The oxidation number of hydrogen in sodium hydride (NaH)

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.

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

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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

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:

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Answer:

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

Explanation: I just answered this on USA test prep.

Answer:

a) 5.40 g B : [tex]3.012\times 10^{23}[/tex] atoms

b) 0.250 mol K : [tex]1.51\times 10^{23}[/tex] atoms

c) 0.0384 mol K : [tex]0.23\times 10^{23}[/tex] atoms

d) 0.02550 g Pt: [tex]7.8\times 10^{19}[/tex] atoms

e)  [tex]1.00\times 10^-{10} g[/tex] Au: [tex]0.03\times 10^{13}[/tex] atoms

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]

a) 5.40 g B

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{5.40g}{11g/mol}=0.5moles[/tex]

1 mole of boron contains = [tex]6.023\times 10^{23}[/tex] atoms

Thus 0.5 moles  of boron contain =[tex]\frac{6.023\times 10^{23}}{1}\times 0.5=3.012\times 10^{23}[/tex] atoms

b) 0.250 mol K

1 mole of potassium (K) contains = [tex]6.023\times 10^{23}[/tex] atoms

Thus 0.250 moles of potassium contain =[tex]\frac{6.023\times 10^{23}}{1}\times 0.250=1.51\times 10^{23}[/tex] atoms

c) 0.0384 mol K

1 mole of potassium (K) contains = [tex]6.023\times 10^{23}[/tex] atoms

Thus 0.0384 moles of potassium (K) contain =[tex]\frac{6.023\times 10^{23}}{1}\times 0.0384=0.23\times 10^{23}[/tex] atoms

d) 0.02550 g Pt

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{0.02550 g}{195g/mol}=1.3\times 10^{-4}moles[/tex]

1 mole of platinum contains = [tex]6.023\times 10^{23}[/tex] atoms

Thus [tex]1.3\times 10^{-4}moles[/tex] of platinum contain=[tex]\frac{6.023\times 10^{23}}{1}\times 1.3\times 10^{-4}=7.8\times 10^{19}[/tex] atoms

e) [tex]1.00\times 10^-{10}g[/tex] Au,

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{1.00\times 10^{-10}g}{197g/mol}=0.005\times 10^{-10}moles[/tex]

1 mole of gold contains = [tex]6.023\times 10^{23}[/tex] atoms

Thus [tex]0.005\times 10^{-10}moles[/tex] of platinum contain=[tex]\frac{6.023\times 10^{23}}{1}\times 0.005\times 10^{-10}=0.03\times 10^{13}[/tex] atoms

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%.

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.

Given:

C= 58.8% ,

H= 9.8% ,

and O=31.4 %

molar mass is = 102 g/mol.

empirical formula=

58.8/C(12.01)= 4.9

9.8/H(1.01)= 9.7

31.4/O(15.99)= 1.96

divide it by the least value so it is 1.96 which shows,

4.9/1.96= 2.5

9.7/1.96= 5

(C2.5H₅O)₂= C₅H₁₀O₂

molecular formula= C₁₀H₂₀O₄

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 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.

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:

-∆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.

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.