Answer:
o
Explanation:
it is not a gas because the particles do not move freely it may be a liquid or a solid partly and mostly liquidized.
If 16 grams of O2 react with excess C2H6, how many grams of CO2 will be formed? The formula mass of O2 = 32 amu and the formula mass of CO2 = 44amu. The balanced chemical equation is 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)
Answer:5.17g of Co2
Explanation:
Determine the molecular formula of a compound if it is composed of 40.92% carbon, 4.58% hydrogen, and 54.50% oxygen. The molar mass is 176.12 g/mol.
A) C3H8O3.
B) CH2O.
C) C2H3O2.
D) C3H4O3.
Answer:
No correct answer listed. See explanation for defense.
Explanation:
Given
C: 40.92% => 40.92g/100wt => (40.92/12)moles C = 3.41 moles O
H: 4.58% => 4.58g/100wt => (4.58/1)moles H = 4.58 moles H
O: 54.50% => 54.5g/100wt => (54.5/16)moles O = 3.41 moles O
Empirical ratio => C : H: O => (3.41/3.41) : (4.58/3.41) : (3.41/3.41) => 1 : 1.34 : 1
=> C : H : O => 3(1 : 1.34 : 1) => 3 : 4 : 3 => Empirical Formula C₃H₄O₃
Molecular Weight = Empirical Formula Wt x N
176.12 = 88 x N
N = whole number multiple of empirical formula = 176.12/88 = 2
∴ Molecular Formula => (C₃H₄O₃)₂ => C₆H₈O₆
Note => Only ionic compounds (salts) have subscripts reduced to lowest whole number ratios. Molecular compounds as C₆H₈O₆ are not reduced to lowest whole number ratios. Therefore, there is no correct answer in the answer choice list for the 'Molecular Formula'. Doc :-)
the -OH group cannot exhibit inductive effect? true/false, and reason for ur choice
Answer:
false
Explanation:
The inductive effects are know as the ability of the atom or a group to create polarization and electronic density long the covalent bond and it needs a higher density. The -OH group cannot exhibit the indictive effects as it becomes --O.Consider the reaction: CaCO3(s)CaO(s) CO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.58 moles of CaCO3(s) react at standard conditions.
Answer:
the entropy change for the system when 1.58 moles of CaCO3(s) react at standard conditions is 253.748 J/K
Explanation:
Given the data in the question;
CaCO₃(s) → CaO(s) + CO₂(g)
1.58 moles 1.58 moles 1.58 moles
Since 1 mole of CaCO₃ gives 1 mole of CaO and 1 mole of CO₂
Thus, 1.58 mole of CaCO₃ gives 1.58 moles of CaO and 1.58 moles of CO₂.
Now,
At 298 K, standard entropy values are;
ΔS° ( CaCO₃ ) = 92.9 J/mol.K
ΔS° ( CaO ) = 39.8 J/mol.K
ΔS° ( CO₂ ) = 213.7 J/mol.K
So,
ΔS°[tex]_{system[/tex] = ∑ΔS°( product ) - ∑ΔS°( reactant )
ΔS°[tex]_{system[/tex] = [ ΔS°(CaO) + ΔS°( CO₂ ) ] - ΔS°( CaCO₃ )
we substitute
ΔS°[tex]_{system[/tex] = [ 39.8 J/mol.K + 213.7 J/mol.K ] - 92.9 J/mol.K
ΔS°[tex]_{system[/tex] = 160.6 J/mol.K
i.e, for 1 mol CaCO₃, ΔS°[tex]_{system[/tex] = 160.6 J/mol.K
Now, for 1.58 mol CaCO₃,
ΔS°[tex]_{system[/tex] = 1.58 mol × 160.6 J/mol.K
ΔS°[tex]_{system[/tex] = 253.748 J/K
Therefore, the entropy change for the system when 1.58 moles of CaCO3(s) react at standard conditions is 253.748 J/K
Flag A solution of the weak acid, HF, and a solution of the strong acid, HCl, have the same pH. Which solution will require the most sodium hydroxide, NaOH, to neutralize
Answer:
C) Both will require the same amount because the concentrations are equal.
Explanation:
The pH of a solution is defined as:
pH = -log [H+]
This H+ is the ion that reacts with OH- (From NaOH) as follows:
H+ + OH- → H2O
When all H+ reacts, we can say the solution was neutralized.
Now, as both, the solution with the weak acid and the solution with strong acid have the same pH, we can say that their [H+] is the same. Assuming the volume of both solutions is the same:
Both will require the same amount because the concentrations are equal.
Categorize the following reaction as an acid-base neutralization, precipitation, combination, decomposition, combustion, displacement, or disproportionation reaction.
Ba(C2H3O2)2(aq) + Na2CO3(aq) → BaCO3(s) + 2 NaC2H3O2(aq)
Answer:
Precipitation
Explanation:
Let's consider the balanced chemical equation between barium acetate and sodium carbonate to form barium carbonate and sodium acetate.
Ba(C₂H₃O₂)₂(aq) + Na₂CO₃(aq) → BaCO₃(s) + 2 NaC₂H₃O₂(aq)
Both products and reactants are salts. But, among the products, barium carbonate is solid. This allows us to classify it as a precipitation reaction.
For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data :
ΔH∘rxn 180.5kJ/mol
ΔS∘rxn 24.80J/(mol⋅K)
Required:
a. Calculate the temperature in kelvins above which this reaction is spontaneous
b. Calculate the equilibrium constant for the following reaction at room temperature, 25°C
Answer:
a. 7278 K
b. Kc = 4.542 × 10⁻³¹
Explanation:
a.
The reaction is spontaneous when ΔG° < 0. We can calculate ΔG° using the following expression.
ΔG° = ΔH° - T × ΔS°
Then, the reaction will be spontaneous when,
ΔH° - T × ΔS° < 0
T > ΔH°/ΔS
T > (180.5 × 10³ J/mol)/(24.80J/mol⋅K)
T > 7278 K
b.
First, we will calculate ΔG° at 25 °C (298 K)
ΔG° = ΔH° - T × ΔS°
ΔG° = (180.5 × 10³ J/mol) - 298 K × (24.80J/mol⋅K) = 1.731 × 10⁵ J/mol
Then, we will calculate the equilibrium constant (Kc) using the following expression.
ΔG° = - R × T × ln Kc
-ΔG°/R × T = ln Kc
-(1.731 × 10⁵ J/mol)/(8.314 J/mol.K) × 298 K = ln Kc
Kc = 4.542 × 10⁻³¹
why beta carbon hydrogen is easily replaced
Answer:
γ−Hydrogen is easily replacable during bromination reaction in presence of light , because Allylic substitution is being preferred.
Explanation:
that's all
γ−Hydrogen is easily replacable during bromination reaction in presence of light , because Allylic substitution is being preferred.
What is the concentration (M) of Ch3OH a solution prepared by dissolving of CH3OH sufficient water to give exactly 230 of solution?
Answer:
1.59 M
Explanation:
What is the concentration (M) of CH₃OH a solution prepared by dissolving 11.7 g of CH₃OH sufficient water to give exactly 230 mL of solution?
Step 1: Given data
Mass of CH₃OH: 11.7 gVolume of solution: 230 mL (0.230 L)Step 2: Calculate the moles corresponding to 11.7 g of CH₃OH
The molar mass of CH₃OH is 32.04 g/mol.
11.7 g × 1 mol/32.04 g = 0.365 mol
Step 3: Calculate the molarity of the solution
M = moles of solute / liters of solution
M = 0.365 mol / 0.230 L = 1.59 M
If 225 cal of heat is added to 15.0 g of 2-propanol at room temperature, 25ºC, the alcohol will reach a final temperature of ________ ºC. The specific heat of 2-propanol is 0.612 cal/(g•ºC).
Explanation:
I have no idea about this
Balance the following chemical equation by
providing the correct coefficients.
C3H8 + O2 + CO2 + H2O
A 2.8.4.2
B 1.5.3.4
Answer:
C3H8 + 502+3CO2 + 4H20
Explanation:
In order to get this answer you would have to would balance out the carbons on the products side by multiplying CO2 by 3. The new equation will be
C3H8 + 02 ----> 3CO2 + H20
Now, that the carbons are balanced, we will look at the hydrogens. We can multiply the number of waters on the product side by 4, to make 4 H20 molecules. This gives both sides of the equation 8 hydrogens. Next we can check our oxygens. There will be 10 oxygens on the products side and 2 on the reactants so to balance these out, we multiply the 02 on the reactants side by 5.
1. When 6.0 grams of zinc are dropped into excess hydrochloric acid, how many grams of zinc chloride will be produced?
2. When 45.0 grams of copper (II) carbonate are decomposed with heat, how many grams of carbon dioxide will be produced? (Teachers note: Other product is copper (II) oxide.)
Please explain as well if possible! Thanks.
Explanation:
here are the answers for your questions
I basically converted the given grams to moles, and then multiplied that by the product-to-reactant ratio in the equation, and then convert that to grams
Answer:
Explanation:
1.
The reaction can be represented by the equation: Zn + 2 HCl -> ZnCl2 + H2
From the equation, molar ratio of Zn and ZnCl2 is 1:1.
Molar mass of Zn = 65.38
Molar mass of ZnCl2 = 65.38 + 35.45*2 = 136.28
So 6.0 grams of Zn will produce 6 / 65.38 * 136.28 = 12.5 grams
2.
As the only other product is copper (II) oxide, the reaction can be represented by the equation: CuCO3 -> CuO + CO2
From the equation, molar ratio of CuCO3 and CuO is 1:1.
Molar mass of CuCO3 = 123.55
Molar mass of CuO = 79.55
So 45.0 grams of CuCO3 will produce 45 / 123.55 * 79.55 = 28.96 grams
Enzyme catalyzing breakdown of atp to adp
Answer:
ATP hydrolase
Explanation:
Enzymes are biological catalysts which perform diverse functions in the body. Enzymes are specific in their mode of action because an enzyme fits into its substrate as a key fits into a lock.
The particular enzyme that catalyzes the breakdown of ATP to ADP is ATP hydrolase. The phosphate released by the action of this enzyme is used in the phosphorylation of other compounds thereby making them more reactive.
1. Why is it necessary to equalize the pressure(i.e, have the water level the same in each tube) before taking a volume reading?
2. Why is it important to use water that has been pre-saturated with CO2 in the gas burettes?
3.If your antacid sample had been contaminated by moisture, what effect(if any )would you expect this to have on your result
4.Explain why an'antacid is called as such,what is the role of the NAHCO3 or CACO3 in reactions?
Answer:
If you contact water with a gas at a certain temperature and (partial) pressure, the concentration of the gas in the water will reach an equilibrium ('saturation') according to Henry's law.
Explanation:
This means: if you increase the pressure (e.g. by keeping the vial closed), the CO2 concentration will increase. So it simply depends what concentration you need for your assay: 'CO2-saturated' water at low pressure or 'CO2-saturated' water at high pressure.
Predict the Normality of H2SO4 if 75 ml of 96.6 % pure H2SO4 added to 425 ml water. The density of H2SO4 is 1.83 g/cm3?
Explanation:
Normality is one of the concentration terms.
It is expressed as:
[tex]N=\frac{mass of the substance}{equivalent mass}* \frac{1}{volume of solution in L.}[/tex]
The volume of the solution is 425 mL.
Mass of sulfuric acid given is:
[tex]mass=volume * purity* density\\ = 75 mL * 0.966 * 1.83 g/mL\\\\=132.5 grams\\[/tex]
The equivalent mass of sulfuric acid is 49.0g/equivalents
Hence, the normality of the given solution is:
[tex]N=\frac{132.5g}{49.0g/equi.} *\frac{1000}{425mL} \\Normality=6.36N[/tex]
Answer is: 6.36N.
Write the separation scheme for the isolation of triphenylmethanol from the reaction mixture once the reaction is complete. The separation begins after the addition of HCl and water to the reaction and includes the column chromatography procedure to further purify crude triphenylmethanol isolated in the day 1 procedure.
````````````````````
What is the oxidation state of nitrogen in N ?
Answer:
+5
Explanation:
The diagram above shows the hydrides of groups 14, 15, 16, and 17 elements. Why does H20, HF, and NH3 have much higher boiling points than the rest of the molecules in their groups?
Answer:
Hydrogen Bonding
Explanation:
Hydrogen Bonding occurs when a hydrogen atom is bonded to N, O, and F atoms.
The molecules H₂O, HF, and NH₃ all experience hydrogen bonding, which is a relatively strong IMF, causing the molecules to have stronger attraction to each other. Having a stronger attraction between molecules results in more energy required to separate them, thus these molecules will have a higher boiling point than the rest of the molecules in their group.
Name the following molecule
Answer:
It is a Biological Molecule
3. A mixture of gases- oxygen, carbon dioxide and nitrogen - exerts a total pressure of 107.6 kPa. If the oxygen
exerts 45.8 kPa and the carbon dioxide exerts 37.1 KPA, then what is the partial pressure of the nitrogen?
Answer: 23.7kPa
Explanation: Given:
P total= Sum of partial P of all gases in the mix P total = PO2 +PN2 +PCO2 107.6 = 45.8 + PN2 +37.1 PN2= 107.6- (45.8+ 37.1) = 107.6- 83.1 = 23.7kPA Therefore Partial pressure of Nitrogen is 23.7 kPAA solution is made by dissolving 20 ml of acetic acid in 180ml of water. Calculate its volume concentration
Answer:
water is 9/10
chemical would be 1/10
Explanation:
180/200 would be water concentration in solution
and 20/200 would be chemical solution concentration in solution (if the chemical were to be polar and mix)
how does iron I differ from iron II
Answer:
Metals tend to form positive oxidation states. Here, Iron (I) has an oxidation state of +1 while Iron (II) has an oxidation state of +2. Similarly, Lead (I) has an oxidation state of +1 while Lead(II) has an oxidation state of +2. A change in oxidation state can rather cause significant changes in the compound.
The absorption spectrum of neon has a line at 633 nm. What is the energy of this line? (The speed of light in a vacuum is 3.00 x 108 m/s, and Planck's constant is 6.626 x 10-34 J·s.)
Answer:
B
Explanation:
E = hc/[tex]\lambda[/tex]. Remember, it is in meters not nanometers so you have to convert. You end up with B.
A researcher is attempting to produce ethanol using an enzyme catalyzed batch reactor. The ethanol is produced from corn starch by first-order kinetics with a rate constant of 0.05 hr-1. Assuming the concentration of ethanol initially is 1 mg/L, what will be the concentration of ethanol (in mg/L) after 24 hours
Answer:
The correct solution is "3.32 gm/L".
Explanation:
Given:
Rate constant,
[tex]K = 0.05 \ hr^{-1}[/tex]
Time,
[tex]t = 24 \ hours[/tex]
Concentration of ethanol,
[tex]C_o= 1 \ mg/L[/tex]
Now,
The concentration of ethanol after 24 hours will be:
⇒ [tex]C_o=C\times e^{-K\times t}[/tex]
By putting the values, we get
[tex]1=C\times e^{-0.05\times 24}[/tex]
[tex]1=C\times 0.30119[/tex]
[tex]C= 3.32 \ gm/L[/tex]
A rigid, sealed container that can hold 26 L of gas is filled to a pressure of
5.97 atm at 374 °C. The pressure suddenly decreases to 3.64 atm. What is
the new temperature inside the container, in units of °C?
Answer:
121 °C
Explanation:
From the question given above, the following data were obtained:
Initial pressure (P₁) = 5.97 atm
Initial temperature (T₁) = 374 °C
Final pressure (P₂) = 3.64 atm
Final temperature (T₂) =?
NOTE: Volume = constant
Next, we shall convert 374 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T₁) = 374 °C
Initial temperature (T₁) = 374 °C + 273
Initial temperature (T₁) = 647 K
Next, we shall determine the final temperature. This can be obtained as follow:
Initial pressure (P₁) = 5.97 atm
Initial temperature (T₁) = 647 K
Final pressure (P₂) = 3.64 atm
Final temperature (T₂) =?
P₁ / T₁ = P₂ / T₂
5.97 / 647 = 3.64 / T₂
Cross multiply
5.97 × T₂ = 647 × 3.64
5.97 × T₂ = 2355.08
Divide both side by 5.97
T₂ = 2355.08 / 5.97
T₂ = 394 K
Finally, we shall convert 394 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
Final temperature (T₂) = 394 K
Final temperature (T₂) = 394 – 273
Final temperature (T₂) = 121 °C
Thus, the new temperature is 121 °C
Suppose we have two rock samples, A and B. Rock A was subject to both physical and chemical weathering while rock B was subject to chemical weathering only. Which rock would experience more chemical weathering? Why? (2pts) (Hint: consider the effect of surface area on the rate of chemical weathering)
Answer:
Rock A will have far more chemical weathering than Rock B due to the rise in area effect
Explanation:
Rock A undergoes both Physical and Chemical weathering. So, thanks to physical weathering there'll appear cracks within the rock, which can, in turn, increase the area of rock on which weathering is occurring. So, Chemical weathering will happen much faster now as there's a rise in the area. within the case of Rock B, there's only chemical weathering therefore the increase in the area won't be that very much like compared to Rock A.
Based on the "Reactivity in Substitution Reactions" experiment, which molecule would be expected to react the fastest using AgNO3 in water-ethanol ?
Answer:
C) EtOH 1% AgNO3
How many molecules (or formula units) are in 138.56 g C4H10 Express your answer using four significant figures.
Answer:
dont buy cheap and off we went
How much BaSO4 can be formed from 196.0 g of H2SO4?
Answer:
a) You can form 466 g of BaSO₄.
Explanation:
a) Mass of BaSO4
196 g H₂SO4 × 1 mol H₂SO4
98.08 g H₂SO4
1 mol BaSO 1 mol H₂SO4 X X
466 g BaSO4
233.39 g BaSO4
1 mol BaSO4
how does lead resemble chromium?
