Answer:
The correct answer is "654.54 nm".
Explanation:
According to the question,
⇒ [tex]\frac{1}{\lambda}= Rh(\frac{1}{n1^2} -\frac{1}{n2^2} )[/tex]
By substituting the values, we get
[tex]=1.1\times 10^7(\frac{1}{4} -\frac{1}{9} )[/tex]
[tex]=1.1\times 10^7(\frac{9-4}{36} )[/tex]
[tex]=1.1\times 10^7(\frac{5}{36} )[/tex]
[tex]=654.54\ nm[/tex]
Thus the above is the right solution.
A quantity of 0.27 mole of neon is confined in a container at 2.50 atm and 298 Kand then allowed to expand adiabatically under two different conditions: (a) reversibly to 1.00 atm and (b) against a constant pressure of 1.00 atm. Calculate the final temperature in each case.
Answer:
a) Hence, T = 207 K.
b) Hence, T2 = 226 K.
Explanation:
Now the given,
n = 0.27 moles ; P = 2.5 atm ; T = 298 K
a) γ = 5/3 since Ne is a monoatomic gas.
[tex](1 - \gamma )/\gamma = -2/5\\T1 P1^{(1-\gamma)/\gamma}=T2 P2^{(1-\gamma)/\gamma}\\T2 = T1(P1/P2)^{(1 - \gamma)/\gamma}\\T2 = 298 (2.5/1)^{-2/5}= 207 K\\[/tex]
Hence, T = 207 K
b) We know that,[tex]U = W = n Cv (T2 - T1) = -P (V2 - V1)[/tex]
[tex]n(3/2)R(T2 - T1) = -P( n R T2/P2 - n R T1/P1)\\3/2(T2 - T1) = -P (T2/P2 - T1/P1)[/tex]
But P = P2
[tex]3/2(T2 - T1) = -P2(T2/P2 - T1/P1)\\3/2(T2 - T1) = -T2 + P2T1/P1[/tex]
This gives us:
[tex]T2 = 2/5(P2/P1 + 3/2)T1\\T2 = 2/5 x (1 /2.5 + 3/2)/(298)\\T2 = 19/25 x 298 = 226 K[/tex]
Hence, T2 = 226 K
time is direct propotinally to rate of chemical reaction .explain if time is negretted and temperature remain costant?
Answer:
the constant temperature will be 12 .F bec in places it is cold
For a given fluorophore, select the choice that correctly lists the processes of fluorescence, absorption, and phosphorescence in order from shortest to longest wavelength.
a. absorption, fluorescence, phosphorescence
b. Fluorescence = phosphorescence, absorption
c. fluorescence, phosphorescence, absorption
d. phosphorescence, fluorescence, absorption
e. absorption, phosphorescence, fluorescence
f. absorption, fluorescence = phosphorescence
Answer:
absorption, fluorescence = phosphorescence
Explanation:
Given a particular fluorophore, the wavelength of absorption of the fluorophore is always shorter. Both fluorescence and phosphorescence are kinds of photoluminescence.
Recall that both fluorescence and phosphorescence occur at a longer wavelength. The difference between the two is only in the time taken during the process. While fluorescence takes a shorter time to occur, phosphorescence takes a longer time to occur.
The major difference between fluorescence and phosphorescence is that change of spin occurs during phosphorescence but not fluorescence.
A student attempts to separate 4.656 g of a sand/salt mixture just like you did in this lab. After carrying out the experiment, she recovers 2.775 g of sand and 0.852 g of salt.a. What was the percent composition of sand in the mixture according to the student's data? b. What was the percent recovery?
Answer:
Explanation:
a ) Total mixture = 4.656 g
Sand recovered = 2.775 g
percent composition of sand in the mixture
= (2.775 g / 4.656 g ) x 100
= 59.6 % .
b )
Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .
Total mixture = 4.656 g
percent recovery = (3.627 / 4.656 ) x 100
= 77.9 % .
repining of fruits is which type of change
Answer:
irreversible.
I hope this will help you
Given the chemical equation: KI +Pb(NO3)2—>KNO3 + Pbl2
Balance this chemical equation.
Indicate the type of reaction. How do you know?
Thoroughly discuss how your balanced chemical equation agrees with the law of conservation of mass.
Answer:
[tex]2KI +Pb(NO_3)_2\rightarrow 2KNO_3 + Pbl_2[/tex]
Double replacement reaction.
It is in agreement with the law of conservation of mass because we have two potassium atoms, two iodine atoms, one lead atom, two nitrogen atoms and six oxygen atoms on both sides of the chemical equation (count them).
Explanation:
Hello there!
In this case, according to the given information, it turns possible for us to solve this problem by firstly considering that this reaction occurs between potassium iodide and lead (II) nitrate to yield potassium nitrate and lead (II) iodide which is clearly not balanced since we have one iodine atom on the reactants and two on the products, that is why the balance implies the placement of a coefficient of 2 in front of both KI and KNO3 as shown below:
[tex]2KI +Pb(NO_3)_2\rightarrow 2KNO_3 + Pbl_2[/tex]
Thus, we infer this is a double replacement reaction due to the exchange of both cations, K and Pb with both anions, I and NO3. Moreover, we can tell this balanced reaction is in agreement with the law of conservation of mass because we have two potassium atoms, two iodine atoms, one lead atom, two nitrogen atoms and six oxygen atoms on both sides of the chemical equation (count them).
Regards!
Each week CapeChem, a manufacturer of fine chemicals, uses up of Compound and of Compound in a reaction with only one product, Compound .What is the maximum theoretical mass of Compound that CapeChem could ship each week
Answer: Hello your question is poorly written attached below is the complete question
answer:
450 kg
Explanation:
mass of product formed = mass of reactants that reacted
hence :
mass of compound C that can be formed and shipped
= mass of A + mass of B
= 200 kg + 250 kg = 450 kg ( theoretical mass of compound formed )
The combustion of hydrogen-oxygen mixtures is used to produce very high temperatures (ca. 2500 °C) needed for certain types of welding operations. Consider the reaction to be
H₂(g) + ½O₂(g) → H₂O(g) ∆H° = -241.8 kJ/mol
What is the quantity of heat evolved, in kilojoules, when a 180 g mixture containing equal parts of H₂ and O₂ by mass is burned?
in kj
Answer:
1360kJ are evolved
Explanation:
When 1mole of H2 reacts with 1/2 moles O2 producing 1 mole of water and 241.8kJ.
To solve this question we need to find the limiting reactant knowing were added 90g of H2 and 90g of O2 as follows:
Moles H2 -Molar mass: 2g/mol-
90g H2 * (1mol / 2g) = 45 moles
Moles O2 -Molar mass: 32g/mol-
90g * (1mol / 32g) = 2.81moles
For a complete reaction of 2.81 moles of O2 are needed:
2.81 moles O2 * (1mol H2 / 1/2 mol O2) = 5.62 moles H2
As there are 45 moles, H2 is the excess reactant and O2 the limiting reactant.
As 1/2 moles O2 produce 241.8kJ, 2.81 moles will produce:
2.81 moles O2 * (241.8kJ / 1/2moles O2) =
1360kJ are evolvedThe quantity of heat evolved when 180 g mixture containing equal parts of H₂ and O₂ burned is
The equation for the combustion of hydrogen is given as:
[tex]\mathbf{H_2 + \dfrac{1}{2}O_2 \to H_2O \ \ \ \ \Delta H_r^0 = -241.8\ kJ/mol}[/tex]
Recall that:
number of moles = mass/molar mass:Since the mass of 180 g is equally shared by H₂ and O₂, then:
mass of H₂ = 90 gmass of O₂ = 90 gThe number of moles of the reactant can be determined as follows:
For H₂:
[tex]\mathbf{no \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]
[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{2.016 \ g/mol}}[/tex]
no of moles = 44.6 mol
For O₂:
[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{32 \ g/mol}}[/tex]
no of moles = 2.8 mol
Here, since O₂ has a lesser amount of mole, then O₂ is regarded as the limiting reagent here:
If 1/2 moles of O₂ produces -241.8 kJ/mol of water;
Then, the quantity of heat that will evolve when 180 g mixture containing equal parts of H₂ and O₂ burned is:
[tex]\mathbf{= \Big (\dfrac{2.81 \ mol}{\dfrac{1}{2 } \ mol }\Big) \times (-241.8 \ kJ) }[/tex]
[tex]\mathbf{= \Big (5.62\Big) \times (-241.8 \ kJ) }[/tex]
= - 1358.91 kJ
≅ - 1360 kJ
Therefore, we can conclude that the quantity of heat evolved is - 1360 kJ
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How many grams of nitrogen was combined hydrogen to form 6moles of Ammonia
A voltaic cell is constructed with an Ag/Ag half-cell and a Pb/Pb2 half-cell. The silver electrode is positive. Write the balanced half-reactions and the overall reaction. Include the phases of each reactant and product.
Answer:
Following are the chemical equation to the given question:
Explanation:
The Electrode is a silver film that is covered with such a thin coating of silver chloride, either by dipping its wire directly into silver-molten chloride, plating the wire using hydrogen peroxide, or oxidation silver in a chloride. In the given silver electrode, this anode acts as a cathode and thus reduces.
Half of the response reduction: [tex]Ag^+(aq)+e^-\rightarrow Ag(s)[/tex]
Half-effect oxidation: [tex]Pb(s)\rightarrow Pb^{2+}(aq)+2e^-[/tex]
Complete reaction: [tex]Pb(s)+2Ag^+(aq) \rightarrow Pb^{2+}(aq)+2Ag(s)[/tex]
Monomers that each contain a 5-carbon sugar, a phosphate group, and a nitrogenous base combine and form which type of polymer?
A. Amino acid
B. Carboxylic acid
C. Nucleic acid
D. Fatty acid
Answer:
The correct answer is C. Nucleic acid
Explanation:
Nucleic acids are biological polymers which play an important role in the storage and expresion of genetic information. There are two types of nucleic acids: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Both are basically composed of:
- a 5-carbon sugar: deoxyribose in DNA and ribose in RNA
- phosphate group
- a nitrogenous base: adenine, cytosine, guanine and thymine in DNA; while RNA contains adenine, cytosine, guanine and uracil.
A Chef fills out a 50mL container with 43.5g of cooking oil, What is the density of the oil?
Answer:
.87
Explanation:
p = m/V
43.5/50
.87
What does the term "basic unit of matter" refer to?
O A.
Atoms
ОВ.
Elements
O c. Molecules
Explanation:
The term "basic unit of matter "refers to atom
A Atom
3. Which of the following statements is not correct?
A. All hybridization must involve an s-orbital
B. Excitation of Carbon atom in CH, involves promotion of one is electron to the empty 2p-orbital
C. Hybridization in transition elements can take the form dsp*
D. Hybridization involves only the valence electrons and orbitals
E. None of the above
Answer:
Excitation of Carbon atom in CH, involves promotion of one is electron to the empty 2p-orbital
examples s name of thosse food items we can store for a month?
Answer:
1. Nuts
2. Canned meats and seafood
3. Dried grains
4. Dark chocolate
5. Protein powders
Consider the following data on some weak acids and weak bases:
Acid Base Ca
Name Formula Name Formula
Hydrocyanic acid HCN 4.9 x 10^-10 Ammonia NH3 1.8x 10^-5
Hypochlorous acid HCIO 3.0x10^-8 Ethylamine C2H5NH2 6.4 x 10^-4
Use this data to rank the following solutions in order of increasing pH.
Solution pH
0.1 M NaCN
0.1M C2H5NH3Br
0.1 M Nal
0.1 M KCIO
Answer:
0.1 M Nal
0.1M C2H5NH3Br
0.1M KClO
0.1M NaCN
Explanation:
The strongest acid is the one that has the higher Ka. Now, the weakest conjugate base is the conjugate base of the strongest acid and vice versa:
In the problem, we have only conjugate bases, as the HCN is the weakest acid, the strongest conjugate base is NaCN, then KClO and as last C2H5NH3Br and NaI (The conjugate base of a strong acid, HI).
The strongest base has the higher pH, that means. Thus, the rank in order of increasing pH is:
0.1 M Nal
0.1M C2H5NH3Br
0.1M KClO
0.1M NaCN
The mass of a crucible and lid is 23.422 g. After adding a sample of hydrate compound the crucible, cover, and contents weigh 24.746 g. After heating with a Bunsen burner to remove the water of hydration, the mass of the crucible, cover, and sample was 24.213 g. How many moles of water did the hydrate compound contain
Answer:
0.030 mole
Explanation:
Mass of crucible + lid = 23.422 g
Mass of crucible + lid + compound = 24.746 g
Mass of crucible + lid + compound - water = 24.213
Mass of water = Mass of crucible + lid + compound + heat
= 24.746 - 24.213
= 0.533 g
Mole of water in the hydrated compound = mass of water in the compound/molar mass of water
= 0.533/18
= 0.0296 mole = 0.030 mole
If the solvent front moves 8.0 cm and the two components in a sample being analyzed move 3.2 cm and 6.1 cm from the baseline, calculate the Rf values.
Answer:
Rf₁ = 0.40Rf₂ = 0.76Explanation:
We can calculate the Rf values by using the following formula:
Rf = Distance from the baseline / Solvent front distance
With that in mind we now proceed to calculate the Rf value for both components:
Rf₁ = 3.2 cm / 8.0 cm = 0.40Rf₂ = 6.1 cm / 8.0 cm = 0.76Choose the correct answer to make the statement true.
a. An exothermic reaction has a positive ΔH and absorbs heat from the surroundings.
b. An exothermic reaction feels warm to the touch. a positive ΔH and gives off heat to the surroundings.
c. An exothermic reaction feels warm to the touch. a negative ΔH and absorbs heat from the surroundings.
d. An exothermic reaction feels warm to the touch. a negative ΔH and gives off heat to the surroundings.
e. An exothermic reaction feels warm to the touch.
If you reacted 88.9 g of ammonia with excess oxygen, what mass of water would you expect to make? You will need to balance the equation first.
NH3(g) + O2(g) -> NO(g) + H2O(g)
Explanation:
here's the answer to your question
Which of the following is not organic compound?
a. CH4
b. H2CO3
c. CCl4
d. CH3-OH
Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant figures in calculations. Please show work!
Convert 192 grams of phosphorus pentabromide to molecules.
Convert 3.42 kilograms of table sugar (C₁₂H₂₂O₁₁) to molecules.
Answer:
1) 2.69 * 10²³ PBr₅
2) 6.02 * 10²⁴ C₁₂H₂₂O₁₁
Explanation:
Question 1)
We want to convert 192 grams of phosphorus pentabromide to molecules. Note that 192 is three significant figures.
Phosphorus pentabromide is given by PBr₅.
To convert from grams to molecules, we can convert from grams to moles first, and then from moles to molecules.
To convert from grams to moles, we will find the molar mass of PBr₅.
Since the molar mass of P is 30.974 g/mol and the molar mass of Br is 79.904 g/mol, the molar mass of PBr₅ is:
[tex](30.974)+5(79.904) = 430.494\text{ g/mol}[/tex]
And since we want to convert from grams to moles, we can write the following ratio:
[tex]\displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}[/tex]
Where grams is in the denominator, which allows us to cancel them out, leaving us with only moles.
To convert from moles to molecules, we can use the definition of the mole: a mole of one substance has 6.022 * 10²³ amount of that substance.
So, a mole of PBr₅ has 6.022 * 10²³ molecules of PBr₅. Since we want to cancel out the moles, we can write the ratio:
[tex]\displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}[/tex]
In combination, starting with 192 grams of PBr₅, we will acquire:
[tex]\displaystyle 192\text{ g PBr$_5$} \cdot \displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}[/tex]
Cancel like units:
[tex]\displaystyle = 192 \cdot \displaystyle \frac{1 }{430.494}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1}[/tex]
Multiply. Hence:
[tex]=2.6858...\times 10^{23}\text{ PBr$_5$}[/tex]
Since the final answer should have three significant digits, our final answer is:
[tex]= 2.69\times 10^{23} \text{ PBr$_5$}[/tex]
So, there are about 2.69 * 10²³ molecules of PBr₅ in 192 grams of the substance.
Question 2)
We want to convert 3.42 kilograms of table sugar (C₁₂H₂₂O₁₁) to molecules. Note that this is three significant figures.
3.42 kilograms is equivalent to 3420 grams of table sugar.
Again, we can convert from grams to moles, and then from moles to molecules.
First, we will find the molar mass of table sugar. The molar mass of carbon is 12.011 g/mol, hydrogen 1.008 g/mol, and oxygen 15.999 g/mol. Thus, the molar mass of table sugar will be:
[tex]12(12.011)+22(1.008)+11(15.999) = 342.297\text{ g/mol}[/tex]
To cancel units, we can write our ratio as:
[tex]\displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}[/tex]
With grams in the denominator.
And by definition:
[tex]\displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}[/tex]
Combining the two ratios and the starting value, we acquire:
[tex]3420 \text{ g C$_{12}$H$_{22}$O$_{11}$}\cdot \displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}[/tex]
Cancel like units:
[tex]=3420 \cdot \displaystyle \frac{1}{342.297}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1}[/tex]
Multiply:
[tex]\displaystyle = 60.1677... \times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}[/tex]
Rewrite:
[tex]\displaystyle = 6.01677... \times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}[/tex]
The resulting answer should have three significant digits. Hence:
[tex]=6.02\times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}}[/tex]
So, there are about 6.02 * 10²⁴ molecules of table sugar in 3.42 kilograms of the substance.
Answer:
2.69×10²³ molecules of PBr₅
6.02×10²⁴ molecules of C₁₂H₂₂O₁₁
Explanation:
To solve the first problem, we want to first find formula for phosphorus pentabromide, which is PBr₅. Now, we need to know the molar mass of PBr₅, which is about 430.49 g/mol. To get to molecules, we need to use Avogadro's number, which is 6.022×10²³ molecules/mol.
[tex]192g*\frac{1mol}{430.49g} *\frac{6.022*10^{23}molecules}{1mol} =2.69*10^{23} molecules[/tex]
Now, we know that there are about 2.69×10²³ molecules of PBr₅.
To solve the second problem, we need to use Avogadro's number, along with finding the molar mass of C₁₂H₂₂O₁₁, and converting kilograms to grams.
[tex]3.42 kg*\frac{1000g}{1kg} *\frac{1mol}{342.3g} *\frac{6.022*10^{23} molecules}{1mol} =6.02*10^{24} molecules[/tex]
Now, we know that there are about 6.02×10²⁴ molecules of C₁₂H₂₂O₁₁.
Write the equation for the reaction between sulfuric acid solution and solid aluminum hydroxide. (Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
Answer:
3 H₂SO₄(aq) + 2 Al(OH)₃(s) ⇒ Al₂(SO₄)₃(aq) + 6 H₂O(l)
Explanation:
Let's consider the balanced chemical equation that takes place when sulfuric acid solution and solid aluminum hydroxide react to form aluminum sulfate and water. This is a neutralization equation.
3 H₂SO₄(aq) + 2 Al(OH)₃(s) ⇒ Al₂(SO₄)₃(aq) + 6 H₂O(l)
Based on their acid and base properties, naphthalene, benzoic acid and ethyl 4-aminobenzonate mixed in ether solution could be separated using separation funnel. One student initially used HCl solution for this separation. Which compound he may find in the lower layer upon the addition of HCl solution
Answer:
See explanation
Explanation:
When using a separating funnel to separate a mixture of substances, we expect that the system will consist of an organic phase and an aqueous phase. The organic phase is usually found on top while the aqueous phase is found below.
When naphthalene, benzoic acid and ethyl 4-aminobenzonate mixed in ether solution is separated using HCl, the ethyl 4-aminobenzonate is found in the lower aqueous layer.
This is because, the ethyl 4-aminobenzonate can react with HCl to form an ionic hydrochloride salt which dissolves in the lower aqueous layer.
If the specific heat of methanol is 32.91 J/K. g. how many joules are necessary to raise the temperature of 120 g of methanol from 24 0C to 98 0C?
Answer:
[tex]Q=292240.8J=292.2kJ[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to use the general heat equation:
[tex]Q=mC(T_2-T_1)[/tex]
For us to plug the given mass, specific heat and temperature change to obtain the required heat:
[tex]Q=120g*32.91\frac{J}{g*K} (98\°C-24\°C)\\\\Q=292240.8J=292.2kJ[/tex]
Regards!
Which of the following blackbody curves is representative of stars like our Sun?
C
A
B
Answer:
its letter b
Explanation:
it represents the spectrum of stars
What is the relative formula mass of Ca(NO3)2? Ar of Ca = 40. Ar of N = 14 and Ar of O = 16.
Answer:
164
Explanation:
40+2(14)+6(16)
40+28+96=164
The formula mass of a compound is the sum of masses individual elements in that compound. The formula mass calcium nitrate (Ca (NO₃)₂) is 164 g/mol.
What is formula mass ?Formula mass of a compound is calculated from the empirical formula of the compound. The mass of one mole of an element is called its atomic mass. The mass of an element in a compound is its subscripts times the atomic mass.
The relative formula mass of a compound is the sum of the relative masses of each element present in the compound multiplied with their number of atoms.
For the compound Ca (NO₃)₂, there are one calcium atom and two nitrate groups.
atomic mass of Ca = 40 g/mol
mass of oxygen = 16 g/mol
mass of 3 O = 48 g.
mass of N = 14 g/mol
Then, mass of two nitrate groups = 2× (48 + 14) =124 g
total mass = 124 g + 40 = 164 g/mol
Therefore, the formula mass of the compound Ca (NO₃)₂ is 164 g/mol.
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A 0.334 g sample of an unknown compound occupies 245 ml at 298 K and 1.22 atm. What is the molar mass of the unknown compound.
Answer:
27.4 g/mol
Explanation:
Assuming the compound is a gas and that it behaves ideally, we can solve this problem by using the PV=nRT formula, where:
P = 1.22 atmV = 245 mL ⇒ 245 mL / 1000 = 0.245 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 298 KInputting the data:
1.22 atm * 0.245 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 KSolving for n:
n = 0.0122 molWith the calculated number of moles and given mass, we calculate the molar mass:
0.334 g / 0.0122 mol = 27.4 g/molHow many molecules of Iron(II)oxide are present in 35.2*10^-23 g of Iron (II)oxide?
Answer:
R.F.M of Iron (II) oxide :
[tex]{ \tt{ = (56 \times 2) + (16 \times 3)}} \\ = 160 \: g[/tex]
Moles :
[tex]{ \tt{ \frac{35.2 \times {10}^{ - 23} }{160} }} \\ = 2.2 \times {10}^{ - 24} \: moles[/tex]
Molecules :
[tex]{ \tt{ = 2.2 \times {10}^{ - 24} \times 6.02 \times {10}^{23} }} \\ = 1.3244 \: molecules[/tex]
The number of molecules of Iron(II) oxide present in 35.2 ×10⁻²³ g of Iron(II) oxide is equal to 2.95.
What is Avogadro's number?Avogadro’s number can be described as the proportionality constant that is used to represent the number of entities or particles in one mole of any substance. Generally, it is used to count atoms, molecules, ions, electrons, or protons, depending upon the chemical reaction or reactant and product.
The value of Avogadro’s constant can be represented as numerically approximately equal to 6.022 × 10²³ mol⁻¹.
Given, the mass of the iron oxide = 35.2 ×10⁻²³ g
The molar mass of the Iron(II) oxide, FeO = 71.84 g/mol
71.84 g of Iron (II) oxide have molecules = 6.022 × 10²³
35.2 ×10⁻²³ g of FeO have molecules = 6.022 × 10²³ × (35.2 ×10⁻²³ /71.84)
The number of molecules of FeO in a given mass = 2.95 molecules
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In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by
A) NaF
B) MgF₂
C) MgBr₂
D) AlF₃
E) AlBr₃
In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct
What factors affect the magnitude of energy of ionic crystalline solids ?For an ionic compound, there are two main terms that this magnitude depends upon: ion size and ion charge.
Ion size: the smaller the ionic radii, the shorter the internuclear distance and, therefore, the closer the ions. This factor makes lattice enthalpy increase
Ion charge: the greater the charge on ions, the greater the attractive forces between them and, therefore, the larger the lattice enthalpy.
The lattice enthalpy of AlF₃ (5215 kJ/mol) is indeed greater than that of other given solids
Therefore , In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct
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