Answer:
oxidizer: one used to support the combustion of a rocket propellant
word: a single distinct meaningful element of speech or writing
Explanation:
In a reaction, there are multiple ways to discuss the amount of compound produced. Identify the definition of each term related to the reaction yield. The comparison of the experimental amount of product to the calculated amount The calculated amount of product under ideal conditions The amount of product experimentally produced
Answer: The comparison of the experimental amount of product to the calculated amount : Percentage yield
The calculated amount of product under ideal conditions : Theoretical yield
The amount of product experimentally produced : Experimental yield or actual yield
Explanation:
Theoretical yield is to determine how much of the product should be formed based on stoichiometry of the given reaction. It is the maximum amount of product that could be formed from the given amounts of reactants.
Experimental yield or actual yield is the amount of product that is actually formed when the reaction is carried out in the laboratory.
Percentage yield is the ratio of the experimental yield to the theoretical yield, expressed as a percentage.
What determines the change of an ion formed by an atom
Answer:
Ions are formed when the number of protons in an atom does not equal the number of electrons. If more protons are present, the ion is positive and is known as a cation; if more electrons are present, the ion is negative and referred to as an anion. Ions are highly reactive species.
Explanation:
Answer:
An atom becomes charged when the number of protons does not equal the number of electrons. For example, if an element has six protons but only five electrons, the net charge of the element is +1. Conversely, if an element has six protons but seven electrons, then the net charge of the element is -1.
Explanation:
2.Your culture media recipe says you need to add 2.625g of Sodium Bicarbonate per liter of media. You realize you only have a 7.5% solution of Sodium Bicarbonate you can use. What volume of that solution can you add, while making your media, to have the necessary mass of NaHCO3
Answer:
35mL of the 7.5% solution are required to meake the media
Explanation:
To prepare the culture media there are required 2.625g of sodium bicarbonate. We have a 7.5% solution of sodium bicarbonate (That is, 7.5g of sodium bicarbonate in 100mL of solution). Thus, to obtain 2.625g we need:
2.625g Sodium bicarbonate * (100mL / 7.5g) =
35mL of the 7.5% solution are required to meake the mediaANSWER QUICK !!!
How many grams of hydrogen gas will be produced if you start with 15.0 grams of zinc and an excess amount of HCI?
balanced equation:
Zn + 2HCI = ZnCl2 + H2
Answer:
uhh
ExplanatioUHHHHHHHHHHHHHHHHHHH
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How many moles are in 25g of NaCI?
Nacl = (. 23+35. 5)
= 58.5g
.
1 mol of Nacl = 58.5g
X mol Of Nacl = 25g
X Mol of Na Cl =25 ÷ 58.5
X mol of Nacl = 0.4 mol
Write a balanced equation for the reaction of sodium phosphide with potassium chloride.
Answer:
Na3PO4 + KCl = NaCl + K3PO4
Explanation:
You have to know the Chemical Formula for Sodium phosphide, = Na3PO4 and Potassium chloride. which is = KCI
Which of the following is a list of the minimum amount of data needed for determining the molar enthalpy of solution of KCl(s) in pure H2O(1) ? (Assume that the KCl(aq) has the same specific heat capacity as pure water and that the initial temperatures of the KCl(s) and the water are the same.)
(A) Mass of KCl(s), initial temperature of the water, and final temperature of the solution
(B) Mass of H20, initial temperature of the water, and final temperature of the solution
(C) Mass of KCl(s), mass of H20, initial temperature of the water, and final temperature of the solution
(D) Mass of KCl(s), mass of H20, initial temperature of the water, final temperature of the solution, and atmospheric pressure
Answer:
(C) Mass of KCl(s), mass of H20, initial temperature of the water, and final temperature of the solution
Explanation:
molar enthalpy of solution of KCl(s) is heat evolved or absorbed when one mole of KCl is dissolved in water to make pure solution . The heat evolved or absorbed can be calculated by the following relation.
Q = msΔt where m is mass of solution or water , s is specific heat and Δt is change in temperature of water .
So data required is mass of water or solution , initial and final temperature of solution , specific heat of water is known .
Now to know molar heat , we require mass of solute or KCl dissolved to know heat heat absorbed or evolved by dissolution of one mole of solute .
The correct option is option C.
To determine the molar enthalpy of the solution of [tex]Kcl(s)[/tex] is pure [tex]H_2O(l)[/tex]
We need to know that the,
grams of [tex]Kcl(s)[/tex] grams of [tex]H_2O[/tex] Initial temperature of [tex]H_2O[/tex] The final temperature of the solution.Learn More:https://brainly.com/question/25758173
is microorganisms bigger or particles bigger
Answer:
particles bigger
Explanation:
Even when compared with animal cells, microorganisms become smaller. They are about 1/10 the size of a typical human cell. Therefore, compared with human-sized animal cells, the size of bacteria (such as bacterial cells) will be the size of a cat or puppy. The size of the virus is about 1/10 of that of other microorganisms such as bacteria.
Hope it helps. If not, I'm so sorry.
Answer:
particles are bigger than microorganism
So I saw this question: If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + NaI → PbI2 + NaNO3
I saw an answer that question but I need one part explained. The answer balanced the equation, but I do not know how they did that or why. They added a 2 to both sides and I do not know where it came from or why it made it balanced.
mass of PbI₂ = 27.6606 g
Further explanationGiven
Pb(NO₃)₂ + NaI → PbI₂ + NaNO₃
28.0 grams of Pb(NO₃)₂ react with 18.0 grams of NaI
Required
mass of PbI₂
Solution
Balanced equation
Pb(NO₃)₂ + 2NaI → PbI₂ + 2NaNO₃
The principle of a balanced reaction is the number of atoms in the reactants = the number of atoms in the product
mol Pb(NO₃)₂ :
= 28 : 331,2 g/mol
= 0.0845
mol NaI :
= 18 : 149,89 g/mol
= 0.12
Limiting reactant : mol : coefficient
Pb(NO₃)₂ : 0.0845 : 1 = 0.0845
NaI : 0.12 : 2 = 0.06
NaI limiting reactant (smaller ratio)
mol PbI₂ based on NaI
= 1/2 x 0.12 = 0.06
Mass PbI₂ :
= 0.06 x 461,01 g/mol
= 27.6606 g
In his experiment, Balmer introduced hydrogen gas in a discharged tube in order to excite electrons on higher energy levels. Then, as the electrons dropped down to lower energy levels, photons were emitted by the atoms. What is the energy of a photon emitted by a electron dropping from level n
Answer:
2.56 eV
Explanation:
Using the formula;
ΔE= -RH(1/n^2final - 1/n^2initial)
RH = -2.18 * 10-18 J
nfinal = 2
ninitial =1
Substituting values;
ΔE= -2.18 * 10-18 J(1/4^2 - 1/2^2)
ΔE= -2.18 * 10-18 J(0.0625 - 0.25)
ΔE= 4.09 * 10^-19 J
To convert to eV
4.09 * 10^-19 J/1.6 * 10^-19
=2.56 eV
In addition to liquid precipitation, which are present when freezing rain occurs?
ice pellets
snowflakes
cold surfaces
warm temperatures
Answer:
I believe the correct answer is cold surfaces.
Explanation:
Answer:
C
Explanation:
Which MOLECULES do cells need to get in order to function properly?
Answer:
they get animal cells
Explanation:
If the flour mixture is then taken and put into the oven for about 20 minutes, it turns into something very tasty. What kind of change has happened now?
Answer:
physical change
Explanation:
i got the same question sorry for the late response but that's the answer good luck
HELPPp now plsss I really need help !!!!!
Answer:
The Forces that form each star would be your answer :)
Draw the electron configuration for a neutral atom of nitrogen.
Answer:
1s22s22p3.
Explanation:
Electronic configuration of a neutral atom is 1s22s22p3.
Please see the image attached
Neutral atom of nitrogen will have equal number of proton and electron i.e equal to 7. 7 electron of the nitrogen are placed into the s and p orbitals in the ground state.
Name the intermolecular force that corresponds to: an attraction between a partially positive region in one molecule and a partially negative region in another molecule. an attraction between two temporarily polarized molecules. an attraction between a negatively charged particle and a partially positive region in a molecule. an attraction between a partially positive hydrogen atom in a molecule and a partially negative and highly electronegative atom on another molecule.
Answer:
an attraction between a partially positive region in one molecule and a partially negative region in another molecule....dipole-dipole interaction
an attraction between two temporarily polarized molecules...dispersion forces
an attraction between a negatively charged particle and a partially positive region in a molecule....ion dipole interaction
an attraction between a partially positive hydrogen atom in a molecule and a partially negative and highly electronegative atom on another molecule....Hydrogen bonding
Explanation:
Some molecules have permanent dipole. As a result of this, the positive part of one molecule may attract the negative part of the other molecule leading to dipole-dipole interaction.
Dispersion forces occur in all molecules and is as a result of temporary polarization of a molecule due to instantaneous dipole–induced dipole attractions.
If a charged particle is attracted by a dipole in a molecule, we call it ion-dipole interaction.
When hydrogen is bonded to a highly electronegative element, the positive end of the dipole is on hydrogen while the negative end of the dipole is on the electronegative element. Molecular associations often result from this permanent dipole and is called hydrogen bonding, e.g, HF.
[10 pts]Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to approximately 20 wt.% V at room temperature. Determine the concentration in weight percent of V that must be added to iron to yield a unit cell edge length of 0.289 nm.
Answer:
The answer is below
Explanation:
The unit cell edge length (a) = 0.298 nm = 0.289 * 10⁻⁷ cm
The unit volume (V) = a³ = (0.289 * 10⁻⁷ cm)³ = 24 * 10⁻²⁴ cm³
There are 2 atoms per cell, hence N = 2. Also, [tex]n_a=avogadro\ constant=6.02*10^{23}\ mol^{-1}[/tex]
[tex]atomic\ weight\ of\ iron(A_f)=55.85,atomic\ weight\ of\ vanadium(A_V)=50.94,[/tex]
[tex]density\ of\ iron(\rho_f)=7.87,density\ of\ vanadium(\rho_v)=6.1.\\\\V=\frac{nA_{av}}{n_a\rho_{av}} \\\\A_{av}=\frac{100}{C_v/A_v+C_f/A_f} \\\\\rho_{av}=\frac{100}{C_v/\rho_v+C_f/\rho_f} \\\\C_v=concentration\ of\ vanadium,C_f=concentration\ of\ iron.\\\\V=\frac{nA_{av}}{n_a\rho_{av}}=\frac{n*\frac{100}{C_v/A_v+C_f/A_f}}{n_a*\frac{100}{C_v/\rho_v+C_f/\rho_f}}\\\\[/tex]
[tex]24*10^{-24}=\frac{2*\frac{100}{C_v/50.94+C_f/55.85}}{6.02*10^{23}*\frac{100}{C_v/6.1+C_f/7.87}}\\\\7.2=\frac{C_v/6.1+C_f/7.87}{C_v/50.94+C_f/55.85}\ \ \ (1) \\\\Also:\\\\C_v+C_f=100\%\ \ \ (2)\\\\\\solving\ equation\ 1\ and\ 2\ simultaneously\ gives:\\\\C_v=10\% \ and\ C_f=90\%\\\\[/tex]
Draw a Lewis structure for SO2 in which all atoms have a formal charge of zero. Explicitly showing the zero charges is optional. Do not consider ringed structures.
Answer:
See explanation and image attached
Explanation:
"A formal charge (FC) is the charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity" (chemlibretext).
We can obtain the formal charge from the formula;
Formal Charge = [number of valence electrons on atom] – [non-bonded electrons + 1/2number of bonds].
A structure in which SO2 has a zero formal charge is attached to this answer.
Image credit: Chemtopper
A series of dilute NaCl solutions are prepared starting with an initial stock solution of 0.100 M NaCl. Solution A is prepared by pipeting 10 mL of the stock solution into a 250-mL volumetric flask and diluting to volume. Solution B is prepared by pipeting 25 mL of solution A into a 100-mL volumetric flask and diluting to volume. Solution C is prepared by pipeting 20 mL of solution B into a 500-mL volumetric flask and diluting to volume. What is the molar concentration of NaCl in solutions A, B and C
Answer:
Solution A: 0.00400M
Solution B: 0.00400M
Solution C: 4.00x10⁻⁵M
Explanation:
Solution A is diluting the 0.100M NaCl from 10mL to 250mL. That is:
250mL / 10mL = 25 times.
That means molar concentration of sln A is:
0.100M / 25 = 0.00400M
Solution B is obtained diluting 25mL to 100mL:
100mL / 25mL = 4 times
0.00400M / 4 times = 0.00100M
And solution C is obtained diluting the solution C from 20mL to 500mL:
500mL / 20mL = 25 times
Solution C:
0.00100M / 25 times = 4.00x10⁻⁵M
The formula for serial dilution can be used to obtain the molarity of solution A, B , C.
For solution AM1V1 = M2V2
M2 = 0.100 M × 10 mL/250-mL
M2 = 0.004 M
For solution BM1V1 = M2V2
M2 = 0.004 M × 25 mL/100-mL
M2 = 0.001 M
For solution CM1V1 = M2V2
M2 = 0.001 M × 20 mL/500-mL
M2 = 0.00004 M
Learn more about serial dilution: https://brainly.com/question/2167827
The heat of vaporization of water at the normal boiling point, 373.2 K, is 40.66 kJ/mol. The molar heat capacity of liquid water is 75.37 J K-1 mol-1 and that of gaseous water is 36.4 J K-1 mol-1. Assume that these values are independent of temperature. What is the heat of vaporization of water at 300.2 K?
Answer:
[tex]\Delta _{vap}H(300.2K)=43,658\frac{J}{mol}=43.66\frac{kJ}{mol}[/tex]
Explanation:
Hello!
In this case, according to the Kirchhoff's law for the enthalpy change, it is possible to compute the heat of vaporization at 300.2 K by considering the following thermodynamic route:
[tex]\Delta _{vap}H(300.2K)=Cp_{liq}(T_b-T\°)+\Delta _{vap}H\°+Cp_{vap}(T-T_B)[/tex]
Whereas the first term stands for the effect of taking the liquid from 298.15 K to 373.15 K, the second term stands for the standard enthalpy of vaporization and the last term that of the vapor from the boiling point to 300.2 K; thus we plug in to obtain:
[tex]\Delta _{vap}H(300.2K)=75.37\frac{J}{mol*K} (373.2K-298.15K)+40,660\frac{J}{mol} +36.4\frac{J}{mol*K}(300.2K-373.2K)\\\\\Delta _{vap}H(300.2K)=43,658\frac{J}{mol}=43.66\frac{kJ}{mol}[/tex]
Best regards!
The combustion of gasoline produces carbon dioxide and water. assume gasoline to be pure octane (c8h18) and calculate how many kilograms of carbon dioxide are added to the atmosphere per 3.8 kg of octane burned. ( hint: begin by writing a balanced equation for the combustion reaction.) express your answer using two significant figures.
Answer:
11.73 Kg of CO₂
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2C₈H₁₈ + 25O₂ —> 16CO₂ + 18H₂O
Next, we shall determine the mass of C₈H₁₈ that reacted and the mass of CO₂ produced from the balanced equation. This can be obtained as follow:
Molar mass of C₈H₁₈ = (12×8) + (18×1)
= 96 + 18
= 114 g/mol
Mass of C₈H₁₈ from the balanced equation = 2 × 114 = 228 g
Convert 228 g to kg.
1000 g = 1 Kg
Therefore,
228 g = 228 g × 1 Kg / 1000 g
228 g = 0.228 Kg
Molar mass of CO₂ = 12 + (16×2)
= 12 + 32
= 44 g/mol
Mass of CO₂ from the balanced equation = 16 × 44 = 704 g
Convert 704 g to Kg
1000 g = 1 Kg
Therefore,
704 g = 704 g × 1 Kg / 1000 g
704 g = 0.704 Kg
SUMMARY:
From the balanced equation above,
0.228 Kg of C₈H₁₈ reacted to produce 0.704 Kg of CO₂.
Finally, we shall determine the mass of carbon dioxide, CO₂, that will be produced by the reaction of 3.8 kg of octane, C₈H₁₈. This can be obtained as follow:
From the balanced equation above,
0.228 Kg of C₈H₁₈ reacted to produce 0.704 Kg of CO₂.
Therefore, 3.8 kg of C₈H₁₈ will react to produce = (3.8 × 0.704) / 0.228 = 11.73 Kg of CO₂
Thus, 11.73 Kg of CO₂ is added to the atmosphere per 3.8 kg of C₈H₁₈.
If snails and crayfish die, what happens to the other species?
Answer:
Other species will slowly die, because the species that eat snails and crayfish will not have anything to eat, then the thing that eats them will not have anything. This process will go on and on. So long story short populations will decrease, then off.
A certain molecular compound has a solubility in acetonitrile of at . Calculate the greatest mass of that could be dissolved in of acetonitrile at this temperature. You may assume the volume of the solution doesn't change as the is dissolved. Be sure your answer has the correct unit symbol and number of significant digits.
Answer:
Amount = 2250 g
Explanation:
Note: This question is incomplete and lacks each and every data to solve. However, I have found the similar question on the internet and will be solving this question for you. Cheers!
Data Missing:
Organic Compound O
Solubility = 0.225 g/ml
Temperature = 10°C
Calculate the greatest mass of O that can be dissolved in 10.0 L acetonitrile at 10°C temperature.
Solution:
As we know the volume of the acetonitrile = 10.0 L
but first, we need to convert this into mL
We know that,
1 L contains 1000 mL
So,
10 L will contain = 10 x 1000 mL
10 L = 10000 mL
We, can write, 10000 mL as:
1 x [tex]10^{4}[/tex] mL
Now, the formula to calculate the greatest amount of O that can be dissolved in 1 x [tex]10^{4}[/tex] mL acetonitrile at 10°C temperature is:
Amount = Solubility x volume
Here,
Solubility = 0.225 g/ml
and
Volume = 1 x [tex]10^{4}[/tex] mL
Plugging in the values, we get:
Amount = 0.225 g/ml x 1 x [tex]10^{4}[/tex] mL
Amount = 0.225 x [tex]10^{4}[/tex] g
Amount = 2250 g
The reaction that occurs in a Breathalyzer, a device used to determine the alcohol level in a person's bloodstream, is given below. If the rate of appearance of Cr2(SO4)3 is 1.24 mol/min at a particular moment, what is the rate of disappearance of C2H6O at that moment
Answer:
The rate of disappearance of C₂H₆O = 2.46 mol/min
Explanation:
The equation of the reaction is given below:
2 K₂Cr₂O₇ + 8 H₂SO₄ + 3 C₂H₆O → 2 Cr₂(SO₄)₃ + 2 K₂SO₄ + 11 H₂O
From the equation of the reaction, 3 moles of C₂H₆O is used when 2 moles of Cr₂(SO₄)₃ are produced, therefore, the mole ratio of C₂H₆O to Cr₂(SO₄)₃ is 3:2.
The rate of appearance of Cr₂(SO₄)₃ in that particular moment is given 1.64 mol/min. This would than means that C₂H₆O must be used up at a rate which is approximately equal to their mole ratios. Thus, the rate of of the disappearance of C₂H₆O can be calculated from the mole ratio of Cr₂(SO₄)₃ and C₂H₆O.
Rate of disappearance of C₂H₆O = 1.64 mol/min of Cr₂(SO₄)₃ * 3 moles of C₂H₆O / 2 moles of Cr₂(SO₄)₃
Rate of disappearance of C₂H₆O = 2.46 mol/min of C₂H₆O
Therefore, the rate of disappearance of C₂H₆O = 2.46 mol/min
When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2.7 C lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride (NH CI) are dissolved in the same mass of X, the freezing point of the solution is 9.9 °C lower than the freezing point of pure X.
Required:
Calculate the van't Hoff factor for ammonium chloride in X.
Answer:
1.62
Explanation:
From the given information:
number of moles of benzamide [tex]=\dfrac{70.4 \ g}{121.14 \ g/mol}[/tex]
= 0.58 mole
The molality = [tex]\dfrac{mass \ of \ solute (i.e. \ benzamide )}{mass \ of \ solvent }[/tex]
[tex]= \dfrac{0.58 }{0.85 }[/tex]
= 0.6837
Using the formula:
[tex]\mathbf {dT = l \times k_f \times m}[/tex]
where;
dT = freezing point = 27
l = Van't Hoff factor = 1
kf = freezing constant of the solvent
∴
2.7 °C = 1 × kf × 0.6837 m
kf = 2.7 °C/ 0.6837m
kf = 3.949 °C/m
number of moles of NH4Cl = [tex]\dfrac{70.4 \ g}{53.491 \ g /mol}[/tex]
= 1.316 mol
The molality = [tex]\dfrac{1.316 \ mol}{0.85 \ kg}[/tex]
= 1.5484
Thus;
the above kf value is used in determining the Van't Hoff factor for NH4Cl
i.e.
9.9 = l × 3.949 × 1.5484 m
[tex]l = \dfrac{9.9}{3.949 \times 1.5484 \ m}[/tex]
l = 1.62
What is the electron configuration of phosphorus
1s², 2s², 2p⁶, 3s², 3p⁵
Further explanationGiven
Phosphorus element
Required
The electron configuration
Solution
The energy level is expressed in the form of electron configurations.
Writing electron configurations starts from the lowest to the highest sub-shell energy level. There are 4 sub-shells in the shell of an atom, namely s, p, d and f.
Charging electrons in the sub shell uses the following sequence:
1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.
Electron configurations are based on the number of electrons in the atomic number.
The atomic number of the phosphorus is 15, so the number of electrons is 15.
The configuration:
1s², 2s², 2p⁶, 3s², 3p⁵
The camel stores the fat tristearin (C57H110O6) in its hump. As well as being a source of energy, the fat is also a source of water because, when it is used, the reaction 2 C57H110O6(g) + 163 O2(g) 114 CO2(g) + 110 H2O(ℓ) takes place. What mass of water is available from 1.6 pound of this fat? Answer in units of g.
Answer:
8.1 × 10² g
Explanation:
Step 1: Write the balanced equation
2 C₅₇H₁₁₀O₆ + 163 O₂ ⇒ 114 CO₂ + 110 H₂O
Step 2: Convert 1.6 lb of C₅₇H₁₁₀O₆ to g
We will use the conversion factor 1 lb = 453.592 g.
1.6 lb × 453.592 g/1 lb = 7.3 × 10² g
Step 3: Calculate the moles corresponding to 7.3 × 10² g of C₅₇H₁₁₀O₆
The molar mass of C₅₇H₁₁₀O₆ is 890.83 g/mol.
7.3 × 10² g × 1 mol/890.83 g = 0.82 mol
Step 4: Calculate the moles of water produced from 0.82 moles of C₅₇H₁₁₀O₆
The molar ratio of C₅₇H₁₁₀O₆ to H₂O is 2:110. The moles of H₂O produced are 110/2 × 0.82 mol = 45 mol
Step 5: Calculate the mass corresponding to 45 moles of H₂O
The molar mass of H₂O is 18.02 g/mol.
45 mol × 18.02 g/mol = 8.1 × 10² g
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Answer:
Silver Lake Picnic Area, Or C
Explanation:
______ is required for making a scientific inquiry
Calculate the number of oxygen atoms in a 50.0g sample of scheelite CaWO4
Answer:
0.696 atoms of oxygen
Explanation:
We'll begin by calculating the number of mole in 50 g of scheelite CaWO₄. This can be obtained as follow:
Mass of CaWO₄ = 50 g
Molar mass of CaWO₄ = 40 + 184 + (4×16)
= 40 + 184 + 64
= 288 g/mol
Mole of CaWO₄ =?
Mole = mass / Molar mass
Mole of CaWO₄ = 50 / 288
Mole of CaWO₄ = 0.174 mole
Finally, we shall determine the number of oxygen atom in 50 g (i.e 0.174 mole) of CaWO₄. This can be obtained as follow:
1 mole of CaWO₄ contains 4 atoms of oxygen.
Therefore, 0.174 mole of CaWO₄ will contain = 0.696 atoms of oxygen.
Thus, 50 g (i.e 0.174 mole) of CaWO₄ contains 0.696 atoms of oxygen.