Answer: The rate increases 3 times on raising the temperature from 20degree to 30 degree
Explanation:
According to Arrhenius equation with change in temperature, the formula is as follows.
[tex]ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}][/tex]
where [tex]k_2[/tex] = rate constant at temp [tex]T_2[/tex]
[tex]k_1[/tex] = rate constant at temp [tex]T_1[/tex]
[tex]E_a[/tex]= activation energy
R= gas constant
[tex]T_1[/tex]= temperature = [tex]20^0C=(20+273)K=293K[/tex]
[tex]T_2[/tex]= temperature = [tex]30^0C=(30+273)K=303K[/tex]
[tex]ln \frac{k_{2}}{k_{1}} = \frac{-85\times 1000J/mol}{8.314J/Kmol}[\frac{1}{303} - \frac{1}{293}][/tex]
[tex]ln \frac{k_{2}}{k_{1}}=1.15[/tex]
[tex]\frac{k_{2}}{k_{1}}=3[/tex]
Thus rate increases 3 times on raising the temperature from 20degree to 30 degree
What solid figure has two bases, but no lateral faces?
cylinder
rectangular prism
triangular prism
cone
Answer:
it is cylinder as it has two bases but no lateral faces.
Is liquid ammonia a household acid?
Yes it is an cleaning product
Answer:
yes
Explanation:
it cleans in the house
When an organism eats a plants,the energy and nutrients stored in the consumed organism flows up to the____level
Answer:
Energy is transferred between trophic levels when one organism eats another and gets the energy-rich molecules from its prey's body.
Explanation:
The flow of energy in an ecosystem is always linear ie uni direction. At each energy step in food chain, the energy received by the organisms is used for its own metabolism and maintenance. The left over energy is passed to next higher trophic level. Thus the energy flow decreases with successive trophic level.
what is reduction potential?
Answer:
A reduction potential measures the tendency of a molecule to be reduced by taking up new electrons. ... Standard reduction potentials can be useful in determining the directionality of a reaction. The reduction potential of a given species can be considered to be the negative of the oxidation potential.
Explanation:
Part A. Classify each of these soluble solutes as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. Solutes Formula Nitric acid HNO3 Potassium hydroxide KOH Formic acid HCOOH Ethyl amine CH3CH2NH2 Sodium bromide NaBr Butanol C4H9OH Sucrose C12H22O11
Part B. Enter a molecular equation for the reaction that occurs between aqueous HBr and aqueous LiOH.
Answer:
1. Strong electrolytes = Nitric acid HNO3,Potassium hydroxide KOH, Sodium bromide NaBr
Weak electrolytes = Formic acid HCOOH Ethyl amine CH3CH2NH2, Butanol C4H9OH
Non-electrolyte = Sucrose C12H22O11
2. The reaction between aqueous HBr and aqueous LiOH is shown innthe equation below:
HBr + LiOH ----> NaBr + H₂O
Explanation:
Electrolytes are substances that when in molten state or in aqueous solution dissociate into ions both positively-charged and negatively-charged ions known as actions and anions, which are then able conduct electricity.
Strong electrolytes are aqueous or molten solutions of compounds which ionize completely to produce anions and actions. Ionic compounds such as organic salts, acids and bases are strong electrolytes.
Weak electrolytes only ionize partially, that is, they produce few ions. Some organic salts, acids and bases are weak elctrolytes
From the compounds given above;
Strong electrolytes = Nitric acid HNO3,Potassium hydroxide KOH, Sodium bromide NaBr
Weak electrolytes = Formic acid HCOOH Ethyl amine CH3CH2NH2, Butanol C4H9OH
Non-electrolyte = Sucrose C12H22O11
2. The reaction between aqueous HBr and aqueous LiOH is shown innthe equation below:
HBr + LiOH ----> NaBr + H₂O
PLEASEEEE HELLPPP I CANT FAILL PLSS
Please give me the answer please
Answer:
A. 30cm³
Explanation:
Based on the chemical reaction:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂
To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:
Moles CaCO₃ -Molar mass: 100.09g/mol-
1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles
Moles HCl:
50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles
For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:
2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.
The moles produced of CO₂ are:
2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂
Using PV = nRT
Where P is pressure = 1atm assuming STP
V volume in L
n moles = 1.25x10⁻³ moles CO₂
R gas constant = 0.082atmL/molK
T = 273.15K at STP
V = nRT / P
1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V
0.028L = V
28cm³ = V
As 28cm³ ≈ 30cm³
Right option is:
A. 30cm³How many grams are in 32.2 L of CO2?
Answer:
63.25 grams of CO₂
Explanation:
To convert from liters to grams, we first need to convert from liters to moles. To do this, we divide the liters by 22.4, the amount of liters of a gas per mole.
32.2 / 22.4
= 1.4375 moles of CO₂
Now we want to convert from moles to grams. To do this, we multiply the moles by the molar mass of CO₂. The total molar mass can be found on the periodic table by adding up the molar mass of carbon (12) and two oxygen (32).
12 + 32 = 44
Now we want to multiply the moles by the molar mass.
1.4375 • 44
= 63.25 grams of CO₂
This is your answer.
Hope this helps!
can someone help me with 5?
Answer:28 g
Explanation:
On the line provided, write the formula for the compound made by each of the following. (3 pts each) 20. Mgand NO3-1, 21. CIO3-1 and K 22. Ca and So4-2, 23. NH4+1 and SO4-2, 24. C2H30-1, ' and H 25. Hand CO, t
Answer:
20. Mg(NO3)2
21. KClO3
22. CaSO4
23. (NH4)2SO4
24. HC2H3O
25. CO is carbon dioxide, which is neutral and would not combine with a hydrogen ion. Did you mean H and CO3(2-)? That would be H2CO3 (carbonic acid)
Explanation:
You have to match the charges of the Cations and Anions.
What volume of 0.215 M HCl is required to neutralize 50.0 mL of
0.800 M NaOH?
Answer:
186 mL HCl
Explanation:
M1V1 = M2V2
M1 = 0.215 M HCl
V1 = ?
M2 = 0.800 M NaOH
V2 = 50.0 mL
Solve for V1 --> V1 = M2V2/M1
V1 = (0.800 M)(50.0 mL) / (0.215 M) = 186 mL HCl
How does the entropy change in the reaction 2C3H6(g) + 9O2(g) → 6C02(g) + 6H2O (g)?
I will mark brainliest!! Thank you so much!!
Answer:
The entropy increases!!!
Explanation:
a pex
The entropy increases in the reaction.
What is entropy?Entropy is defined as the measure of the disorder of a system.Entropy is an extensive property of a thermodynamic system, to put it in simple words, its value changes depending on the amount of matter that is present.Entropy is denoted by the letter S and has units of joules per kelvin (JK−1)The entropy increases in the reaction if the total number of product molecules are greater than the total number of reactant molecules.
2C3H6(g) + 9O2(g) → 6C02(g) + 6H2O (g)
In the above reaction, the product molecules are greater than the reactant molecules. Hence, entropy increases.
Hence, we can conclude that option A is the answer.
To learn more about entropy here
https://brainly.com/question/22861773
#SPJ2
What scientifically goes on with a plasma ball?
Answer:
the electrode at the center of a plasma ball emits a high-frequency,high-voltage alternating electric current. This current flows through the plasma filaments to create colorful tendrils of light.
Explanation:
what colors depend upon the gases used inside the plasma ball? common gases include neon, argon, xenon, and krypton.
A 50.0 mL sample of buffer solution contains 0.120 M acetic acid and 0.150 M sodium acetate. If 5.55 mL of 0.092 M NaOH is added to this solution, identify the resulting number of moles of acetic acid, sodium acetate, and NaOH.
Answer:
see calculations below
Explanation:
Given: HOAc ⇄ [H⁺] + [OAc⁻]
C(i) 0.12M 0M 0.15M
mix => 5.55ml(0.092M NaOH) / (50ml + 5.55ml)
= 0.00555(0.092)mole NaOH / 0.0555 L Soln
= 0.0092M in NaOH is added into the initial buffer solution
= 0.0092M in OH⁻ (NaOH is a strong base => 100% ionized)
Rxn => Addition of 0.0092M OH⁻ will react with 0.0092M H⁺ shifting buffer . equilibrium to the right decreasing [HOAc] and increasing [OAc⁻] by . 0.0092M each.
Therfore ...
Given: HOAc ⇄ [H⁺] + [OAc⁻]
C(i) 0.12M 0M 0.15M
ΔC - 0.0092M +x +0.0092M
C(f) 0.1108M x 0.1592M => New Concentrations . after adding 0.0092M . NaOH
Substituting new acid and ion concentrations into Ka expression ...
Ka = [H⁺][OAc⁻]/[HOAc] = (x)(0.1592M)/(0.1108M) = 1.75 x 10⁻⁵M
=> x = [H⁺](new) = (1.75 x 10⁻⁵M*)(0.1108M)/(0.1592M) = 1.22 x 10⁻⁵M in H⁺ ions
*units of Ka are Molar
FYI => Adding a strong base to a buffer solution will shift pH to more basic.
Adding a strong acid to a buffer solution will shift pH to more acidic.
=> (such is a good way to check that your buffer calculations are correct.)
NOTE => Question asks for moles of HOAc, Na⁺OAc⁻ & NaOH after adding base. Giving answers in terms of Molarity (moles/Liter) is same as moles. Therefore ...
[HOAc] = 0.1108M
[NaOAc] = 0.1592M
[NaOH] = ∅M (from rxn of H⁺ + OH⁻ => H₂O, all NaOH was consumed in acid/base reaction. Remaining are only Na⁺ as a spectator ion and OH⁻ as a function of the new concentration of H⁺ => [OH⁻] = Kw/[H⁺] = 1 x 10⁻¹⁴/1.22 x 10⁻⁵ = 8.2 x 10⁻¹⁰M.
Hope this helps. :-)
The pH of an acidic solution is 5.59. What is [H+]?
Answer:
[H⁺] = 2.57x10⁻⁶ M
Explanation:
This problem can be solved using the definition of pH:
pH = -log[H⁺]
Now we isolate [H⁺] in the equation:
-pH = log[H⁺]
[tex]10^{-pH}[/tex]=[H⁺]
As we are given the pH by the problem, we can now proceed to calculate the [H⁺]:
[H⁺] = [tex]10^{-5.59}[/tex]
[H⁺] = 2.57x10⁻⁶ M
Thus, when the pH of a solution is 5,59; the molar concentration of H⁺ species is 2.57x10⁻⁶.
what are possible source of error for rusting of a nail
Answer:
A nail can rust when exposed to oxygen. the molecules of iron on the surface of the nail exchange atoms with the oxygen in the air and produce a new substance, the reddish brown ferrous oxide i.e rust.
2 NaClO3→ 2 NaCl + 3 O2
How many moles of O2 are produced when 40g of NaCl are formed?
Answer:
75.6
Explanation:
Which example is a mixture? alcohol water mercury orange soda
Answer:
water is an example of mixture