x)(x)/0.50 = (9.55 x [tex] 10^{-3})^{2}[/tex] /0.50= 1.82 x [tex] 10^{-4}[/tex]

Now, calculating the percent ionization for this problem.  

which will represent the relative number of acid molecules which dissociate. It is calculated as :

[tex] [H^{+}]_{eq} [/tex] x  100 /[tex] [HCOOH]_{i}[/tex]                                                        

∴ percent ionization = {(9.55 x [tex] 10^{-3})[/tex]/ (0.50)}x 100 = 1.91 %

This value of 1.91 % indicates that very little of this acid dissociates (ionizes) under these conditions.  

For strong acids and bases, the percent ionization is 100%.

2 moles of hydrogen molecules forms 2 moles of water or 36 g of water. 2 molecule of hydrogen contains 1.2 × 10²⁴. Thus, 1.11× 10²⁴ hydrogen molecules will give 33.3 g of water.

One mole of any compound contains 6.022 × 10²³ molecules. This number is called Avogadro number. Thus,one mole water contains 6.022 × 10²³  water molecules.

2 moles of hydrogen molecules will give 2 moles of water. One mole of hydrogen molecule contains 6.022 × 10²³ molecules.

molar mass of water = 18 g/mol

2 moles of water = 36 g.

Then, 2 moles of H₂ = 2 × 6.022 × 10²³ =  1.2 × 10²⁴ molecules

These much hydrogen molecules produces 36 g of water. Thus, mass of water produced by 1.11× 10²⁴ hydrogen molecules is :

(1.11× 10²⁴ × 36 )/1.2 × 10²⁴  = 33.3 g.

Therefore, the mass of water produced from 1.11× 10²⁴ hydrogen molecules is 33.3 g.

To find more on Avogadro number, refer here:

https://brainly.com/question/11907018

#SPJ5

To calculate the mass percent, molality, and mole fraction of sulfuric acid in a lead storage battery, you need to perform calculations involving the solution's molarity, density, and the molar mass of H₂SO₄. You must find the mass of the solution, then determine the mass and moles of the solute and solvent.

To calculate the mass percent, molality, and mole fraction of sulfuric acid (H₂SO₄) in an automobile lead storage battery with 3.75 M sulfuric acid and a density of 1.230 g/mL, you need to perform several calculations.

Mass Percent.The mass percent of the solution is the mass of the solute (in this case, H₂SO₄) divided by the total mass of the solution multiplied by 100. Based on the volume of the solution and its molarity, you can find the number of moles of H₂SO₄, which helps to calculate mass, as you know the molar mass of H₂SO₄ is 98.079 g/mol.

Molality. Molality is the number of moles of solute per kilogram of solvent (water in this case). To find the molality, you would calculate the number of moles of sulfuric acid based on its molarity and then calculate the mass of the solvent in kilograms. Remember that the density and the volume of the solution can lead you to find the mass of the solution itself, from which you subtract the mass of the solute to get the mass of the solvent.

Mole Fraction.The mole fraction is the ratio of the number of moles of the solute to the total number of moles of all components in the solution. This requires you to know the moles of both the solute and the solvent.

Answer:

-∆G and +Eocell

Explanation:

For an electrochemical reaction to be spontaneous, the change in free energy must be negative. This applies to all chemical reactions as a basic condition for spontaneity of chemical reactions.

More particular to an electrochemical cell is the value of the standard cell voltage. A positive value of standard cell voltage implies a spontaneous electrochemical reaction.

Answer:

Electrons exchange creating ions to form an ionic bond by electrostatic attraction.

Explanation: I just answered this on USA test prep.

The partial pressure of oxygen when the atmospheric pressure is 770 mm Hg and oxygen represents 13% of the total is 100.1 mm Hg. Since the choices don't include decimals, the closest answer and the correct one would be 100 mm Hg.

If the atmospheric pressure is 770 mm Hg and oxygen represents 13% of the total, then the partial pressure of oxygen is calculated by multiplying the total atmospheric pressure by the percentage of oxygen present. So, the partial pressure of oxygen (Po₂) can be found using the formula:

Po₂ = Total atmospheric pressure x (percent content in mixture)

In this case:

Po₂ = 770 mm Hg x 0.13 = 100.1 mm Hg

Since the values provided in the question for the partial pressure of oxygen don't include decimals, the most appropriate answer is 100 mm Hg, which is the value closest to our calculated result.

Answer:

See explanation

Explanation:

First, let's write the equation again.

CH4 + NH3 + O2 -----> HCN + H2O

In order to balance this equation, we need to see both sides of the reaction (Reactants and products), and count the elements and atoms there. it's usually begin with the metals, then non metals, and finally hydrogen and oxygen.

Let's begin with the Carbon and nitrogen.

In reactant we have 1 C and product the same, so theorically speaking, we don't need to balance, same thing happen with N (1 and 1).

However, when we look at H, we can see we have 7 in reactants (4 and 3) and only 3 in products (1 and 2), so we need to balance them. As we have those atoms in different compounds, we need to put a number in both compound so the sum gives an even number. In this case if we put a 2 in CH4, we'll get 8 hydrogen and the other 3, will be 11.

2CH4 + NH3 + O2 -----> HCN + H2O

To get an even number, let's put a 2 in NH3 too, the total is now 14.

2CH4 + 2NH3 + O2 -----> HCN + H2O

Let get 14 in the products, to do that we simply put a 2 in HCN and 6 in H2O.

2CH4 + 2NH3 + O2 -----> 2HCN + 2H2O

With that, we balance Hydrogen, and also Carbon and nitrogen were balanced too with this.

Now only the oxygen needs to be balanced. The 6 in H2O put the oxygen with 6, so we need to put a 3 in reactants, and with that, the equation is balanced:

2CH4 + 2NH3 + 3O2 -----> 2HCN + 6H2O

Now that we have the balanced equation, we can calculate the rest of the questions.

The limiting reagent is the reagent that it gets consumed first and completely in the reaction. To do that, we need to do stechiometry of the reagents, and the easier way to do that is with the number of moles.

moles are calculated:

n = m/MM

The molecular mass of CH4 is 16 g/mol and NH3 is 17 g/mol. According to the balanced equation, we have at least 2 moles of CH4 and 2 moles of NH3. The limiting reagent will be the reagent with the lower moles so:

nCH4 = 8/16 = 0.5 moles

nNH3 = 10/17 = 0.59 moles

We have more moles of NH3 than moles CH4, so the limiting reagent is CH4.

Now to get the mass of HCN, we need the moles. But we already know which is the limiting reagent and we know (according to the balanced reaction) that 2 moles of CH4 produces 2 moles HCN, so, the moles of CH4 are the same of HCN produced therefore:

moles CH4 = moles HCN = 0.5 moles

To get the mass we need the molecular mass of HCN which is 27 g/mol, so the mass:

m = 27 * 0.5

m = 13.5 g

This is the mass formed of HCN

Final answer:

The volume of gas produced by two antacid tablets is correctly identified as 400 cm3, making option A correct. For the moles of oxygen inhaled under given conditions, 1.21 moles, option B, is also correctly identified using the ideal gas law.

Explanation:

Chemistry Problem Solutions

The first question deals with the stoichiometry of a reaction between vinegar and antacid tablets, specifically focusing on the volume of gas produced in reaction to the quantity of antacid tablets used. Given that three tablets produce 600 cm3 of gas and assuming the reaction proceeds in a linear fashion (due to being the limiting reagent), the volume of gas produced by two tablets would indeed be 400 cm3 which is option A. Hence, your answer to the first question is correct.

For the second question about calculating moles of oxygen inhaled, we use the ideal gas law formula, PV=nRT, where P is the pressure (converted to atm), V is the volume in liters, n is the moles of gas, R is the universal gas constant (0.0821 L·atm·K-1·mol-1), and T is the temperature in Kelvin. By plugging in the values (and converting 1000 kPa to approximately 9.869 atm), we find that the moles of oxygen inhaled are roughly 1.21 moles, and so your answer B is also correct.

The heat of reaction for the combustion of a mole of titanium in the given calorimeter setup is approximately 14765.89 kJ/mol, as determined by calculating the total heat absorbed using the calorimeter's heat capacity and the mass and molar mass of titanium.

When 1.640 g of titanium is combusted in a bomb calorimeter, the resulting temperature increase from 25.00 °C to 76.50 °C can be used to calculate the heat of reaction for the combustion of a mole of titanium. Given that the heat capacity of the calorimeter is 9.84 kJ/°C, we can determine the total heat absorbed by the calorimeter during the reaction.

First, we calculate the total heat absorbed (q) by multiplying the heat capacity (C) by the change in temperature (ΔT):
q = C × ΔT
q = 9.84 kJ/°C × (76.50 °C - 25.00 °C)
q = 9.84 kJ/°C × 51.50 °C
q = 506.34 kJ

Next, to find the heat of reaction per mole of titanium, we convert the mass of titanium to moles using its molar mass (47.87 g/mol):
Number of moles = mass / molar mass
Number of moles = 1.640 g / 47.87 g/mol
Number of moles ≈ 0.0343 mol

Now, we can determine the heat of reaction per mole:
Heat of reaction per mole = q / Number of moles
Heat of reaction per mole ≈ 506.34 kJ / 0.0343 mol
Heat of reaction per mole ≈ 14765.89 kJ/mol

The heat of reaction for the combustion of a mole of Ti in this calorimeter is approximately 14765.89 kJ/mol.

This problem requires using the concept of pH and pOH. The “p” stands for the negative log so pH and pOH represent the negative log of the concentration of hydrogen or hydroxide ions.

Here is the solution:

pOH = -log [OH-]

pOH = -log [1.83x10^-7 M]

pOH = 6.74

pH + pOH = 14

pH = 14 - 6.74

pH = 7.26

pH = -log [H3O+]

7.26 = -log[H3O+]

[H3O+] = 5.46 x 10^-8 M

Answer: Option (b) is the correct answer.

Explanation:

Atomic number is the sum of only total number of protons present in an element. Whereas mass number or atomic mass is the sum of total number of both protons and neutrons present in an element.

For example, given atom has atomic number is 8.  And, it is known that atomic mass of O is 16 g/mol.

Therefore, number of neutrons present in it will be calculated as follows.

                    Mass number = no. of protons + no. of neutrons

                         16 = 8 + no. of neutrons

                    no. of neutrons = 16 - 8  

                                               = 8

Hence, we can conclude that in the given isotope of oxygen-16 there are 8 protons and 8 neutrons.

The wavelength of a radiation beam with an energy of 2.39×10² kJ/mol can be calculated using the Planck's formula. Convert the energy to J/photon, then rearrange the formula to solve for wavelength. The wavelength for this energy is approximately 501 nm.

The energy of a beam of radiation can be used to determine its wavelength using Planck's formula, E=hv, where E is energy, h is Planck’s constant, and v is frequency. However, since we want to find the wavelength, we can modify the formula to make use of the speed of light: E=hc/λ. Here, c denotes the speed of light, and λ is wavelength.

First, convert the energy from kJ/mol to J/photon. As Avogadro's number (6.022 × 10²³ molecules/mol) is the number of photons in one mole, the energy per photon is (2.39×10⁵ J/mol) / (6.022×10²³ photon/mol) = 3.97×10⁻¹⁹ J/photon. Now, use Planck's formula to calculate the wavelength:

λ = hc/E = (6.63×10⁻³⁴ J.s × 3.0×10⁸ m/s) / 3.97×10⁻¹⁹ J = 5.01×10⁻⁷ m = 501 nm

https://brainly.com/question/34184937

#SPJ12

The relative atomic mass of copper on a newly discovered planet is [tex]\boxed{63.54{\text{ amu}}}[/tex] .

Further explanation:

An atom is also written as [tex]_{\text{Z}}^{\text{A}}{\text{X}}[/tex] , where A is the atomic mass or mass number, Z is the atomic number, and X is the letter symbol of the element.

Isotopes:

Atoms of the same element with same value of atomic number but different value of mass numbers are called isotopes. Isotopes generally have the same number of protons but the number of neutrons is different.

[tex]_{\mathbf{6}}^{{\mathbf{11}}}{\mathbf{C}}[/tex]  and [tex]_{\mathbf{6}}^{{\mathbf{12}}}{\mathbf{C}}[/tex]  are examples of isotopes. Both are the atoms of the same element carbon. The mass number of [tex]_{\mathbf{6}}^{{\mathbf{11}}}{\mathbf{C}}[/tex]  is 11 while that of  [tex]_{\mathbf{6}}^{{\mathbf{12}}}{\mathbf{C}}[/tex] is 12. But both of these have the same atomic number, i.e, 6.

Relative Atomic Mass:

It is defined as the average mass of an atom of the element compared to 1/12th of the mass of carbon-12 atom. It is represented by [tex]{{\text{A}}_{\text{r}}}[/tex] . The formula to calculate the relative atomic mass of an element is as follows:

[tex]{{\text{A}}_{\text{r}}} = \frac{{\sum {\left\{{\left({{\text{Relative isotopic mass}}}\right)\left({{\text{\%  abundance}}}\right)}\right\}}}}{{100}}[/tex]

Here,

[tex]{{\text{A}}_{\text{r}}}[/tex]  is the relative atomic mass.

It is given that copper (Cu) exists in two isotopic forms, one is 63 Cu, and the other is 65 Cu. So the relative atomic mass of Cu is calculated as follows:

[tex]{{\text{A}}_{\text{r}}}{\text{ of Cu}}=\frac{{\left[\begin{gathered}\left({{\text{Mass of 63 Cu}}}\right)\left({{\text{\%  abundance of 63 Cu}}}\right)+\hfill\\\left({{\text{Mass of 65 Cu}}}\right)\left({{\text{\%  abundance of 65 Cu}}}\right)\hfill\\\end{gathered}\right]}}{{100}}[/tex]                       …… (1)

The mass of 63 Cu is 62.9300 amu.

The percentage abundance of 63 Cu is 69.15 %.

The mass of 65 Cu is 64.9200 amu.

The percentage abundance of 65 Cu is 30.85 %.

Substitute these values in equation (1).

[tex]\begin{aligned}{{\text{A}}_{\text{r}}}{\text{ of Cu}}&=\frac{{\left[{\left( {{\text{62}}{\text{.9300 amu}}}\right)\left({{\text{69}}{\text{.15}}}\right)+\left({{\text{64}}{\text{.9200 amu}}}\right)\left({{\text{30}}{\text{.85}}}\right)}\right]}}{{100}}\\&=\frac{{{\text{4351}}{\text{.6095 amu}}+{\text{2002}}{\text{.782 amu}}}}{{100}}\\&={\text{63}}{\text{.543915 amu}}\\&\approx{\text{63}}{\text{.54 amu}}\\\end{aligned}[/tex]

So the relative atomic mass of copper on a newly discovered planet is 63.54 amu.

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: https://brainly.com/question/3064603

2. Calculate the moles of ions in the solution: https://brainly.com/question/5950133

Answer details:

Grade: Middle School

Subject: Chemistry

Chapter: Mole Concept  

Keywords: percentage abundance, 63 Cu, 65 Cu, 62.9300 amu, 64.9200 amu, 30.85 %, 69.15 %, 63.54 amu, copper, isotopes, carbon-12, Ar, Cu.

The common name to the family member of phylum Echinodermata of marine family is echinoderm. They are usually characterized by a five-fold symmetry, and possess an internal skeleton of calcite plates.  They are found at every ocean depth.

The features of all adult echinoderm are:

- They have a five-fold symmetry.

- Body without segmentation.

- Spiny skin.

- Internal skeleton.

- found at every ocean depth.