Identifying the particles of an atom

Answer:

3.65 mol O₂

Explanation:

Step 1: RxN

2H₂O₂ → 2H₂O + O₂

Step 2: Define

Given - 7.30 mol H₂O₂

Solve - x mol O₂

Step 3: Stoichiometry

[tex]7.30 \hspace{3} mol \hspace{3} H_2O_2(\frac{1 \hspace{3} mol \hspace{3} O_2}{2 \hspace{3} mol \hspace{3} H_2O_2} )[/tex] = 3.65 mol O₂

Answer:

The theoretical yield of CaS is 6.01 g.

Explanation:

The balanced reaction is given as:

[tex]Ca+S\rightarrow CaS[/tex]

The molar mass of Ca and S is 40.08 and 32.065 g/mol respectively.

Number of moles = [tex]\frac{Mass}{Molar Mass}[/tex]

So, 7.19 g of Ca contains [tex](\frac{7.19}{40.08})[/tex] mol of Ca or 0.179 mol of Ca

Also, 2.67 g of S contains [tex](\frac{2.67}{32.065})[/tex] mol of S or 0.0833 mol of S

According to the balanced equation:

1 mol of Ca produces 1 mol of CaS

So, 0.179 mol of Ca produces 0.179 mol of CaS

According to the balanced equation:

1 mol of S produces 1 mol of CaS

So, 0.0833 mol of S produces 0.0833 mol of CaS

As the least number of mol of CaS (product) is produced from S , therefore, S is the limiting reactant.

So, thoretically, 0.0833 mol of CaS is produced.

The molar mass of CaS is 72.143 g/mol.

So, the mass of 0.0833 mol of CaS is [tex](0.0833\times 72.143)[/tex] g or 6.01 g

Hence, the theoretical yield of CaS is 6.01 g.

Answer:

B

Explanation:

I did the question before and got it right.

Answer:

1 ➡️ Cells

2 ➡️ Arteries

3 ➡️ Veins

4 ➡️ Heart

Explanation:

The parts of the electric circuit that best correspond to the heart, arteries, veins, and cells have been properly labeled.

The circulatory system involves the transportation of nutrients, oxygen and water by blood to other the parts of the body.

From the electric circuit, we see that arteries transport blood away from the heart to the other cells in the body. The veins actually return the blood back to the heart from the cells. The heart pumps the blood

The electric circuity diagram has the label 1 bulb analogous to cell, label 2 analogous to arteries, label 3 analogous to veins, and label 4 cell analogous to heart.

The electric circuit has been given as the power source and the conducting wires that allows the flow of the current in the circuit.

In the human body, the heart has been transported the oxygenated blood through the arteries to the cell and carried the deoxygenated blood from the cells back to the heart via veins.

In the circuit, the battery has been the source of the power/blood. The current has been carried from the heart to the cell/bulb through the arteries labeled, 2, and transported back to the battery via veins labeled 3.

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Answer:

380

Explanation:

Answer:

The molecular formula of the hydrocarbon is C10H10

Explanation:

Here, we are interested in finding the molecular formula for the hydrocarbon.

Since there are equal number of moles of carbon and hydrogen, then we have the molecular formula looking like;

CnHn

Kindly recall that the atomic mass of carbon is 12 a.m.u while that of hydrogen is 1 amu

so calculating the atomic mass of the compound, we have;

12(n) + 1(n) = 130.18

13n = 130.18

n = 130.18/13

n = 10.01

So the molecular formula will be C10H10

Answer:

C10H10

Explanation:

A hydrocarbon is a binary compound of carbon and hydrogen. Hence a hydrocarbon is a compound of the general formula (CH)n

Thus;

(12 + 1) n = 130.18

n= 130.18/13

n = 10

Hence the molecular formula of the compound is C10H10

Answer:

A chemical formula tells us the number of atoms of each element that is in a compound. It contains the symbols of the atoms for the elements present in the compound as well as how many there are for each element in the form of subscripts.

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Answer: The answer is 5.15x10^22

Explanation:

The formula unit present in a teaspoon of salt [tex]NaCl[/tex] having a mass of about 5.0 g is [tex]5.15 \times10^{22}[/tex] formula units.

Molar mass, also known as molecular weight, is the mass of one mole of a substance. It is calculated by summing up the atomic masses of all the atoms in a molecule. The unit of molar mass is grams per mole (g/mol).

Now, to determine the number of formula units in a teaspoon of salt (NaCl), we need to use Avogadro's number and the molar mass of NaCl.

Avogadro's number [tex](N_a)[/tex] is approximately. [tex]6.022 \times10^{23}[/tex] formula units per mole.

The molar mass of [tex]NaCl[/tex] is the sum of the atomic masses of sodium (Na) and chlorine ([tex]Cl[/tex]), which are approximately 22.99 g/mol and 35.45 g/mol, respectively.

To calculate the number of formula units in 5.0 g of [tex]NaCl[/tex], we can follow these steps:

Now, calculate the number of moles of [tex]NaCl[/tex] using its molar mass:

Moles = Mass / Molar mass

Moles = [tex]5.0 g[/tex] / [tex](22.99 g/mol + 35.45 g/mol)[/tex]

Calculate the number of formula units using Avogadro's number:

Formula units = [tex]Moles \times Avogadro's number[/tex]

Let's perform the calculation:

Molar mass of [tex]NaCl[/tex]= [tex]22.99 g/mol + 35.45 g/mol = 58.44 g/mol[/tex]

Moles of [tex]NaCl[/tex] = [tex]5.0 g[/tex] / [tex]58.44 g/mol[/tex] ≈ [tex]0.0856 mol[/tex]

Formula units = [tex]0.0856 mol \times (6.022 \times 10^{23})[/tex] formula units/mol ≈ [tex]5.15 \times10^{22}[/tex]formula units.

Therefore, there are approximately [tex]5.15 \times10^{22}[/tex] formula units in a teaspoon of salt ([tex]NaCl[/tex]) having mass [tex]5.0 g[/tex].

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Answer:

ΔU = −55.45 kJ

Explanation:

From first law of thermodynamics in chemistry, we have;

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the net heat transfer

W is the net work done

We are given;

Q = 74.6 kJ

But Q will be negative since heat is released

Thus;

ΔU = -74.6 kJ + W

We are given;

Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²

Volume before reaction; Vi = 8.2 L = 0.0082 m³

Volume after reaction; V_f = 2.8 L = 0.0028 m³

Now,

W = -P(V_f - V_i)

W = - 3546375(0.0028 - 0.0082)

W = 19.15 KJ

Thus;

ΔU = Q + W

ΔU = -74.6 kJ + 19.15 KJ =

ΔU = −55.45 kJ

Answer:

K loses one electron to CI

Explanation:

The lewis electron dot notation shows only the chemical symbol of the element surrounded by dots to represent the valence electrons.

  We have atom of K with one valence electrons

   Cl with 7 valence electrons

For an electrostatic attraction to occur, both particles must be charged. To do this, one of the species must lose an electron, and the other gains it.

This will make both species attain a stable octet;

   Hence, K will lose 1 electron and Cl will gain the electrons.

Answer:

C: K loses one electron to CI

Explanation:

I took the test and got it correct!!

Answer:

[tex]\rm Sn[/tex], [tex]\rm Se[/tex], [tex]\rm S[/tex], [tex]\rm Si[/tex].

Explanation:

The relative atomic mass of an element is numerically equal to the mass (in grams) of one mole of its atoms. This quantity can help estimate the number of moles of atoms in each of these four [tex]10.0\; \rm g[/tex] samples.  

Look up the relative atomic mass for each of these four elements (on a modern periodic table.)

The relative atomic mass of [tex]\rm Si[/tex] is (approximately) [tex]28.085[/tex]. Therefore, the each mole of silicon atoms would have a mass of approximately [tex]28.085\; \rm g[/tex]. How many moles of silicon atoms would there be in a [tex]10.0\; \rm g[/tex] sample?

Given:

Number of mole of silicon atoms in the sample: [tex]\displaystyle n(\mathrm{Si}) = \frac{m(\mathrm{Si})}{M(\mathrm{Si})} = \frac{10.0\; \rm g}{28.085\; \rm g \cdot mol^{-1}}\approx 0.356\; \rm mol[/tex].

Similarly:

[tex]\displaystyle n(\mathrm{S}) = \frac{m(\mathrm{S})}{M(\mathrm{S})} = \frac{10.0\; \rm g}{32.06\; \rm g \cdot mol^{-1}}\approx 0.312\; \rm mol[/tex].

[tex]\displaystyle n(\mathrm{Se}) = \frac{m(\mathrm{Se})}{M(\mathrm{Se})} = \frac{10.0\; \rm g}{78.971\; \rm g \cdot mol^{-1}}\approx 0.127\; \rm mol[/tex].

[tex]\displaystyle n(\mathrm{Sn}) = \frac{m(\mathrm{Sn})}{M(\mathrm{Sn})} = \frac{10.0\; \rm g}{118.710\; \rm g \cdot mol^{-1}}\approx 0.0842\; \rm mol[/tex].

Therefore, among these [tex]10.0\; \rm g[/tex] samples:

[tex]n(\mathrm{Sn}) < n(\mathrm{Se}) < n(\mathrm{S}) < n(\mathrm{Si})[/tex].

It is not a coincidence that among these four samples, the one with the fewest number of atoms corresponds to the element with the largest relative atomic mass.

Consider two elements, with molar mass [tex]M_1[/tex] and [tex]M_2[/tex] each. Assume that [tex]n_1[/tex] moles and [tex]n_2[/tex] moles of atoms of each element were selected, such that the mass of both samples is [tex]m[/tex]. That is:

[tex]m = n_1\cdot M_1[/tex].

[tex]m = n_2\cdot M_2[/tex].

Equate the right-hand side of these two equations:

[tex]n_1 \cdot M_1 = n_2\cdot M_2[/tex].

[tex]\displaystyle \frac{n_1}{n_2} = \frac{M_2}{M_1} = \frac{1/M_1}{1/M_2}[/tex].

In other words, the number of moles atoms in two equal-mass samples of two elements is inversely proportional to the molar mass of the two elements (and hence inversely proportional to the formula mass of the two elements.) That explains why in this question, the sample containing the smallest number of atoms corresponds to element with the largest relative atomic mass among those four elements.

The heat of vaporization of benzene is required.

The heat of vaporization of benzene is 33009 J/kg.

[tex]T_0[/tex] = Normal boiling point = 80.1+273.15 K

[tex]T_B[/tex] = Boiling point at given pressure = 26.1+273.15 K

[tex]R[/tex] = Gas constant = 8.314 J/mol K

[tex]P[/tex] = Pressure at given [tex]T_B[/tex] = 100 torr

[tex]\Delta H[/tex] = Heat of vaporization

From the Clausius–Clapeyron equation

[tex]\dfrac{1}{T_B}=\dfrac{1}{T_0}-\dfrac{R\ln(\dfrac{P}{P_0})}{\Delta H}\\\Rightarrow \Delta H=\dfrac{R\ln\dfrac{P}{P_0}}{\dfrac{1}{T_0}-\dfrac{1}{T_B}}\\\Rightarrow \Delta H=\dfrac{8.314\times \ln\left(\frac{100}{760}\right)}{\frac{1}{80.1+273.15}-\frac{1}{26.1+273.15}}\\\Rightarrow \Delta H=33008.99\ \text{J/kg}[/tex]

The heat of vaporization of benzene is 33009 J/kg.

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Answer:

this

Explanation:

What is the unit rate of the better deal? (Number answers only.)

LaTeX: \frac{160\:miles}{5\:gallons}\:or\:\frac{315\:miles}{9\:gallons}160 m i l e s 5 g a l l o n s o r 315 m i l e s 9 g a l l o n s

The better deal has a unit rate of [m] miles per gallon.

Answer:

u measure how much power it has

Explanation:

for example u can power a light bulb woth it if u can it was 100eg energy

Answer:

K and ba

Explanation:

Answer: K and Ba

Explanation:

Answer : The correct match is:

1 positive charge = 1 negative charge

2 positive charges = 2 negative charges

3 positive charges = 3 negative charges

Explanation :

As we now that there are three subatomic particles which are protons, electrons and neutrons.

The protons and neutrons are located inside the nucleus and electrons are located around the nucleus.

The protons are positively charged, the electrons are negatively charged and neutrons are neutral.

As we know that all the things are made up of charges and opposite charges attract to each other.

In a neutral atom, the positive charges and negative charges are balanced in an object. That means, in neutral atom the number of positive charges are equal to the negative charges.

So we can say that:

1 positive charge = 1 negative charge

2 positive charges = 2 negative charges

3 positive charges = 3 negative charges

Answer:

1 positive---- 1 negative

2 positive---- 2 negative

3 positive---- 3 negative

Explanation:

edge 2023

Answer:

molar mass of unknown gas = 1.987 g/mol

Explanation:

First, the number of moles of the unknown gas is found

Using the ideal gas equation: PV = nRT

P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹

n = PV/RT

n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)

n = 0.2109 moles

Molar mass = mass/ number of moles

molar mass = 0.419 g/ 0.2109 mols

molar mass of unknown gas = 1.987 g/mol

The molar mass of unknown gas by using ideal gas equation = 1.987 g/mol.

This equation gives the relation between pressure, volume, temperature as given below:

[tex]PV = nRT[/tex]

P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹

Substitute the above values in the above equation as follows:

n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)

n = 0.2109 moles

[tex]Molar mass = mass/ number of moles[/tex]

Calculate molar mass by using the above equation,

molar mass = 0.419 g/ 0.2109 moles

The molar mass of unknown gas = 1.987 g/mol

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Answer. After cytokinesis is completed at end of meiosis - I two haploid cells are formed.on:

Answer:

C. TWO HAPLOID CELLS ARE FORMED

Explanation:

I TOOK THE EDGUNITY TEST AND I GOT IT CORRECT

Answer:

29.8

Explanation:

The formula for volume is mass/ density, so 59/1.98. 29.8 is the answer.

Answer:

The Group 2 metals become more reactive towards the water as you go down the Group.

Explanation:

These all react with cold water with increasing vigour to give the metal hydroxide and hydrogen. ... You get less precipitate as you go down the Group because more of the hydroxide dissolves in the water. Summary of the trend in reactivity.

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Answer:

Explanation:The ether functional group consists of an oxygen atom that forms single bonds with two carbon atoms

Answer:

A=element

B=compond

C=solution

Answer:

0.75 moles of CO2

Explanation:

molar mass of CO2⇒ 44.01 g/mol

molar mass Oxygen ⇒ 31.998 g/mol

divide the mass given by molar mass of oxygen  so

48÷31.998= 1.50009376 moles of O

then you use the ration of oxygen to carbon to find the moles of CO2 which is one mole of CO2 for 2 moles of oxygen

1.50009376 moles of O ×[tex]\frac{1}{2}[/tex] = 0.75004688

with sig figs

0.75 moles of CO2

Answer:

7

Explanation:

Let x represent the number of moles of water in the hydrated salt i.e MgSO₄.xH₂O

The following data were obtained from the question:

Mass of MgSO₄.xH₂O = 1.547 g

Mass of anhydrous MgSO₄ = 0.7554 g

Mole of H₂O = x =?

Next, we shall determine the mass of water, H₂O in the hydrated salt, MgSO₄.xH₂O. This can be obtained as follow:

Mass of MgSO₄.xH₂O = 1.547 g

Mass of anhydrous MgSO₄ = 0.7554 g

Mass of H₂O =?

Mass of H₂O = (Mass of MgSO₄.xH₂O) – (Mass of anhydrous MgSO₄)

Mass of H₂O = 1.547 – 0.7554

Mass of H₂O = 0.7916 g

Finally, we shall determine the value of the x as illustrated below:

Mass of MgSO₄.xH₂O = 1.547 g

Molar mass of MgSO₄.xH₂O = 24 + 32 + (16×4) + x[(2×1) + 16]

= 24 + 32 + 64 + x(2 + 16)

= 120 + 18x

Mass of H₂O = 0.7916 g

Molar mass of xH₂O = 18x

Molar Mass of xH₂O/ Molar mass of MgSO₄.xH₂O = mass of xH₂O /Mass of MgSO₄.xH₂O

18x/ 120 + 18x = 0.7916/1.547

Cross multiply

0.7916 (120 + 18x) = 18x × 1.547

94.992 + 14.2488x = 27.846x

Collect like terms

94.992 = 27.846x – 14.2488x

94.992 = 13.5972x

Divide both side by 13.5972

x = 94.992 / 13.5972

x = 7

Thus, the formula for the hydrated salt, MgSO₄.xH₂O is MgSO₄.7H₂O

Number of moles of water, H₂O in the hydrated salt MgSO₄.7H₂O is 7.

The number of moles of attached water molecules is 7.

Mass of hydrated MgSO4 = 1.547 grams

Mass of anhydrous MgSO4 = 0.7554 grams

Number of moles of hydrated MgSO4 = 1.547 grams/120 + 18x

Number of moles of anhydrous MgSO4 = 0.7554 grams /120

Number of moles of anhydrous salt = Number of moles of hydrated salt

0.7554 grams /120 = 1.547 grams/120 + 18x

0.7554(120 + 18x) = 1.547 × 120

90.6 + 13.6x = 185.6

185.6 - 90.6 /13.6 = x

x = 7

The number of moles of attached water molecules is 7.

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Answer:

Explanation:

Caffeine is a weak base with pKb = 10.4

Kb = 10⁻¹⁰°⁴ = 3.98 x 10⁻¹¹

molecular weight of caffeine = 194.2

455 x 10⁻³ g / L = 455 x 10⁻³ / 194.2 moles / L

concentration of given solution a = 2.343 x 10⁻³ M

Let the caffeine be represented by B .

B    +   H₂O =  BH + OH⁻

a - x                   x        x  

x² / ( a - x ) = Kb

x² / ( a - x ) = 3.98 x 10⁻¹¹

x is far less than a so a -x is almost equal to a

x² = 3.98 x 10⁻¹¹ x 2.343 x 10⁻³ = 9.32  x 10⁻¹⁴

x = 3.05 x 10⁻⁷

[ OH⁻ ] = 3.05 x 10⁻⁷

pOH = - log ( 3.05 x 10⁻⁷ )

= 7 - log 3.05

= 7 - 0.484 = 6.5

pH = 14 - 6.5 = 7.5  

The pH of 455 mg/L of caffeine is 7.5

Using the formula;

Mass concentration = molar concentration × molar mass

Molar mass of caffeine = 194 g/mol

Mass concentration of caffeine = 455 mg/L

Molar concentration = Mass concentration/molar mass

Molar concentration = 455 × 10^-3g/L/194 g/mol

= 0.00235 M

Let Caffeine by depicted by the general formula BH

We can now set up the ICE table as follows;

           :B + H2O   ⇄ BH       +       OH^-

I      0.00235              0                      0

C           - x                  +x                    +x

E    0.00235  - x           x                      x

Note that water is present in large excess

Again; pKb of caffeine =10.4

Kb = Antilog[-pKb]

Kb = Antilog [-10.4]

Kb = 3.98 × 10^-11

Kb = [BH] [OH^-]/[:B]

3.98 × 10^-11 = [x] [x]/[ 0.00235  - x  ]

3.98 × 10^-11 [ 0.00235  - x  ] = [x] [x]

9.4 × 10^-14 - 3.98 × 10^-11x = x^2

x^2 + 3.98 × 10^-11x  - 9.4 × 10^-14 = 0

x = 3.1 × 10^-7 M

Recall  [BH] = [OH^-] = 3.1 × 10^-7 M

Now;

pOH = - log [OH^-]

pOH = log [3.1 × 10^-7 M]

pOH = 6.5

But;

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 6.5

pH = 7.5

The pH of 455 mg/L of caffeine is 7.5

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Missing parts

Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4.  Calculate the pH of a solution containing a caffeine concentration  of 455 mg/L.

Answer:

Explanation:

In my opinion the answer should be SO2

Answer:

a should be answer i think.