H3BNH3(s) - NH3(g) + BH3 is the correct option
what is enthalpy of formation?The standard enthalpy of production is the enthalpy change that occurs when 1 mole of a substance is made up of its constituent elements in their standard states. In its natural condition, a pure element has a standard enthalpy of production of zero.
The physical states of the reactants and products (gas, liquid, solid, or solution), the pressure of any gases present, and the reaction temperature all affect how much hydrogen (H) is created during a reaction.
The change in a system's enthalpy that happens when one mole of a substance burns completely in oxygen or air at a certain temperature is known as the enthalpy of combustion.
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Help pls! I need help im just writing to use the character limit!
Answer:
se que la imaje es un poco confusa con la materia pero
desiaria ayudarte pero CREO que con mi TÉCNICA es la A
A student dissolves 12.1 g of potassium chloride (KCl) in 250. g of water in a well-insulated open cup. He then observes the temperature of the water fall from 21.0 °C to 17.1 °C over the course of 6.9 minutes. Use this data, and any information you need from the ALEKS Data resource, to answer the questions below about this reaction:
KCl(s) → K (aq) + Cl (aq)
a. Is this reaction exothermic, endothermic, or neither?
b. Calculate the reaction enthalpy ΔH.rxn per mole of KCI.
Answer:
a. Endothermic
b. 26.37kJ/mol
Explanation:
a. As we can see, the temperature of the water is decreasing when the reaction is occurring, that means the reaction is absorbing heat and is endothermic
b. To find the enthalpy we must find the change in heat when 12.1g of KCl are dissolved. Using the equation:
Q = -m*ΔT*C
Where Q is change in heat
m the mass of solution (250g + 12.1g = 262.1g)
ΔT is change in heat (17.1°C - 21.0°C = -3.9°C)
And C is specific heat of the solution (4.184J/g°C assuming is the same than the specific heat of water).
Replacing:
Q = -262.1g*-3.9°C*4.184J/g°C
Q = 4277J = 4.28kJ
As reaction enthalpy is the change in heat per mole of reaction, we must find the moles of 12.1g of KCl:
Moles KCl -Molar mass: 74.55g/mol-:
12.1g KCl * (1 mol / 74.55g) = 0.1623 moles KCl
The reaction enthalpy us:
4.28kJ / 0.1623mol = }
26.37kJ/mol
Which of the following is the best example of water changing from a liquid to a gas?
A. clouds on a summer day
B. morning dew on orange trees
C. fog rising from a pond on a cold day
D. ice melting back to water when it is left out at room temperature
Answer:
c
Explanation:
hope this helps! ^-^
Answer:
C
Explanation:
It's the answer
A Period 2 element has the following successive ionization energies. Identify the element...
1st: 1087 kJ/mol
2nd: 2353 kJ/mol
3rd: 4621 kJ/mol
4th: 6223 kJ/mol
5th: 37832 kJ/mol
6th: 47279 kJ/mol
7th: 55261 kJ/mol
8th: 69875 kJ/mol
Answer:
9th:08473 kJ/mol
Explanation:
HAHAHAHAHAHAHAHAHAHA
I HAVE 25 MINUTES TO FINISH True or false the deeper you go into Earth the temperature goes up ?
Answer:
true
Explanation:
you are getting closer to the core hope this helps!
which type of rock is rhyolite A.Intusive igneous B.Sedimentary C.Extrusive igneous D.Metamorphic
Answer:
C - Extrusive igneous
Explanation:
Rhyolite is an extrusive igneous due to the high silica content, the lava is very dangerous.
Answer:
Its C
Explanation:
A P E X
What is the formula for Decaoxygen pentasulfide
Answer:
Compound Formula. 1, Carbon dioxide, CO2. 2, Carbon monoxide, CO. 3, Diphosphorus pentoxide, P2O5. 4, Dinitrogen monoxide, N2O.
Explanation:
dont even know by thoAnswer:
Explanation:
What is one of the major conclusions made from the study of line spectra of various elements?
A)
the movement of electrons outside the nucleus
B)
absorption and emission of energy is quantized
C)
the presence of energy beyond the visible spectrum
D)
the presence of electrons outside the nucleus of an atom
Answer:
B
Explanation:
450. grams of water are heated from 20.0 °C to 37.0 °C. How many kcal were absorbed by the water? Note that this is the amount of heat absorbed when a glass of cold water is consumed.
Answer:
Q = 32,007.6 J
Explanation:
Hello there!
In this case, since the energy involved during a heating/cooling process is:
[tex]Q=mC(T_f-T_i)[/tex]
Thus, given the mass, specific heat of water, initial temperature and final one, we plug in obtain:
[tex]Q=450g*4.184\frac{J}{g\°C} (37.0\°C-20.0\°C)\\\\Q=32,007.6J[/tex]
Best regards!
what should i do the next time i see someone that i havent seen in a year and they said they want to kiss me
Answer:
Well if you want to kiss them go for it, but if you don't then say sorry but I don't feel comfortable with that.
Hope everything goes well <3
For each of the following compounds, decide whether the compound's solubility in aqueous solution changes with pH.
a. Ca(OH)2
b. CuBr
c. Ca3(PO4)2
Answer:
a. pH raises
b. pH neutral
c. pH raises
Explanation:
First of all we need to dissociate these compounds.
Ca(OH)₂ → Ca²⁺ + 2OH⁻
This is the calcium hydroxide, a strong base.
In this case, we are giving OH⁻ to medium, so the pH will be increased.
OH⁻ + H⁺ ⇄ H₂O
Hydroxides will neutralize the protons in order to make water, but if we have many OH, we talk about a basic solution where pH is lower than 7.
b. CuBr → Cu⁺ + Br⁻
This is similar to NaCl, a ionic salt which is neutral.
The ions from this salt, do not make hydrolisis, that's why pH does not change.
c. Ca₃(PO₄)₂ → 3Ca²⁺ + 2PO₄⁻³
Calcium cathion comes from a strong base. It means, that the ion can not make hydrolisis because it is the conjugate weak acid.
Phosphate anion comes from a weak acid, it can gives OH⁻ to medium so pH will increase; the PO₄⁻³ anion is the conjugate strong base of the phosphoric acid, that's why it can react:
PO₄⁻³ + H₂O ⇄ H₂PO⁻² + OH⁻
When running an experiment, it is essential to record data and observations within your lab notebook. The data and observations are then discussed and analyzed, results and discussion. However, as you write a conclusion, some data are more important to include than others. Which of the following data would you include if you had to write a conclusion in for outcome of this week's LLE experiment?
a. MP range of product
b. mass of product
c. mass of flask and cork support
d. instrument used to evaporate the solvent
e. volume of crude reaction mixture used
f. % recovery
g. volume of acid and base used
h. theoretical MP
i. mass of flask, cork support, and product
Answer:
% recovery
MP range of product
mass of product
Explanation:
Liquid–liquid extraction (LLE) is a process of transferring one (or more) solute(s) which are present in a feed solution to another immiscible liquid (solvent). The other solvent that becomes enriched in the target solute(s) is called extract. The original feed solution that is depleted in solute(s) is subsequently referred to as the raffinate.
This method is used to purify compounds and separate mixtures of compounds. This is very important when we want to isolate a product from a reaction mixture.
The percent recovery is the amount of solute that is transferred to the extract. This is the most important data to be recorded in an LLE experiment.
The melting point range necessarily helps us to identify the product and the mass of solid tells us the quantity of the solid obtained after extraction.
Write True if the statement is correct and False if it is incorrect.
1. Mosquitos are flies.
2. All large invertebrates live in the
sea or at least in water.
3. An arthropod is a form of arachnid.
4. The giant squid is the world's largest
invertebrate.
5. Earthworms have skeleton
6. The octopus is the most advance of
the mollusks.
7. Insects are vertebrates animals.
8. Starfish have shells.
9. Snails have spiral protective shell.
10. Ants and termites both live in colonies
with a complex social structure.
A per the statement the true and false have been stated.
The mosquitos are flies is True. All large invertebrates live in the sea or at least in water is False. An arthropod is a form of arachnid. False. The giant squid is the world's largest invertebrate is true. Earthworms have skeleton is false. An octopus is the most advance of the mollusks is True. Insects are vertebrates animals is false . Most starfish have shells is true. All Snails have a spiral protective shell is true. The Ants and the termites both live in their colonies with a complex social structure true.Learn more about the statement is correct and False if it is incorrect.
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The lobes are important for speech and language. O A. parietal O B. temporal o o Ο Ο C. frontal O D. occipital
Answer:
Explanation:
D or b
Sorry if wrong
Answer:
temporal
Explanation:
Which of the following may add salt to ocean waters?
volcanic eruptions
melting glaciers
rain
biological precipitation
Answer:
rain
Explanation:
The volcanic eruptions may add salt to ocean waters. option A is correct.
What is a volcanic eruption?
Volcanic eruption produces or releases direct minerals into the water or in the ocean and salt domes are also responsible to increase the salt concentration in the water.
Two of the most commonly found ions in the water of oceans are chloride ions and sodium ions. 85% of the concentration found remaining are the magnesium and sulfate salts.
This process also happens with the mountains but in oceans it invertases the concentration of salt.
Therefore, volcanic eruptions may add salt to ocean waters. option A is correct.
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Salt of a weak acid with strong base When
dissolved in water gives?
Answer:
it gives base, it is process of neutralization,when acid reacts with base it gives salt and water
What is the density of an object with mass 80 kg and volume 05 cubic meters
Answer:
mesure it
Explanation:
Examining the equations or equilibrium constants related to a base, salt, or an acid is an indirect way to determine strength of an electrolyte. The strength of an electrolyte can be examined directly by placing a solution into a circuit so that the voltage or amount of current can be measured. Although conductivity refers to the flow of charged species, we usually examine conductivity with respect to resistance. As the name implies, a solution that does not conduct electricity very well also has a very high resistance. If solutions containing various acids, bases, and salts were prepared and connected to a circuit that powers a light bulb, the strength of the electrolyte could be estimated by examining the intensity of the light bulb. Complete the following sentences regarding the experimental determination of the electrolyte strength for various molecules dissolved in solution.
a. strong electrolyte
b. weak electrolyte
c. non-electrolyte
1. A 20 mL solution containing 2 mmol of Ca3(PO4)2(s) was integrated into to a circuit that powers a light but, When the power supply was turned on, the light bulb produced a glow. Ca3(PO4)2(s) is__.
2. A 20 mL, solution containing 2 mmol of C12H22O11(aq) was integrated into a circuit that powers a light bulb. When the power supply was turned on the light bulb remained off. C12H22O11(aq) is___.
3. A 20 mL, solution containing 2 mmol of HF(g) was integrated into to a circuit that power a light bulb. When the power supply was turned on the light bulb faintly flickered. HF(g) is an____.
4. A 20 mL solution containing 2 mmol of LiOH(g) was integrated into to a circuit that power a light bulb. When the power supply was turned on the light bulb produced a bright glow LiOH(s) is a____.
Answer:
1 - Weak electrolyte
2- Non electrolyte
3- Weak electrolyte
4- Strong electrolyte
Explanation:
A strong electrolyte refers to an electrolyte that decomposes completely in solution. This means that there are more charge carriers in solution when a strong electrolyte is dissolved in water. A strong electrolyte produces a strong glow. LiOH is a strong electrolyte.
A weak electrolyte is not completely dissociated in water. Only a small amount dissociates in water. HF is a weak electrolyte. A weak electrolyte does not produce a bright light.
A non-electrolyte does not dissociate in solution at all hence it does not power a bulb E.g C12H22O11.
A 50.0 mL solution of 0.129 M KOH is titrated with 0.258 M HCl. Calculate the pH of the solution after the addition of each of the given amounts of HCl.
A) 0.00ml
B)7.00ml
C)12.5ml
D)18.0ml
E)24.0ml
F)25.0ml
G)26.0ml
H)29.0ml
please show work with steps .
Answer:
A- pH = 13.12
B- pH = 12.91
C- pH = 12.71
D- pH = 12.43
E- pH = 11.55
F- pH = 7
G- pH = 2.46
H- pH = 1.88
Explanation:
This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) → H₂O(l) + KCl(aq)
Our pH at the equivalence point is 7, because we have made a neutral salt.
To determine the volume at that point we state the formula for titration:
mmoles of base = mmoles of acid
Volume of base . M of base = Volume of acid . M of acid
50mL . 0.129M = 0.258 M . Volume of acid
Volume of acid = (50mL . 0.129M) / 0.258 M → 25 mL (Point F)
When we add 25 mL of HCl, our pH will be 7.
A- At 0 mL of acid, we only have base.
KOH → K⁺ + OH⁻
[OH⁻] = 0.129 M
To make more easy the operations we will use, mmol.
mol . 1000 = mmoles → mmoles / mL = M
- log 0.129 = 0.889
14 - 0.889 = 13.12
B- In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺
Initially we have 0.129 M . 50 mL = 6.45 mmoles of OH⁻
1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:
6.45 mmol - 1.81 = 4.64 mmoles of OH⁻
This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.
[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M
- log 0.0815 M = 1.09 → pOH
pH = 14 - pOH → 14 - 1.09 = 12.91
C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺
Our initial mmoles of OH⁻ would not change through all the titration.
Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.
6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻
Total volume is: 50 mL of base + 12.5 mL = 62.5 mL
[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M
- log 0.0517 = 1.29 → pOH
14 - 1.11 = 12.71
D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺
6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.
6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻
Total volume is: 50 mL of base + 18 mL = 68 mL
[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M
- log 0.0265 = 1.57 → pOH
14 - 1.57 = 12.43
E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺
6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.
6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻
Total volume is: 50 mL of base + 24 mL = 74 mL
[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M
- log 3.51×10⁻³ = 2.45 → pOH
14 - 2.45 = 11.55
F- This the equivalence point.
mmoles of OH⁻ = mmoles of H⁺
We add (25 mL . 0.258M) = 6.45 mmoles of H⁺
All the OH⁻ are neutralized.
OH⁻ + H⁺ ⇄ H₂O Kw
[OH⁻] = √1×10⁻¹⁴ → 1×10⁻⁷ → pOH = 7
pH → 14 - 7 = 7
G- In this case we have an excess of H⁻
We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺
We neutralized all the OH⁻ but some H⁺ remain after the equilibrium
6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺
[H⁺] = 0.26 mmol / Total volume
Total volume is: 50 mL + 26 mL → 76 mL
[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M
- log 3.42×10⁻³ = 2.46 → pH
H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺
We neutralized all the OH⁻ but some H⁺ remain after the equilibrium
7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons
Total volume is 50 mL + 29 mL = 79 mL
[H⁺] = 1.03 mmol / 79 mL → 0.0130 M
- log 0.0130 = 1.88 → pH
After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.
The pH value is given by 14 less the logarithm of the hydronium ion
concentration or the hydrogen ion concentration in the solution.
Responses (approximate values):
The pH values are;
A) 13.11
B) 12.91
C) 12.71
D) 12.42
E) 11.54
F) 7
G) 2.469
H) 1.884
Which is used to find the pH of the solution?A) Concentration of the KOH = 0.129 M
Amount of HCl added = 0.00 ml
The pH = -log[H⁺] = 14 - pOH
pOH = -log[OH⁻]
Which gives;
pH = 14 - (-log[OH⁻] )
pH = 14 - (-log(0.129)) ≈ 13.11B) Volume of acid added = 7.00 mL = 0.007 L
Concentration of the acid = 0.258 M HCl
Number of moles of acid, H⁺ = 0.007 × 0.258 moles = 0.001806 moles
Number of moles of KOH remaining, OH⁻= 0.05 × 0.129 - 0.001806 = 0.004644
Number of moles of OH⁻ = 0.004644 moles
[tex]Concentration, \ [OH^-] = \dfrac{0.004644 \, moles}{0.05 7 \, L} \approx \mathbf{ 0.0815 \, M}[/tex]
pH of solution = 14 - (-log(0.0815)) ≈ 12.91C) 12.5 mL HCl contains, 0.0125 × 0.258 moles = 0.003225 moles
OH⁻ remaining = 0.05 × 0.129 - 0.003225 = 0.003225 moles
[tex]Concentration \ of \ [OH^-]= \dfrac{0.003225\, moles}{(0.05 + 0.0125) \, L} = \mathbf{0.0516 \, M}[/tex]
pH of solution = 14 - (-log(0.0516)) ≈ 12.71D) 18.0 mL HCl contains, 0.018 × 0.258 moles = 0.004644 moles
OH⁻ remaining = 0.05 × 0.129 - 0.004644 = 0.001806 moles
[tex]Concentration \ of \ [OH^-] = \dfrac{0.001806\, moles}{(0.05 + 0.018) \, L} \approx \mathbf{0.0266\, M}[/tex]
pH of solution = 14 - (-log(0.0266)) ≈ 12.42E) 24.0 mL HCl contains, 0.024 × 0.258 moles = 0.006192 moles
OH⁻ ion remaining = 0.05 × 0.129 - 0.006192 = 0.000258 moles
[tex]Concentration \ of \ [OH^-] = \dfrac{0.000258\, moles}{(0.05 + 0.024) \, L} \approx \mathbf{0.0035\, M}[/tex]
pH of solution = 14 - (-log(0.0035)) ≈ 11.54F) 25.0 mL HCl contains, 0.025 × 0.258 moles = 0.00645 moles
[OH⁻] remaining = 0.05 × 0.129 - 0.00645 = 0 moles
[tex]Concentration \ of \ [OH^-] = \dfrac{0\, moles}{(0.05 + 0.024) \, L} \approx 0\, M[/tex]
Therefore;
Number of moles of KOH = 0, or the solution is neutralized
[OH⁻] = [H⁺]
Which gives;
pH = pOH = 7G) 26.0 mL HCl contains, 0.026 × 0.258 moles = 0.006708 moles
[OH⁻] remaining = 0.05 × 0.129 - 0.006708 = -0.000258 moles
Therefore
Number of moles of H⁺ = 0.000258
[tex]\mathbf{Concentration} \ of \ \mathbf{[H^+] }= \dfrac{0.000258\, moles}{(0.05 + 0.026) \, L} \approx 0.003395\, M[/tex]
pH of solution = (-log(0.003395)) ≈ 2.469H) 29.0 mL HCl contains, 0.029 × 0.258 moles = 0.007482 moles
H⁺ remaining = 0.007482 - 0.05 × 0.129 = 0.001032 moles
Therefore
Number of moles of H⁺ = 0.001032
[tex]Concentration \ of \ [H^+] = \mathbf{ \dfrac{0.001032\, moles}{(0.05 + 0.029) \, L}} \approx 0.01306\, M[/tex]
pH of solution = (-log(0.01306)) ≈ 1.884Learn more about the pH of a solution here:
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Two identical spoons are electroplated with Ag or Cd through the use of the electrolytic cells. A current of 5.00A was supplied to each cell for 600. seconds, and the masses of the spoons before and after the electroplating were recorded. Write down the mathematical equations can best be used to account for the much larger increase in mass of the spoon electroplated with Ag compared with the spoon electroplated with Cd.
Answer:
Explanation:
One farad of charge is capable of depositing one gram equivalent of a metal
One gram equivalent of Ag = 108 grams
One gram equivalent of Cd = 112 / 2 grams
= 56 grams . [ for cadmium equivalent mass = atomic mass / s ]
electric charge flowing = current x time = 5 x 600 = 3000 coulomb
one farad = 96500 coulomb
96500 coulomb deposits 108 gram of Ag
3000 coulomb deposits 108 x 3000 / 96500 gram
= 3.35 grams of Ag
Similarly ,
96500 coulomb deposits 56 gram of Cd
3000 coulomb deposits 56 x 3000 / 96500 gram
= 1.74 grams of Cd
So there will be much larger increase in the spoon of Ag due to larger deposit of Ag by charge .
How does ice float on water?
Why does solid ice float on liquid water?
Answer:
Explanation:
Solid ice floats over water because ice is less dense than liquid water, or weighs less, by about 10%, than liquid water
In this stage of team building, the team is running smoothly and they are comfortable in their roles.
norming
performing
forming
storming
Answer:
performing
Explanation:
A metal ion (X) with a charge of 4+ is attracted to non metal ion (Z) with a charge of 3-. Which of these formulas represents the resulting compound pls answer asap options in photo
Answer:
D
Explanation:
the charges need to balence out
so finding the LCM which is 12 we find we need 3x's
and 4 zs
so that makes the formula X3Z4 which is D
Carbon monoxide is a colorless, odorless gas that binds irreversibly to hemoglobin in our blood, causing suffocation and death. CO is formed during incomplete combustion of carbon. One way to represent this equilibrium is: CO(g)C(s) 1/2 O2(g) 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.
The question is missing some parts. Here is the complete question.
Carbon monoxide is a colorless, odorless gas that binds irreversibly to hemoglobin in our blood, causing suffocation and death. CO is formed during incomplete combustion of carbon. One way to represent this equilibrium is:
[tex]2CO_{(g)}[/tex] ⇄ [tex]2C_{(s)}+O_{2}_{(g)}[/tex]
we could also write this reaction three other ways listed below. The equilibrium constant 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) [tex]2C_{(s)}+O_{2}_{(g)}[/tex] ⇄ [tex]2CO_{(g)}[/tex] K₁ =
2) [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex] ⇄ [tex]CO_{(g)}[/tex] K₂ =
3) [tex]CO_{(g)}[/tex] ⇄ [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex] K₃ =
Answer: 1) [tex]K_{1}=\frac{1}{K}[/tex]
2) [tex]K_{2}=\frac{1}{K^{1/2}}[/tex]
3) [tex]K_{3}=K^{1/2}[/tex]
Explanation: A chemical reaction can be reversible, i.e., can proceed in both directions: to the right of the arrow (forward) or towards the left of the arrow (backward).
When the rates of forward and backward reactions are the same, the reaction is in equilibrium. In that state, we can determine the equilibrium constant, [tex]K_{c}[/tex].
For the first way to represent equilibrium of CO formed, the [tex]K_{c}[/tex] is calculated
[tex]2CO_{(g)}[/tex] ⇄ [tex]2C_{(s)}+O_{2}_{(g)}[/tex]
[tex]K=\frac{[O_{2}]}{[CO]^{2}}[/tex]
in which the symbol [ ] is concentration of the compound.
In equilibrium constant, solids are not included.
Equilibrium constants for the other reactions:
1) [tex]2C_{(s)}+O_{2}_{(g)}[/tex] ⇄ [tex]2CO_{(g)}[/tex]
[tex]K_{1}=\frac{[CO]^{2}}{[O_{2}]}[/tex]
Comparing K₁ and K, the first one is the inverse of K, so writing in terms of K
[tex]K_{1}=\frac{1}{K}[/tex]
2) [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex] ⇄ [tex]CO_{(g)}[/tex]
[tex]K_{2}=\frac{[CO]}{[O_{2}]^{1/2}}[/tex]
In terms of K, K₂ is
[tex]K_{2}=\frac{1}{K^{1/2}}[/tex]
3) [tex]CO_{(g)}[/tex] ⇄ [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex]
[tex]K_{3}=\frac{[O_{2}]^{1/2}}{[CO]}[/tex]
This constant in terms of K will be
[tex]K_{3}=K^{1/2}[/tex]
In conclusion, K₁, K₂ and K₃ in terms of K is [tex]\frac{1}{K}[/tex],[tex]\frac{1}{K^{1/2}}[/tex] and [tex]K^{1/2}[/tex], respectively.
1. A solution is always a mixture (true or false)
im not sure what the answer is so if anyone could help me that would be great
(ps im in 5th grade)
Answer:
yes solution is always a mixture but not all mixtures are solution
Explanation:
A solution.is a homogeneous mixture of substance that have uniform composition throughout
And a mixture hVe twoo or more substances that are not chemically.combine
How many pounds of ice are required to absorb 4900 kJ of heat as the ice melts? The heat of fusion of water is 0.334 kJ/g.
help pls
Answer:
Q = ΔH fusion * mass (g)
when we have:
ΔH fusion (the heat (or enthalpy) of fusion = 0.334 kJ/g
and mass of ice = 22.4 g
so by substitution, we can get the energy (Q) required to melt this mass of ice:
∴ Q = 0.334KJ/g * 22.4 g
= 7.48 KJ
∴ the energy required to melt 22.4 g of ice is = 7.48 KJ
Explanation:
what is the name of these 4 compound in isomers
Answer:
C9H19 C5H12 C6H17 C7H20
Animals get energy from the food that they eat. However, when the molecules from the food enter your cells, how do the molecules turn in to energy?
Question 3 options:
The glucose molecules from the food are broken down by the mitochondria in the cell to produce ATP which the cells use as energy.
The ATP is broken down into glucose which the cells use for energy.
Upon hitting the stomach, the food molecules are changed to energy molecules which fuel the body.
The cells absorb ATP from food and use it for energy.
Answer:
The ATP is broken down into glucose which the cells use for energy.
When do Florida plants prepare themselves for the cold winter temperatures? (1 point)
a
Beginning of the spring
b
During the fall
c
End of the spring
d
End of the summer
PLEASE HELP ASAP!!! >.
Answer:
A. Beginning of the spring
Explanation:
Pa like, Pa follow, Pa rate
1) De los siguientes cambios indicar cuál es químico y cual es físico:
• Aplastar una lata:
• Congelar agua:
• Quemar pasto:
• Hervir agua:
• Se descompone fruta:
• Se oxida hierro:
• Derretir chocolate:
• Mezclar agua y sal:
• Pulverizar un ladrillo:
• Quemar pólvora