If an object goes through a chemical reaction but does not change in mass that object has

What happens when water is added to quicklime? Write two observations.

Neporo4naja [7]

Answer:

1.Most metal oxides are insoluble in water but some of these (e.g. Na2O.

Explanation:

2.: (i) A hissing sound is observed.

1.ii) The mixture starts boiling and lime water is obtained.

3 0

3 years ago

"You have a solution of glucose in water that has a concentration of 2.50 M and a volume of 0.442 liters. You dilute this soluti

taurus [48]

Answer:

0.737M

Explanation:

C1V1=C2V2

2.50 x 0.442 = C x 1.5

1.105 = C x 1.5

C = 1.105/1.5

C = 0.737M

7 0

3 years ago

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coeffici

Brut [27]

Question:

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K

Answer:

The mean activity coefficient for HCl solution is 0.78.

Explanation:

Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is

$E=E_{0}-0.0514 \mathrm{V} \ln \gamma$

As we know that $E_{0}$ = 0.22 V and E = 0.342 V, the equation will become

$0.342 V+0.0514 V \ln (0.124)=0.22 V-0.0514 V \ln \gamma$

$0.342 V-0.222 V=-0.0514 V(\ln \gamma+\ln 0.124)$

$0.12 \mathrm{V}=-0.0514 \mathrm{V}(\ln \gamma+\ln 0.124)$

$\frac{0.12}{0.0514}=-\ln (0.124 \gamma)$

$-2.3346=\ln (0.124 \gamma)$

$e^{-2.3346}=0.124 \gamma$

$0.0968=0.124 \gamma$

$\gamma=\frac{0.0968}{0.124}=0.78$

So, the mean activity coefficient is 0.78.

6 0

3 years ago

4NH3 (g)+502(g)->6H20g)+4NOg) What is the enthalpy change for this reaction?

Feliz [49]

Enthalpy is energy of bonds broken - energy of bonds formed. Here, the NH3 and O2 are broken and H2O and NO are formed. So the energy to break the NH3 bonds is 3 times the amount of energy it takes to break a N-H single bond (because there are three of them in a NH3 molecule) and then multiplied by 4 because there are four particles.

So the energy of the bonds broken is 12x the energy to break a N-H single bond plus 5x the amount of energy to break an O—O double bond (you don’t multiply this by anything because in each O2 molecule there is only one bond).

The energy of the bonds formed is 6*2 = 12 Times the amount of energy for a O-H single bond plus 4 times the amount of energy required to break a N—O double bond.

Subtract energy of bonds broken - energy of bonds formed and this is the change in enthalpy.

To know what type of bond it is, draw the Lewis structure.

7 0

4 years ago

The picture shows a contractile vacuole of a unicellular freshwater organism. The contractile vacuole regulates the flow of wate

yuradex [85]

Answer:

A.The concentration of water is greater outside the cell than inside the cell.

Explanation:

The contractile vacuole of certain organisms functions to regulate water flow in and out of the cell. It does this by storing excess water that comes into the cell. In the case of this organism with a filled up contractile vacuole, it means water is flowing into the cell.

Naturally, water will flow into a living cell when an osmotic gradient i.e. difference in concemtration, has been created between intracellular and extracellular solutions. Osmosis involves movement of substances from a region of high water concentration to a region of low water concentration. This means that if water is flowing into the cell, which is stored by the contractile vacuole, the concentration of water must be greater outside the cell than inside.

6 0

3 years ago