Answer:
d. van der waals force
Explanation:
Van der Waals force :
the weakest intermolecular forceand consist of dipole-dipole force and dispersion force.
If a student drops 2.3g pieces of magnesium into a flask of hydrochloric acid, this reaction occurs: Mg + 2HCl= MgCl2 + H2
How many liters of hydrogen can be produced at a pressure of 2 atm and a temperature of 298 K
Answer:
1.2 L
Explanation:
Step 1: Write the balanced equation
Mg + 2 HCl ⇒ MgCl₂ + H₂
Step 2: Calculate the moles corresponding to 2.3 g of Mg
The molar mass of Mg is 24.31 g/mol.
2.3 g × 1 mol/24.31 g = 0.095 mol
Step 3: Calculate the moles of H₂ produced
0.095 mol Mg × 1 mol H₂/1 mol Mg = 0.095 mol H₂
Step 4: Calculate the volume occupied by the hydrogen
We will use the ideal gas equation.
P × V = n × R × T
V = n × R × T/P
V = 0.095 mol × (0.0821 atm.L/mol.K) × 298 K/2 atm = 1.2 L
compound of aspartame is a dipeptide that is often used as a sugar substitute which functional groups are present
Answer:
Carboxyl, primary amine, amide, ester, and phenyl.
Explanation:
The functional groups present in the compound of aspartame are carboxyl, primary amine, amide, ester, and phenyl. Aspartame is an artificial non-saccharide sweetener which is 200 times sweeter than sucrose. This aspartame is commonly used as a sugar substitute in many foods and beverages. It has the trade names such as NutraSweet, Equal, and Canderel.
(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms
Answer:
(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms
Explanation:
Alcohols are poor leaving groups.
To make -OH group a better-leaving group, it should be treated with sulfonyl chlorides.
Then, methane sulfonyl group makes will be substituted on the -OH group and forms sulfonyl esters and makes it a better leaving group.
After that treating with KI proceeds through nucleophilic bimolecular substitution and the final product formed is shown below:.
What is the phase of water at 0.25 atm and 0°C?
Water
(liquid)
Pressure (atm)
0.5-
0.25
Ice
(solid)
Water vapor
(gas)
0
000
Temperature (°C)
O A. Gas
O B. Solid and gas
O C. Solid and liquid
D. Solid
Water is in the solid phase at 0.25 atm and 0°C.
In what phase is water at 25?A pressure of 50 kPa and a temperature of 50 °C correspond to the “water” region—here, water exists only as a liquid. At 25 kPa and 200 °C, water exists only in the gaseous state.
What phase is water in at 0 C?Under standard atmospheric conditions, water exists as a liquid. But if we lower the temperature below 0 degrees Celsius, or 32 degrees Fahrenheit, water changes its phase into a solid called ice.
Learn more about the solid phase here https://brainly.com/question/13396621
#SPJ2
Suppose a 0.042M aqueous solution of phosphoric acid (H3PO4) is prepared. Calculate the equilibrium molarity of HPO4^−2.
Answer:
2.89x10⁻⁵M = [HPO₄²⁻]
Explanation:
The equilibrium of H3PO4 in water occurs H2PO4-:
H3PO4(aq) + H2O(l) ⇄ H3O⁺(aq) + H2PO4⁻(aq)
pKa = 2.16. And as pKa = -log Ka; Ka = 10^-2.16
Ka = 6.9183x10⁻³ = [H3O⁺] [H2PO4⁻] / [H3PO4]
As both [H3O⁺] and [H2PO4⁻] comes from the same equilibrium,
[H3O⁺]=[H2PO4⁻] :
[H3O⁺] = X
[H2PO4⁻] = X
[H3PO4] = 0.042 - X
Where X is reaction coordinate
Replacing:
6.9183x10⁻³ = [X] [X] / [0.042 - X]
6.9183x10⁻³ = X² / 0.042 - X
2.905686x10⁻⁴ - 6.9183x10⁻³X - X² = 0
Solving for X:
X = -0.02M. False solution. There is no negative concentration.
X = 0.014M. Right solution
[H2PO4⁻] = 0.014M
In the second equilibrium:
H2PO4⁻(aq) + H2O(l) ⇄ HPO4-(aq) + H3O+(aq)
Based on the same principles of the last equilibrium:
pKa2 = 7.21
Ka2 = 6.166x10⁻⁸ = [HPO4-] [H3O+] / [H2PO4⁻]
[HPO4-] = X
[H3O+] = X
[H2PO4⁻] = 0.014M - X
6.166x10⁻⁸ = X² / [0.014M - X]
8.3623x10⁻¹⁰ - 6.166x10⁻⁸X - X² = 0
Solving for X:
X = -0.0000289485. False solution.
X =
2.89x10⁻⁵M = [HPO₄²⁻]In a single displacement reaction Zinc can displace ALL but…
Iron
Nickel
Calcium
Lead
Answer:
Calcium
Explanation:
Zinc cannot displace Ca because calcium is above it in the reactivity series
What is the volume of a gas 622.7 mL at 25.1 C if the temperature is increased to 60.7 C without changing the pressure, what is the new volume of the gas?
A. 697
B. 556
C. 1510
D. 9.35
Answer:
A) 697 mL
Explanation:
First convert degrees Celsius to Kelvin.
That gives;
For T1; temp 1 (273 + 25.1)
= 298.1K
T2; temp 2 (273+60.7)
=333.7K
From the formula, (V1/T1) = (V2/T2)
To find V2 = (V1/T1) × T2
= (622.7/298.1) × 333.7.
= 697.06mL
~~ 697mL
Consider reaction AgCIO3(aq)+Mgl2(aq)
Answer:
the product is Mg(Clo3)2 + AgI
An unknown weak acid with a concentration of 0.530 M has a pH of 5.600. What is the Ka of the weak acid
Answer:
Ka = 3.45x10⁻⁶
Explanation:
First we calculate [H⁺], using the given pH:
pH = -log[H⁺][H⁺] = [tex]10^{-pH}=10^{-5.6}[/tex] [H⁺] = 2.51x10⁻⁶ MTo solve this problem we can use the following formula describing a monoprotic weak acid:
[H⁺] = [tex]\sqrt{C*Ka}[/tex]We input the data that we already know:
2.51x10⁻⁶ = [tex]\sqrt{0.530*Ka}[/tex]And solve for Ka:
Ka = 3.45x10⁻⁶