Answer:
-241 kJ/mol
Explanation:
Let's consider the reaction of hydrogen with excess oxygen to form water.
2 H₂ + O₂ ⟶ 2 H₂O
When 2.16g of hydrogen reacts with excess oxygen, 258 kJ of heat are released, that is, Q = -258 kJ. Considering that the molar mass of hydrogen is 2.02 g/mol, the change of enthalpy associated with the reaction of 1.00 mol of hydrogen gas is:
ΔH° = -258 kJ/2.16 g × (2.02 g/1.00 mol) = -241 kJ/mol
Determine each type of reaction. 2 C 2 H 2 ( g ) + 5 O 2 ( g ) ⟶ 4 C O 2 ( g ) + 2 H 2 O ( l ) 2CX2HX2(g)+5OX2(g)⟶4COX2(g)+2HX2O(l) Choose... N H 4 N O 3 ( s ) ⟶ N 2 O ( g ) + 2 H 2 O ( l ) NHX4NOX3(s)⟶NX2O(g)+2HX2O(l) Choose... C O ( g ) + 2 H 2 ( g ) ⟶ C H 3 O H ( l ) CO(g)+2HX2(g)⟶CHX3OH(l) Choose... 2 F e ( s ) + 6 H C l ( a q ) ⟶ 2 F e C l 3 ( a q ) + 3 H 2 ( g ) 2Fe(s)+6HCl(aq)⟶2FeClX3(aq)+3HX2(g) Choose... C a C l 2 ( a q ) + N a 2 C O 3 ( a q ) ⟶ 2 N a C l ( a q ) + C a C O 3 ( s ) CaClX2(aq)+NaX2COX3(aq)⟶2NaCl(aq)+CaCOX3(s) Choose...
Answer:
2 C 2 H 2 ( g ) + 5 O 2 ( g ) ⟶ 4 C O 2 ( g ) + 2 H 2 O ( l )- combustion reaction
N H 4 N O 3 ( s ) ⟶ N 2 O ( g ) + 2 H 2 O ( l )- decomposition reaction
C O ( g ) + 2 H 2 ( g ) ⟶ C H 3 O H ( l ) - combination reaction
2 F e ( s ) + 6 H C l ( a q ) ⟶ 2 F e C l 3 ( a q ) + 3 H 2 ( g )- Redox reaction
C a C l 2 ( a q ) + N a 2 C O 3 ( a q ) ⟶ 2 N a C l ( a q ) + C a C O 3 ( s )- double displacement reaction
Explanation:
We can determine the type of reaction by considering the reactants and products.
Combustion is a reaction between a substance and oxygen which produces heat and light. The first reaction is the equation for the combustion of ethyne.
A decomposition reaction is one in which a single reactant breaks down to form products. The second reaction is the decomposition of ammonium nitrate.
A combination reaction is said to occur when two elements or compounds react to form a single product. The third reaction is the combination of carbon dioxide and methane to form methanol.
An oxidation-reduction reaction is a reaction in which there is a change in oxidation number of species from left to right of the chemical reaction equation. The fourth reaction is the oxidation of iron (0 to +3 state) and reduction of hydrogen (+1 to 0 state).
A double displacement reaction is a reaction in which ions exchange partners from left to right in the reaction equation. The fifth reaction is a double displacement reaction. Both Na^+ and Ca^2+ exchanged partners from left to right of the reaction equation.
Reactions are the formation of the products from the reactant. The types of reactions are combustion, decomposition, combination, Redox and double displacement.
What are the types of reactions?The reaction is a chemical change in the properties of the reactant that forms the products. It can be of various types based on the formation of the product.
The first reaction is combustion as the reactants react and use oxygen to form heat, carbon dioxide and water. The combustion reaction of ethyne can be shown as,
[tex]\rm 2 C _{2} H _{2} ( g ) + 5 O _{2} ( g ) \rightarrow 4 C O _{2} ( g ) + 2 H _{2} O ( l )[/tex]
The second reaction is decomposition in which a single reactant decomposes to form two or more products. The decomposition of ammonium nitrate can be shown as,
[tex]\rm N H _{4} N O _{3} ( s ) \rightarrow N _{2} O ( g ) + 2 H _{2} O ( l )[/tex]
The third reaction is a combination reaction in which two compound or elements combines to form one product. The combination reaction between carbon monoxide and hydrogen to form methanol can be shown as,
[tex]\rm C O ( g ) + 2 H _{2} ( g ) \rightarrow C H _{3} O H ( l )[/tex]
The fourth reaction is redox and includes the oxidation and the reduction of the species of the reaction. In the reaction, iron undergoes oxidation and hydrogen reduction. The redox reaction can be shown as,
[tex]\rm 2 F e ( s ) + 6 H C l ( a q ) \rightarrow 2 F e C l _{3} ( a q ) + 3 H _{2} ( g )[/tex]
The fifth reaction is a double displacement reaction in which the calcium and sodium interchange their position in the product formation. The reaction can be shown as,
[tex]\rm C a C l _{2} ( a q ) + N a _{2} C O _{3} ( a q ) \rightarrow 2 N a C l ( a q ) + C a C O _{3} ( s )[/tex]
Therefore, the type of reactions is 1. combustion, 2. decomposition, 3. combination, 4. redox and 5. double displacement.
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3. Which of the following molecules would want except to have a nonpolar covalent bond
Answer:
polar bonds are caused by different kind of atoms, because almost every atoms have different powers to attract electrons.
the answer will be the two same atoms, F2
20 POINTS PLEASE ANSWER ASAP!!!!
Why is calcium (Ca) in group 2, period 4 on the periodic table?
A Calcium, like all group 2 elements, is nonreactive and a gas at room temperature.
B Calcium, like all period 4 elements, is nonreactive and a gas at room temperature,
C Calcium, like all group 2 elements, is reactive and a solid at room temperature.
D. Calcium, like all period 4 elements, is reactive and a solid at room temperature.
Calcium (Ca) is in group 2 and period 4 on the periodic table be because Calcium has 2 valence electrons and 4 electron shell. Thus, calcium is a metal like all other group 2 element.
The correct answer to the question is Option C. Calcium, like all group 2 elements, is reactive and a solid at room temperature.
Calcium is a group 2 element majorly because it has 2 valence electrons. It is also in period 4 because it has 4 electron shells.
Being a group 2 element, calcium is a solid at room temperature and also reactive. All elements in the group 2 are metals.
There are other elements in period 4 which are not solid. For example krypton is an element in period 4 and it is a gas and not reactive.
From the above information, we can conclude that the correct answer to the question is:
Option C. Calcium, like all group 2 elements, is reactive and a solid at room temperature.
See attached image
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PLS PLS PLS PLS PLS PLS PLS help ASAP!!!!! Scientists call all of the compounds that contain carbon and are found in living things Organic because ________.
WILL DO BRAINLIEST!
SOMEONE PLZ HELP!!!!
Answer:
4.22mL
Explanation:
V=m/d
v= 18.45g/4.37g/mL
Which profile best shows the topography alone line AD
What is the momentum of a 1kg ball moving at 5m/s?
Answer:
5Ns
momentum= mass *velocity
=1*5
=5Ns
Measurements show that unknown compound X has the following composition: element mass % carbon 41.0% hydrogen 4.58% oxygen 54.6% Write the empirical chemical formula of X.
Answer:
CHO
Explanation:
Carbon = 41%, Hydrogen = 4.58%, oxygen = 54.6%
Step 1:
Divide through by their respective relative atomic masses
41/ 12, 4.58/1, 54.6/16
3.41 4.58 3.41
Step 2:
Divide by the lowest ratio:
3.41/3.41, 4.58/3.41, 3.41/3.41
1, 1, 1
Hence the empirical formula is CHO
Answer:
The empirical formula of X is C3H4O3.
Explanation:
Consider the diagram below.
What does C represent?
A) enthalpy of reaction
B) activation energy
C) activated complex
D) energy of the reactants
Answer:
A) enthalpy of reaction
Explanation:
The region C signifies the enthalpy of reaction.
This diagram is the energy profile of an endothermic reaction. In such reaction, heat is absorbed from the surrounding. At the end of the reaction, the heat of product is lesser than that of the reactants.
Enthalpy changes are heat changes accompanying a physical and chemical change. An enthalpy is the difference between the sum of the heat contents of products and sum of the heat contents of reactants.it is indeed A) enthalpy of reaction
Josh heated a certain amount of blue copper sulfate crystals to get 2.1 g of white copper sulfate powder and 1.4 g of water. What is most likely the mass of the blue copper sulfate that he heated and why?
Answer: The mass of blue copper sulfate is 3.5 g
Explanation:
Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.
This also means that total mass on the reactant side must be equal to the total mass on the product side.
The chemical equation for the heating of copper sulfate crystals is:
Let the mass of blue copper sulfate be 'x' grams
We are given:
Mass of copper sulfate powder = 2.1 grams
Mass of water = 1.4 grams
Total mass on reactant side = x
Total mass on product side = (2.1 + 1.4) g
So, by applying law of conservation of mass, we get:
Hence, the mass of blue copper sulfate is 3.5 grams
A major component of gasoline is octane (C8H18). When octane is burned in air, it chemically reacts with oxygen gas (O2) to produce carbon dioxide CO2 and water H2O. What mass of carbon dioxide is produced by the reaction of 7.58 g of octane? Please explain the answer to me like I'm five, I want to understand but the content makes no sense.
Answer:
Mass = 23.232 g
Explanation:
Given data:
Mass of C₈H₁₈ = 7.58 g
Mass of CO₂ produced = ?
Solution:
Chemical equation:
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
Number of moles of octane:
Number of moles = mass/molar mass
Number of moles = 7.58 g/ 114.23 g/mol
Number of moles = 0.066 mol
Now we will compare the moles of CO₂ with octane from balance chemical equation.
C₈H₁₈ : CO₂
2 : 16
0.066 : 16/2×0.066 = 0.528
Mass of CO₂ produced:
Mass = number of moles × molar mass
Mass = 0.528 mol × 44 g/mol
Mass = 23.232 g
Bromine is found above iodine in Group 17 of the periodic table. If an ion formed by bromine has a charge of 1-, what is the charge on an ion
formed by iodine?
A. -7
B. -2
C. -1
D. +1
Answer:
Explanation:
Iodine is the most flexible of the atoms in Group 17 of the periodic table. It can have more than 1 charge, especially when combined with oxygen. The answer that you are expected to give, I think is C. Iodine's dominant charge is - 1
The atomic notation for a particular atom of boron is ' B. The
atomic number is while the mass number is
Answer:
5;11
Explanation:
Problem:
[Ar]4s2
Identify the period (p) , group (g) and valence electrons block of the element
Answer:
it is Calcium (Ca)
4th period, 2nd group, 2 valence electrons
Is nuclear fission exothermic or endothermic? Explain your answer.
Answer:
Exothermic
Explanation:
Nuclear fission means splitting, so there is a lot of energy being released.
Use the Rydberg Equation to calculate the energy in Joules of the transition between n = 7 and n = 3 for the hydrogen atom. Find the frequency in Hz of this transition if the wavelength is 1000nm.
Answer:
The energy of each transition is approximately [tex]1.98\times 10^{-19}\; \rm J[/tex].
The frequency of photons released in such transitions is approximately [tex]3.00\times 10^{14}\; \rm Hz[/tex].
Explanation:
The Rydberg Equation gives the wavelength (in vacuum) of photons released when the electron of a hydrogen atom transitions from one main energy level to a lower one.
Let [tex]\lambda_\text{vac}[/tex] denote the wavelength of the photon released when measured in vacuum.Let [tex]R_\text{H}[/tex] denote the Rydberg constant for hydrogen. [tex]R_\text{H} \approx 1.09678 \times 10^{7}\; \rm m^{-1}[/tex].Let [tex]n_1[/tex] and [tex]n_2[/tex] denote the principal quantum number of the initial and final main energy level of that electron. (Both [tex]n_1\![/tex] and [tex]n_2\![/tex] should be positive integers; [tex]n_1 > n_2[/tex].)The Rydberg Equation gives the following relation:
[tex]\displaystyle \frac{1}{\lambda_\text{vac}} = R_\text{H} \cdot \left(\frac{1}{{n_2}^2}} -\frac{1}{{n_1}^2}\right)[/tex].
Rearrange to obtain and expression for [tex]\lambda_\text{vac}[/tex]:
[tex]\displaystyle \lambda_\text{vac} = \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)}[/tex].
In this question, [tex]n_1 = 7[/tex] while [tex]n_2 = 3[/tex]. Therefore:
[tex]\begin{aligned} \lambda_\text{vac} &= \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)} \\ &\approx \frac{1}{\displaystyle 1.09678 \times 10^{7}\; \rm m^{-1} \cdot \left(\frac{1}{3^2} - \frac{1}{7^2}\right)} \approx 1.0 \times 10^{-6}\; \rm m \end{aligned}[/tex].
Note, that [tex]1.0\times 10^{-6}\; \rm m[/tex] is equivalent to [tex]1000\; \rm nm[/tex]. That is: [tex]1.0\times 10^{-6}\; \rm m = 1000\; \rm nm[/tex].
Look up the speed of light in vacuum: [tex]c \approx 3.00\times 10^{8}\; \rm m \cdot s^{-1}[/tex]. Calculate the frequency of this photon:
[tex]\begin{aligned} f &= \frac{c}{\lambda_\text{vac}} \\ &\approx \frac{3.00\times 10^{8}\; \rm m\cdot s^{-1}}{1.0\times 10^{-6}\; \rm m} \approx 3.00 \times 10^{14}\; \rm Hz\end{aligned}[/tex].
Let [tex]h[/tex] represent Planck constant. The energy of a photon of wavelength [tex]f[/tex] would be [tex]E = h \cdot f[/tex].
Look up the Planck constant: [tex]h \approx 6.62607 \times 10^{-34}\; \rm J \cdot s[/tex]. With a frequency of [tex]3.00\times 10^{14}\; \rm Hz[/tex] ([tex]1\; \rm Hz = 1\; \rm s^{-1}[/tex],) the energy of each photon released in this transition would be:
[tex]\begin{aligned}E &= h \cdot f \\ &\approx 6.62607 \times 10^{-34}\; \rm J\cdot s^{-1} \times 3.00 \times 10^{14}\; \rm s^{-1} \\ &\approx 1.98 \times 10^{-19}\; \rm J\end{aligned}[/tex].
The energy of the transition between n = 7 and n = 3 is 1.96 × 10^-19 J while the frequency is 3 × 10^14 Hz.
Using the Rydberg Equation for energy;
ΔE = -RH(1/n^2final - 1/n^2initial)
Given that;
nfinal = 3
ninitial = 7
RH = 2.18 × 10^-18 J
ΔE = - 2.18 × 10^-18(1/3^2 - 1/7^2)
ΔE = - 2.18 × 10^-18(0.11 - 0.02)
ΔE = - 1.96 × 10^-19 J
For the second part;
Since the wavelength is 1000nm, we have;
λ = 1000nm
c = 3 × 10^8 m/s
f = ?
c = λf
f = c/λ
f = 3 × 10^8 m/s/1000 × 10^-9 m
f = 3 × 10^8 m/s/ 1 × 10^-6 m
f = 3 × 10^14 Hz
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A gas has a volume of 300 mL and a pressure of 2 atm. What volume will the gas occupy when the pressure is
increased to 7 atm (total)?
Answer:
The answer is 85.71 mLExplanation:
The new volume can be found by using the formula for Boyle's law which is
[tex]P_1V_1 = P_2V_2[/tex]
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we are finding the new volume
[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]
We have
[tex]V_2 = \frac{300 \times 2}{7} = \frac{600}{7} \\ = 85.714285...[/tex]
We have the final answer as
85.71 mLHope this helps you
Which 2 main body systems work alongside the digestive system?
ou are given a solution containing a pair of enantiomers (A and B). Careful measurements show that the solution contains 98% A and 2% B. What is the ee of this solution
Complete Question
You are given a solution containing a pair of enantiomers (A and B). Careful measurements show that the solution contains 98% A and 2% B. What is the enantiomeric excess?
Answer:
The value is [tex]k = 96 %[/tex]
Explanation:
From the question we are told that
The percentage of enantiomer A is A = 98%
The percentage of enantiomer B is B = 2%
Generally the enantiomeric excess is mathematically represented as
[tex]k = \frac{A -B}{A+B} * 100[/tex]
=> [tex]k = \frac{98 -2}{98+2} * 100[/tex]
=> [tex]k = 96 %[/tex]
The molar mass of gallium (Ga) is 69.72 g/mol.
Calculate the number of atoms in a 27.2 mg sample of Ga.
Write your answer in scientific notation using three significant figures.
atoms Ga
Answer:
2.35 x 10²⁰ atoms Ga
Explanation:
After converting from mg to g, use the molar mass as the unit converter to convert to moles. Then using Avogadro's number, 6.022 x 10²³ convert from moles to atoms of Ga.
[tex]27.2mgGa*\frac{1g}{1000mg} *\frac{1 mol Ga}{69.72gGa} *\frac{6.022*10^2^3 atoms Ga}{1 molGa} = 2.349 * 10^2^0 atoms Ga[/tex]
Then round to 3 significant figures = 2.35 x 10²⁰ atoms Ga.
The number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms
StoichiometryFrom the question, we are to calculate the number of atoms in a 27.2 mg sample of Ga.
First, we will determine the number of moles of Ga present
Using the formula,
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass} [/tex]
Mass = 27.2 mg = 0.0272 g
Molar mass = 69.72 g/mol
Then,
[tex]Number\ of\ moles \ of\ Ga = \frac{0.0272}{69.72} [/tex]
[tex]Number\ of\ moles \ of\ Ga = [/tex] 0.000390132 moles
Now, for the number of atoms present
From the formula
Number of atoms = Number of moles × Avogadro's constant
Then,
Number of Ga atoms = 0.000390132 × 6.022×10²³
Number of Ga atoms = 2.35 × 10²⁰ atoms
Hence, the number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms
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Which is one way that minerals crystallize from materials dissolved in water?
from the air
from solutions that evaporate
from hot water solutions when water boils
from the soil
Answer:
the second answer its science behind it
Answer:
b
Explanation:
Question 5 of 5
Which two phrases describe the nature of an electromagnetic force?
O A. Acts only when objects touch each other
B. Produced by interactions between magnetic objects
O c. Not a fundamental force of nature
O D. Produced by interactions between electrically charged objects
Answer:
I think it's A and D
Explanation:
I'm not sure if it's right
Answer:
The answer is B and D
Explanation:
trust fr
Peroxyacylnitrate (PAN) is one of the components
of smog. It is a compound of C, H, N, and O.
Determine the percent composition of oxygen and
the empirical formula from the following percent
composition by mass: 19.8 percent C,
2.50 percent
H, 11.6 percent N. What is its molecular formula
given that its molar mass is about 120 g?
C – 19,9%, H – 2,2%, N – 11,6%, O – x%
[tex]M=120\frac{g}{mol}[/tex]
1 percentage
The entire molecule is 100% and all the components of the compound add up to 100%.
100% - 19,9% - 2,5% - 11,6% = 66,1%
The compound contains 66,1% oxygen.
2 molar masses
[tex]M_{C}=12,01\frac{g}{mol}[/tex]
[tex]M_{H}=1,008\frac{g}{mol}[/tex]
[tex]M_{O}=15,999\frac{g}{mol}[/tex]
[tex]M_{N}=14,007\frac{g}{mol}[/tex]
3 masses
The compound has a molar mass of 120g/mol. So one molecule weighs 120 g. To find out how much the percentage of a component weighs, you have to calculate it using the molar mass.
carbon
19,8% of 120g
[tex]m=120g*0,198\\m=23,76g[/tex]
One molecule contains 23,76g of carbon.
hydrogen
2,5% of 120g
[tex]m=120g*0,025\\m=3g[/tex]
One molecule contains 3g of hydrogen.
oxygen
66,1% of 120g
[tex]m=120g*0,661\\m=79,32g[/tex]
One molecule contains 79,32g of oxygen.
nitrogen
11,6% of 120g
[tex]m=120g*0,0,116\\m=13,92g[/tex]
One molecule contains 13,92g of nitrogen.
4 amount of substance
carbon
[tex]n=\frac{23,76g}{12,01\frac{g}{mol} }\\n=1,98mol[/tex]
The compound contains about 2 moles of carbon.
hydrogen
[tex]n=\frac{3g}{1\frac{g}{mol} }\\n=3mol[/tex]
The compound contains about 3 moles of hydrogen.
oxygen
[tex]n=\frac{79,32g}{15,999\frac{g}{mol} }\\n=4,96mol[/tex]
The compound contains about 5 moles of oxygen.
nitrogen
[tex]n=\frac{13,92g}{14,007\frac{g}{mol} }\\n=0,99mol[/tex]
The compound contains about 1 moles of nitrogen.
5. molecular formula
The formula results from the ratio of the amounts of substance.
[tex]n_{C} :n_{H} :n_{O} :n_{N} =2:3:5:1\\C_{2}H_{3}NO_{5}[/tex]
The molecular formula of the given compound is C₂H₃NO₅, and percent composition of oxygen in it is 66.1%.
How do we calculate mass from % composition?Mass of any composition of any compound will be calculated by using the below formula as:
Mass of component = (% composition)×(mass of compound) / 100
Given mass of compound = 120g/mol
Total composition of compound (100%) = Percent composition of all components
% composition of oxygen = 100 - (19.8 + 2.50 + 11.6) = 66.1%
Moles will be calculated as:
n = W/M, where
W = given mass
M = molar mass
For carbon atom:Mass of Carbon component = (0.198)(120g) = 23.76g
Moles of Carbon atom = 23.76g / 12.01g/mol = 1.98mol = 2 moles
For nitrogen atom:Mass of Nitrogen component = (0.116)(120g) = 13.92g
Moles of Nitrogen atom = 13.92g / 14.007g/mol = 0.99mol = 1 moles
For oxygen atom:Mass of Oxygen component = (0.661)(120g) = 79.32g
Moles of Oxygen atom = 79.32g / 15.99g/mol = 4.96mol = 5 moles
For hydrogen atom:Mass of Hydrogen component = (0.025)(120g) = 3g
Moles of Hydrogen atom = 3g / 1g/mol = 3 moles
So, the molecular formula of the compound on the basis of moles of given entities is C₂H₃NO₅.
Hence required molecular formula is C₂H₃NO₅.
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Aluminum metal and bromine liquid (red) react violently to make aluminum bromide (white powder). One way to represent this equilibrium is:
Al(s) + 3/2 Br2(l)AlBr3(s)
We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.
1) 2 AlBr3(s) 2 Al(s) + 3 Br2(l)
2) 2 Al(s) + 3 Br2(l) 2 AlBr3(s)
3) AlBr3(s) Al(s) + 3/2 Br2(l)
Answer:
Explanation:
Al(s) + 3/2 Br₂(l) = AlBr₃(s)
K = [ AlBr₃] / [ Al] [ Br₂]³/²
K² = [ AlBr₃]² / [ Al ] ² [ Br₂]³
2 AlBr₃ = 2 Al(s) + 3 Br₂(l) =
K₁ = [ Al ] ² [ Br₂]³ / [ AlBr₃]²
K₁ = ( 1 / K² ) = K⁻²
2 ) 2 Al(s) + 3 Br₂(l) = 2 AlBr₃(s)
K₂ = [ AlBr₃ ]² / [ Al ]² [ Br₂ ]³
K₂ = K²
3 )
AlBr₃(s) = Al(s) + 3/2 Br₂(l)
K₃ = [ Al ] [ Br₂ ] ³/² / [ AlBr₃ ]
= ( 1 / K ) = K⁻¹
plz help answer both will mark brainest
What concentration of NO−3NO3− results when 897 mL897 mL of 0.497 M NaNO30.497 M NaNO3 is mixed with 813 mL813 mL of 0.341 M Ca(NO3)2?
Answer:
Explanation:
NaNO₃ = Na⁺ + NO₃⁻¹
.497 M .497 M
moles of NO₃⁻¹ = .897 x .497 = .4458 moles
Ca( NO₃)₂ = Ca + 2 NO₃⁻¹
.341 M 2 x .341 M = .682 M
moles of NO₃⁻¹ = .813 x .682 = .5544 moles
Total moles = .4458 moles + .5544 moles
= 1.0002 moles
volume of solution = 897 + 813 = 1710 mL
= 1.710 L
concentration of nitrate ion = 1.0002 / 1.710 M
= .585 M
A student ran the following reaction in the laboratory at 751 K: N2(g) + 3H2(g) 2NH3(g) When she introduced 3.47×10-2 moles of N2(g) and 6.38×10-2 moles of H2(g) into a 1.00 liter container, she found the equilibrium concentration of H2(g) to be 6.25×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.
Answer:
Kc = 4.86×10⁻⁶
Explanation:
We begin from the equation:
N₂ + 2H₂ ⇄ 2NH₃
We start from 3.47×10⁻² moles of N₂(g) and 6.38×10⁻² moles of H₂(g), so when we reach the equilibrium, we get 6.25×10⁻² moles of H₂.
This data means, that in the reaction we made react:
6.38×10⁻² - x = 6.25×10⁻²
x = 1.3×10⁻³ moles of H₂
As stoichiometry is 1:3, we will know that the moles of N₂ that have been reacted were:
1.3×10⁻³ moles / 3 = 4.33×10⁻⁴ moles of N₂
So, in the equilibrium we would have:
3.47×10⁻² moles of N₂ - 4.33×10⁻⁴ moles of N₂ = 0.0343 moles of N₂
How many ammonia, would we have in the equilibrium?
4.33×10⁻⁴ mol . 2 = 8.66×10⁻⁴ moles (from stoichiometry with N₂, 1:2)
(1.3×10⁻³ mol . 2) / 3 = 8.66×10⁻⁴ moles (from stoichiometry with H₂, 2:3)
Let's make the expression for Kc
Kc = [NH₃]³ / [N₂] . [H₂]²
(8.66×10⁻⁴ )³ / (0.0343 . (6.25×10⁻²)² = 4.86×10⁻⁶
pentane or 2,2,3- trimethylhexane has the higher boiling point. why?
Answer:
2,2,3- trimethylhexane because it has more carbon atoms than pentane.
Explanation:
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In this case, since the physical properties of organic compounds are intensified as the number of carbon atoms start increasing on it, the larger the amount of carbon atom, the higher the boiling point since more energy is required to allow the liquid-phase molecules to transcend to the vapor-phase.
In such a way, since pentane has five carbon atoms and 2,2,3- trimethylhexane has nine carbon atoms, 2,2,3- trimethylhexane has the highest boiling point.
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An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 5.50 g of this compound produced 8.07 g of carbon dioxide and 3.30 g of water.
Required:
a. How many moles of carbon, C, were in the original sample?
b. How many moles of hydrogen, H, were in the original sample?
Answer:
a. 0.183 mol C
b. 0.366 mol H
Explanation:
Assuming total combustion, all of the carbon in the unknown compound turned into carbon dioxide, CO₂.
So first we calculate the CO₂ moles produced, using its molecular weight:
8.07 g CO₂ ÷ 44 g/mol = 0.183 mol CO₂This means in the unknown compound there were 0.183 moles of carbon, C.
Conversely, all of the hydrogen in the unknown compound turned into water, H₂O.
Calculating the H₂O moles:
3.30 g ÷ 18 g/mol = 0.183 mol H₂OWe multiply the water moles by two, as there are 2 H moles per H₂O mol:
0.183 * 2 = 0.366 mol H.Which statement best describes the octet rule?
A. When an atom becomes an ion, it gains or loses electrons so that its valence shell holds eight electrons.
B. When an atom becomes an ion, it gains or loses protons so that its nucleus holds eight protons.
C. When an atom becomes an ion, it gains or loses eight electrons.
D. When an atom becomes an ion, it gains or loses eight neutrons.