Explanation:
We are given: density of Al = 2.7g/mL
: mass of Al = 549g
[tex]\begin{gathered} \rho\text{ = }\frac{m}{V} \\ \\ \therefore\text{ V =}\frac{m}{\rho} \\ \\ \text{ =}\frac{549}{2.7} \\ \\ \text{ = 203.33mL} \end{gathered}[/tex]Answer:
The volume is 203.33mL
A volume of 56.0 mL of He at STP. a. How much weight?
We already know that number of moles is 0.0025 mol.
So if we know the volume of the gas at STP and the number of moles we can get the mass of the gas.
V = 0.056 L
1 mol = 22.4 L
Molar mass of He = 4.0026 g/mol
So to calculate the weight we need number of moles and molar mass.
weight = 0.0025 mol x 4.0026 g/mol
= 0.01 g
A solution containing 30 g of potassium chlorate in 100 mL of water at 70 °C isSupersaturated MagasaturadeUnsaturated Saturated
According to the given solubility diagram, the solubility of potassium chlorate in water at 70°C is 30g/100g of water.
It means that if at 70°C there are 30g of this salt, the solution is saturated.
The correct answer is the last choice, saturated.
I need to balance the equation __kciO3➡️__kcio4+__ KCIAnd How many moles of potassium chloride are produced from 3.90 moles of potassium chlorate ?
Explanation:
To balance the equation, we need to have the same number of atoms of each element on both sides of the equation.
So in this case:
4KCIO3 → 3KCIO4 + KCI
where:
reactants:
K - 4
Cl - 4
O - 12
Products:
K - 4
Cl - 4
O - 12
Potassium chloride is KCl and potassium chlorate is KClO3. The balanced reaction equation tells us that:
4 mol of KClO3 produces 1 mol of KCl, we know it because of the equation:
4 KCIO3 → 3 KCIO4 + 1 KCI
So we can do a rule of 3 to find out the quantity in moles produced when we have 3.90 moles of KClO3:
4 mol KClO3 --- 1 mol KCl
3.9 moles KClO3 --- x
4x = 1*3.9
x = 3.9/4
x = 0.975 moles
Answer: 4 KCIO3 → 3 KCIO4 + KCI and 0.975 moles.
Answer question number 11. The question is in the image.
Answer
Explanation
Ethane has a molecular formula of C₂H₆. Its structural formula is CH₃-CH₃ or
What's the volume of 0.2 mol of nitrogen at 20oC and 0.7 atm?
Explanation:
We have to determine the volume of 0.2 mol of nitrogen at 20 °C and 0.7 atm. The ideal gas law describes this situation.
P * V = n * R * T
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant and T is the temperature.
We already know these some of these values.
P = 0.7 atm
n = 0.2 mol
R = 0.082 atm * L/(mol*K)
T = 20 ° C = 293.15 K
We can use the ideal gas law to determine the volume of the sample.
P * V = n * R * T
V = n * R * T/P
V = 0.2 mol * 0.082 atm*L/(mol*K) * 293.15 K/(0.7 atm)
V = 6.86 L
Answer: The volume is 6.86 L.
Infer how resting membrane potential would be affected if the membrane were only permeable to potassium ions.
Option 1) If the cell were only permeable to potassium ions there would be a drastic change in membrane potential to a more positive value as potassium enters the cell
Option 2) If the cell were only permeable to potassium ions there would be a slight change in membrane potential to a more negative value as potassium exits the cell.
Option 3) If the cell were only permeable to potassium ions there would be a drastic change in membrane potential to a more negative value as potassium leaves the cell.
According to the research, the correct answer is Option 2. If the cell were only permeable to potassium ions there would be a slight change in membrane potential to a more negative value as potassium exits the cell.
What is resting membrane potential?It refers to the potential difference on either side of the membrane of a resting cell, that is, already repolarized where the channels are much more permeable to potassium than to sodium.
In this sense, it is determined by the ion concentration and membrane permeability where the inside of the cell has a more negative charge (potassium ions) relative to the outside (sodium ions).
Therefore, we can conclude that resting membrane potential causes the cell membrane inside to become negative because it has sodium ions towards the outside of the cell and potassium ions towards the inside, thus the correct answer is Option 2.
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what is the mass of aluminum used if o.1 moles of aluminum chloride is produced by aluminum chloride reaction?
The mass of aluminum used if 0.1 moles of aluminum chloride is produced by aluminum chloride reaction is 13.33 g.
given that :
number of moles of aluminum chloride = 0.1 moles
The balanced chemical reaction as:
2Al + 3Cl₂ -----> 2AlCl₃
2 moles of moles of AlCl₃ produced by 2 moles of Al
therefor 0.1 moles of AlCl₃ produced by 0.1 moles of Al
number of moles of Al = 0.1 mol
Mass of Al = number of moles × molar mass
= 0.1 × 133.34 g/ mol
= 13.33 g
Thus, The mass of aluminum used if 0.1 moles of aluminum chloride is produced by aluminum chloride reaction is 13.33 g.
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Consider a 2190
lb automobile clocked by law‑enforcement radar at a speed of 85.5 mph (miles per hour). If the position of the car is known to within 5.0 ft at the time of the measurement, what is the uncertainty in the velocity of the car?
The uncertainty in the velocity of the car is 7.85×10⁻³¹
Velocity is the directional speed of an object n motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time
Here given data is
Mass of automobiles = 2190 lb = 993.26kg
Speed of car = v = 85.5 mph = 38.2 m/s
Uncertainty in the position of car = Δx = 5 f = 1.52m
the uncertainty in the velocity of the car is determined by heinsberg uncertainty principle
ΔPΔx ≥ h/4π
mΔvΔx ≥ h/4π
Δv ≥ h/4π×mΔx
Δv ≥ 6.626×10⁻³⁴/4π× 993.26kg× 1.52m
Δv ≥ 7.85×10⁻³¹
The uncertainty in the velocity of the car is 7.85×10⁻³¹
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REACTION; C5H12 + 8O2 5CO2 + 6H2OWhen 35.5 L of C5H12 are consumed in this reaction what volume of CO2 can be produced in liters?
Using the STP (standard temperature and pressure), we can solve this problem. First, let's find the number of moles produce by 35.5 L of C5H12.
Remember that the standard temperature is 0 °C which is the same that 273 K (kelvin) and for pressure is 1 atm and the constant of ideal gas is 0.082 atm*L/(mol*K)
Let's use the formula of an ideal gas to find the number of moles:
[tex]\begin{gathered} PV=\text{nRT,} \\ n=\frac{PV}{RT}, \\ n=\frac{\text{1 atm }\cdot35.5\text{ L}}{0.082\frac{atm\cdot L}{mol\cdot K}\cdot273K}, \\ n(C_5H_{12})=1.58\text{6 moles} \end{gathered}[/tex]Now, using this number of moles, we can find the number of moles produce for CO2 and we can find its volume.
You can see that in the reaction 1 mol of C5H12 produces 5 moles of CO2, so the calculation to find the number of moles of CO2, would be:
[tex]1.586molC_5H_{12}\cdot\frac{5molCO_2}{1molC_5H_{12}_{}}=7.93molCO_2.[/tex]The next and final step is clear V (volume) from the initial formula and replaces the value of moles for CO2, like this:
[tex]\begin{gathered} V=\frac{nRT}{P}, \\ V=\frac{7.93\text{ mol}\cdot0.082\frac{atm\cdot L}{mol\cdot K}\cdot273K}{1\text{ atm}}, \\ V(CO_2)=177.52\text{ L.} \end{gathered}[/tex]So, 25,5 L of C5H12 will produce 177.52L of CO2.
5. You have 2 Liters of a 5 M Solution of H3PO4. How many grams are present?
Answer
979.94 grams
Explanation
Given:
Volume of solution = 2 L
Molarity = 5 M
What to find:
The mass in grams of H3PO4 present in the solution.
Step-by-step solution:
The first step is to calculate the moles of H3PO4 present using the molarity formula below:
[tex]\begin{gathered} Molarity=\frac{Moles}{Volume\text{ }in\text{ }L} \\ \\ \Rightarrow Moles=Molarity\times Volume\text{ }in\text{ }L \end{gathered}[/tex]Put the values of the given parameters into the formula to get the moles:
[tex]Moles=5M\times2L=10\text{ }mol[/tex]The moles of H3PO4 present in the solution = 10 mol.
Therefore, the mass present can be determined using the mole formula:
[tex]\begin{gathered} Moles=\frac{Mass\text{ }in\text{ }grams}{Molar\text{ }mass} \\ \\ \Rightarrow Mass\text{ }in\text{ }grams=Moles\times Molar\text{ }mass \end{gathered}[/tex]From the periodic table, the molar mass of H3PO4 can be known as 97.994 g/mol.
So putting moles = 10 mol and molar mass = 97.994 g/mol, then the mass is:
[tex]Mass=10\text{ }mol\times97.994\text{ }g\text{/}mol=979.94\text{ }grams[/tex]Therefore, the mass in grams of H3PO4 present in the solution = 979.94 grams.
AcellusPerform the following operationand express the answer inscientific notation.7.15x103 x 6.10x10-5[ ? ]x10!??Coefficient (green)Exponent (yellow)Enter
ANSWER
[tex]4.3615\cdot10^{-1}[/tex]STEP-BY-STEP EXPLANATION:
You are given the below expression from the question provided
[tex]7.15\cdot10^3\text{ x 6.10 }\cdot10^{-5}[/tex][tex]\begin{gathered} (7.15\text{ x 6.10) }\cdot10^{3\text{ + (-5)}}\text{ } \\ (43.615)\cdot10^{3-\text{ 5}} \\ 43.615\cdot10^{-2} \\ 4.3615\cdot10^{1\text{ -2}} \\ 4.3615\cdot10^{-1} \end{gathered}[/tex]The molar mass of NO2 is
46.01 g/mol.
What is the mass
of 3.45 moles NO2?
[ ? ] g NO₂
The mass of nitrogen dioxide present in 3.45 moles is equal to 158.73 g.
What is a mole?A mole can be defined as a scientific unit that is used to determine the huge number of quantities of molecules, atoms, ions, etc. The mass of the 1 mole of an element is called atomic mass and the mass of one mole of any compound is called molar mass.
The number of particles in 1 mole was found to be equal to 6.023 × 10 ²³ per mole which is known as Avogadro’s constant.
Given, the number of moles of nitrogen dioxide = 3.45 mol
The molar mass of NO₂ = 46.01 g/mol
It means that 1 mole of NO₂ has mass = 46.01 g
Then 3.45 moles of NO₂ will have mass = 3.45 × 46.01 = 158.73 g
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Based on the equation below, which of the following statements best describes what should be observed as the reaction takes place?Zn(s) + 2HCl(aq) to ZnCl2 (aq)+ H2(g)Solid zinc dissolvingBubbles forming from chlorine gas productionBubbles forming from hydrogen gas productionChlorine forming a solid precipitate
ANSWER
Bubbles forming from hydrogen gas production
EXPLANATION
Given that;
[tex]\text{ Zn}_{(s)}\text{ + 2HCl}_{(aq)}\rightarrow\text{ ZnCl}_{2(aq)}\text{ + H}_{2(g)}[/tex]One of the method used in the preparation of hydrogen gas is the reaction of metals with acid
The above reaction is one of the method of producing hydrogen gas
When Zn reacts with HCl, bubbles formed which is as a result of hydrogen production.
Therefore, the correct answer is Bubbles forming from hydrogen gas production
Mateo has sprayed luminol on a bed spread but sees no evidence of any substances. What mistake is he MOST likely making?
A.
He is spraying too much luminol.
B.
He is diluting the substances with luminol.
C.
He is leaving the lights on in the room.
D.
He is making it too dark in the room.
Answer:b
Explanation:
because i said so
We know that 50.5 grams of sucrose is equal to 0.148 moles of sucrose, we need to use this information to find the molarity (the unit of measuring concentration). We can calculate the molarity (M) or concentration of a solution in moles / L if we know the volume of the solution in liters (L).Let’s say you mixed your 0.148 moles of sucrose into 0.0012 L of solution.What is the molarity of this solution?
123.33mol/L
Explanations:
The formula for calculating the molarity of a solution is expressed as:
[tex]\begin{gathered} \text{Molarity}=\frac{moles\text{ of solute}}{litres\text{ of solution}} \\ \end{gathered}[/tex]Given the following parameters
• Moles, of sucrose (solute) = 0.148 moles
• Litres ,of solution = 0.0012L
Substitute the given parameters into the formula;
[tex]\begin{gathered} \text{Molarity}=\frac{0.148\text{moles}}{0.0012L} \\ \text{Molarity}=123.33\text{mol/L} \end{gathered}[/tex]Therefore the molarity of this solution is 123.33mol/L.
3Ca + 2AICI3 —> 3CaCI2 + 2AIIf you react 100, grams of aluminum chloride, AICI3, with excess calcium, how many grams of calcium chloride,CaCI2, are produced ?
Firstly, we need to convert the 100 g of AlCl₃ to number of moles, using:
[tex]M_{AlCl_3}=\frac{m_{AlCl_3}}{n_{AlCl_{3}}}[/tex]The molar weight of AlCl₃ is calculating consulting the atomic weights:
[tex]\begin{gathered} M_{AlCl_3}=1\cdot M_{Al}+3\cdot M_{Cl} \\ M_{AlCl_3}=(1\cdot26.982+3\cdot35.453)g/mol \\ M_{AlCl_3}=(26.982+106.359)g/mol \\ M_{AlCl_3}=133.341g/mol \end{gathered}[/tex]Thus:
[tex]\begin{gathered} M_{AlCl_3}=\frac{m_{AlCl_3}}{n_{AlCl_3}} \\ n_{AlCl_3}=\frac{m_{AlCl_3}}{M_{AlCl_3}}=\frac{100g}{133.341g/mol}=0.749957\ldots mol \end{gathered}[/tex]For each 2 moles of AlCl₃, we produce 3 moles of CaCl₂. So:
2 mol AlCl₃ --- 3 mol CaCl₂
0.749957... mol AlCl₃ --- x
[tex]\begin{gathered} \frac{2}{0.749957\ldots mol}=\frac{3}{x} \\ 2x=2.24987\ldots mol \\ x=1.12494\ldots mol \end{gathered}[/tex]Now, we convert back to grams, but now we need the molar weight of CaCl₂:
[tex]\begin{gathered} M_{CaCl_2}=1\cdot M_{Ca}+2\cdot M_{Cl} \\ M_{CaCl_2}=(1\cdot40.078+2\cdot35.453)g/mol \\ M_{CaCl_2}=(40.078+70.906)g/mol \\ M_{CaCl_2}=110.984g/mol \end{gathered}[/tex]Thus:
[tex]\begin{gathered} M_{CaCl_{2}}=\frac{m_{CaCl_2}}{n_{CaCl_{2}}} \\ m_{CaCl_2}=M_{CaCl_2}\cdot n_{CaCl_2}=110.984g/mol\cdot1.12494\ldots mol=124.850\ldots g\approx125g \end{gathered}[/tex]So, 100 g of AlCl₃ will produce approximately 125 g of CaCl₂.
Name of element or compoundFormula of element or compound CoCarbon Iron Au KCl CS2Silicon
Answer:
• Formula: ,Co,, Name: ,Cobalt,.
,• Formula: ,C,, Name: ,Carbon,.
,• Formula: ,Fe,, Name:, Iron,.
,• Formula: ,Au,, Name: ,Gold,.
,• Formula: ,KCl,, Name: ,Potassium chloride,.
,• Formula: ,CS2,, Name: ,Carbon disulfide,.
,• Formula: ,Si,, Name: ,Silicon,.
Explanation:
• Co is an element.
,• Carbon is an element.
,• Iron is an element.
,• Au is an element.
,• KCl is a compound with K and Cl. The name has the positive part first (Potassium) and the second part has the name of the negative part (chloride).
,• CS2 is a compound with C and 2 S. The name has the number of each element, carbon is 1 so it is just Carbon, and there are 2 sulfurs (disulfide).
,• Silicon is an element.
The pressure exerted by 1.30 mol of gas in a 13.0 L container at 295 K is
Answer:
The pressure of the gas is 2.42atm.
Explanation:
The given information from the exercise is:
- Number of moles (n): 1.30 moles
- Volume (V): 13.0L
- Temperature (T): 295K
With the Ideal Gases Law formula we can calculate the pressure (P) of the gas, by replacing the values of n, V and T:
[tex]\begin{gathered} P*V=n*R*T \\ P*13.0L=1.30mol*0.082\frac{atm*L}{mol*K}*295K \\ P*13.0L=31.45atm*L \\ P=\frac{31.45atm*L}{13.0L} \\ P=2.42atm \end{gathered}[/tex]So, the pressure of the gas is 2.42atm.
I need help with this. What do I add to make the after reaction
According to the given reaction, one molecule of H2 reacts with one molecule of C2H4 to produce one molecule of C2H6.
In the reactants, there are 7 molecules of C2H4 and 3 molecules of H2, it means that 3 molecules of C2H4 react with the 3 molecules of H2 to form 3 molecules of C2H6, and 4 molecules of C2H4 will remain.
It means that you have to add 3 molecules of C2H6 and 4 molecules of C2H4.
I got 0.009 and I was wondering if that is right
Explanation:
The equation of formation of NO₂
1/2 N₂ (g) + O₂ (g) <----> NO₂ (g)
The equilibrium for the above reaction can be calculated using this equation:
Kc = [NO₂]/([N₂]^(1/2)*[O₂]) where the concentrations are the equilibrium concentrations.
According to the graph the concentration in the equilibrium of each compound is:
[NO₂] = 6 mol/L
[N₂] = 10 mol/L
[O₂] = 20 mol/L
If we replace these values in the given equation:
Kc = [NO₂]/([N₂]^(1/2)*[O₂])
Kc = (6 mol/L)/[(10 mol/L)^(1/2)*(20 mol/L)]
Kc = 0.095
Answer: Kc = 0.095
The reaction mixture, at a certain temperature, contained concentrations of0.31 M of NH3, 0.85 M of N2 and 0.031 M of H2 when it reached equilibrium.Calculate Keq at this temperature.
ANSWER
Keq is 3798.42 or 3.798 x 10^3
EXPLANATION
Given that;
The concentration of NH3 is 0.31M
The concentration of N2 is 0.85M
The concentration of H2 is 0.031M
Follow the steps below to find the chemical equilibrium constant of the reaction
Step 1; Write the balanced equation of the reaction
[tex]\text{ N}_{2(g)}\text{ + 3H}_{2(g)}\text{ }\rightleftarrows\text{ 2NH}_{3(g)}[/tex]Step 2; Write the chemical equilibrium constant for the reaction
[tex]\begin{gathered} \text{ K}_{eq}\text{ = }\frac{K_f}{\text{ K}_b} \\ \text{ Where} \\ \text{ K}_f\text{ = }\lbrack NH_3\rbrack^2 \\ \text{ K}_b\text{ = }\lbrack N_2\rbrack\text{ }\lbrack H_2\rbrack^3 \end{gathered}[/tex]Step 3; Substitute the kf and kb into the formula above
[tex]\text{ K}_{eq}\text{ = }\frac{\lbrack NH_3\rbrack^2}{\lbrack H_2\rbrack^3\times\lbrack N_2\rbrack}[/tex]Recall, that
[NH3] = 0.31M
[H2] = 0.031M
[N2] = 0.85M
[tex]\begin{gathered} \text{ K}_{eq}\text{ = }\frac{(0.31)^2}{(0.85)\times\text{ \lparen0.031\rparen}^3} \\ \\ \text{ K}_{eq}\text{ = }\frac{\text{ 0.0961}}{0.85\text{ }\times\text{ 0.0000298}} \\ \\ \text{ K}_{eq}\text{ = }\frac{\text{ 0.0961}}{0.0000253} \\ \\ \text{ K}_{eq}\text{ = 3798.42 or 3.798 }\times\text{ 10}^3 \end{gathered}[/tex]Therefore, Keq is 3798.42 or 3.798 x 10^3
explain which mgcl2 and alcl3 is stronger bonding?
AlCl3 has a higher charge density which makes the polarization to be high. Therefore AlCl3 has a higher covalent character (stronger bonding).
Covalent character and cationic size are indirectly related to each other. When the size of a cation is smaller, the covalent character would be higher.
In MgCl2, the cation is Mg^2+ and in AlCl3, the cation is Al^3+. The size of Mg^2+ is larger when compared to the size of Al^3+. That is why the covalent character of AlCl3 is larger.
Recalling that a beaker of water is two dimensional, what is the three dimensional shape of the micelle?
The three dimensional shape of the micelle is sphere
micelle is a loosely bound aggregation of several hundreds of atoms, ions which are electrically charged atoms or molecules, which results in forming a colloidal particle. Micelles are basically aggregate or supramolecular assembly of surfactant of amphipathic lipids molecules dispersed in a liquid, forming a colloidal suspension which is also known to be associated with colloidal system.
Micelles are generally composed of hydrophobic and hydrophilic components that are assembled into nanosized spherical, ellipsoid, cylindrical, or uni lamellar structures. Usually the heads are hydrophilic in nature and the tail part is hydrophobic in nature. Therefore their structure in three dimension will be spherical.
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What would the molecular formula be if potassium and chlorine reacted to form a neutral compound?A. KClB. PClC. K2ClD. KCl3
Explanation:
The metal potassium and nonmetal chlorine will react together to form potassium chloride.
The formula for the cation potassium is K+. The anion chloride is Cl-. If we combine them to form a neutral compound we get:
Answer: A. KCl
Please help!! 100pts and brainliest!
Rb+
Fe2+
H-
Al3+
Cl-
O2-
Using the ions above, write the formulas and give the names for 5 possible ionic compounds involving these ions.
Using the ions, the formulas and give the names for 5 possible ionic compounds involving these ions are :
1) Rubidium hydride - RbH
2) Iron(II)chloride - FeCl₂
3) Aluminum chloride - AlCl₃
4) Rubidium oxide - Rb₂O
5) Iron(II)oxide - FeO
1) Rubidium hydride - RbH
In Rubidium hydride : Rubidium forms Rb⁺ ion. hydrogen form H⁻ ion. they combined and form ionic compound. The one positive charge on rubidium neutralizes the one negative on hydrogen.
2) Iron(II)chloride - FeCl₂
In iron(II)chloride : iron form Fe⁺² and Chlorine form Cl⁻ ion. +2 charge are neutralizes by the two chlorine negative ion.
3) Aluminum chloride - AlCl₃
In this Aluminum form Al³⁺ and chlorine forms Cl⁻ and form ionic compound, Aluminum chloride.
4) Rubidium oxide - Rb₂O
In Rubidium oxide : Rubidium for Rb⁺ and oxygen form O²⁻.
5) Iron(II)oxide - FeO
In the ionic compound iron(II)oxide : Iron forms Fe²⁺ ion and oxygen forms O²⁻ ion.
Thus, Using the ions, the formulas and give the names for 5 possible ionic compounds involving these ions are :
1) Rubidium hydride - RbH
2) Iron(II)chloride - FeCl₂
3) Aluminum chloride - AlCl₃
4) Rubidium oxide - Rb₂O
5) Iron(II)oxide - FeO
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The equation for the synthesis of ammonia is N2 + 3 H2 -> 2 NH3. How many moles of H2 I needed to produce 6 mol of NH3? A. 4 B. 6 C. 8 D. 9
The equation for the synthesis of ammonia is:
N₂ + 3 H₂ ---> 2 NH₃
We have to find the number of moles of H₂ needed to produce 6 mol of NH₃. To do that we will use the equation that we were given.
We can read the equation of the reaction as a recipe. When 1 mol of N₂ reacts with 3 moles of H₂, 2 moles of NH₃ are produced. So the relationship between H₂ and NH₃ is that 3 moles of H₂ are needed to produce 2 moles of NH₃. We wiill use that relationship to solve the problem.
6 moles of NH₃ * 3 moles of H₂/(2 moles of NH₃) = 9 moles of H₂
Answer: D) 9 moles of H₂ are needed.
The charge of an election is
The charge of an electron is negative (-1).
If the freezing point of a soda is – 4.5 ºC and you leave it in your car overnight when the temperature dips to 25 ºF, will the soda freeze? (Pay close attention to the units!)
Explanation:
We leave a soda in the car overnight and the temperature dips to 25 °F. To see if it will freeze we have to convert the °F to °C.
T(°C) = ( T(°F) - 32 ) *5/9
T(°C) = (25 - 32) *5/9
T(°C) = -3.9 °C
Since -3.9 °C is a temperature greater than the freezing point (-4.5 °C) the soda won't freeze.
Answer: The soda won't freeze.
In the equation: I2 + 5Cl2 + 6 H2O —>2HIO3 + 10HCL A. Has iodine been oxidized or has it been reduced B. Has chlorine been oxidized or has it been reduced
Answer
A - Iodine is being oxidized
B - Cl2 is being reduced
Explanation
[tex]I_{2(s)}^0\text{ + 5Cl}_{2(g)}^0\text{ + 6H}_2O_{(l)}\rightarrow2HI^VO_{3(aq)}\text{ + HCl}_{(aq)}^{-I}[/tex]7. How many moles of NH3 can be produced from5.00 moles of H, according to the following equation?N2 + 3 H2 → 2 NH3a. 0.67 molb. 2.00 molc. 3.33 mold. 7.50 mol
Answer:
[tex]C\text{ : 3.33 mol}[/tex]Explanation:
Here, we want to get the number of moles of ammonia that could be produced from 5 moles of hydrogen
From the balanced equation of reaction, 3 moles of hydrogen produced 2 moles of ammonia
5 moles of hydrogen will produce x moles of ammonia
Mathematically:
[tex]\begin{gathered} 5\times2\text{ = 3}\times x \\ x\text{ = }\frac{10}{3} \\ x\text{ = 3.33 mol} \end{gathered}[/tex]