Acids are defined as proton donors since they release hydrogen ions. a. True b. False

Match the solutions to the descriptions of the freezing points.a. One mole of the ionic compound Na3PO4 dissolved in 1000 g H2O

rosijanka [135]

Explanation:

Depression in Freezing point

= Kf × i × m

where m is molality , i is Van't Hoff factor, m = molality

Since molality and Kf remain the same

depression in freezing point is proportional to i

i= 2 for CuSO4 ( CuSO4----------> Cu+2 + SO4-2

i=1 for C2h6O

i= 3 for MgCl2 ( MgCl2--------> Mg+2+ 2Cl-)

So the freezing point depression is highest for MgCl2 and lowest for C2H6O

so freezing point of the solution = freezing point of pure solvent- freezing point depression

since MgCl2 has got highest freezing point depression it will have loweest freezing point and C2H6O will have highest freezing point

5 0

3 years ago

PLEASE HELP ME ASAPPP!?

Brrunno [24]

lets guess and say #4???

7 0

3 years ago

Which of the following chemical equations proves that the law of conservation of mass is in effect?

bogdanovich [222]

Answer:

2KCl + F₂ → 2KF + Cl₂

Explanation:

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

2KCl + F₂ → 2KF + Cl₂

In this equation mass of reactant and product is equal. There are 2 potassium 2 chlorine and fluorine atoms on both side of equation it means mass remain conserved.

All other options are incorrect because mass is not conserved.

Mg₂ + LiBr ---> LiMg + Br

In this equation mass of magnesium is more on reactant side.

Na +O₂ ---> Na₂O

In this equation there is more oxygen and less sodium on reactant side while there is more sodium and less oxygen on product side.

H₂O ---> H₂ + O₂

In this equation there is less oxygen on reactant side while more oxygen on product side.

5 0

3 years ago

Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,

Serhud [2]

The question is incomplete. Her eis the complete question.

Steam reforming methane (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: $K_{c}$ = 2. $10^{-2}$

Explanation: The reaction for steam reforming methane is:

$CH_{4} + H_{2}O$ ⇒ $CO_{} + 3H_{2}$

To calculate the concentration equilibrium constant, first calculate the molarity ( $\frac{mol}{L}$ ) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

Molarity of CH4:

CH4 = $\frac{20}{125}$ = 0.16M

Molarity of H20:

H2O = $\frac{10}{125}$ = 0.08M

At final temperature:

Molarity of H2:

H2 = $\frac{18}{125}$ = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

Molarity of CO:

CO = $\frac{0.144}{3}$ = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

Molarity of CH4:

CH4 = 0.16 - 0.048 = 0.112M

Molarity of H2O:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

$K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }$

$K_{c}$ = $\frac{0.048.0.144^{3} }{0.112.0.032}$

$K_{c}$ = 2. $10^{-2}$

The concentration equilibrium constant for the process is $K_{c}$ = 2. $10^{-2}$ .

4 0

3 years ago

Calculate the osmotic pressure of 20 m solution of LiCl at 25C

Soloha48 [4]

Answer:

979 atm

Explanation:

To calculate the osmotic pressure, you need to use the following equation:

π = i MRT

In this equation,

-----> π = osmotic pressure (atm)

-----> i = van't Hoff's factor (number of dissolved ions)

-----> M = Molarity (M)

-----> R = Ideal Gas constant (0.08206 L*atm/mol*K)

-----> T = temperature (K)

When LiCl dissolves, it dissociates into two ions (Li⁺ and Cl⁻). Therefore, van't Hoff's factor is 2. Before plugging the given values into the equation, you need to convert Celsius to Kelvin.

i = 2 R = 0.08206 L*atm/mol*K

M = 20 M T = 25°C + 273.15 = 298.15 K

π = i MRT

π = (2)(20 M)(0.08206 L*atm/mol*K)(298.15 K)

π = 979 atm

5 0

2 years ago