Select all of the following statements which are true of water.

Which of the following processes would you predict to have an increase in entropy?

Anna [14]

C. Melting ice.
It is C because melting ice is a change of state from solid to liquid which requires an addition of energy(or entropy) into the system.
Condensation of water occurs from a gas to a liquid state, which takes energy out of the system(water) and gives it to the surroundings(air around it). Freezing water is the same as condensation except for the state change. Deposition is simply gas to a solid instantaneously so you can again see it as with the other two examples.

6 0

3 years ago

What is the [H+] if the pH is 13.6?

MrRa [10]

Answer:

8???

Explanation:

i honestly don't know???

8 0

3 years ago

Read 2 more answers

Chlorine reacts with sodium and with hydrogen.

Sonbull [250]

Answer:

Sodium Chloride has Ionic bond while Hydrogen Chloride has covalent bond.

Explanation:

Na has 11 electrons (2, 8, 1) and need to give away 1 electron to be stable

Cl has 17 electrons ( 2, 8, 7) and needs 1 electron to be stable.

Na transfers 1 electron to CL to form Ionic bond.

While

Hydrogen has 1 electron and shares with Chlorine to be stable.

Covalent bond involves sharing.

3 0

3 years ago

In this experiment we will be using a 0.05 M solution of HCl to determine the concentration of hydroxide (OH-) in a saturated so

gulaghasi [49]

<u>Answer:</u> The moles of hydroxide ions present in the sample is 0.0008 moles

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl.

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.05M\\V_1=16mL\\n_2=2\\M_2=?M\\V_2=36.0mL$

Putting values in above equation, we get:

$1\times 0.05\times 16=2\times M_2\times 36\\\\M_2=\frac{1\times 0.05\times 16}{2\times 36}=0.011M$

To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of solution = 0.011 M

Volume of solution = 36.0 mL

Putting values in above equation, we get:

$0.011=\frac{\text{Moles of }Ca(OH)_2\times 1000}{36}\\\\\text{Moles of }Ca(OH)_2=\frac{0.011\times 36}{1000}=0.0004mol$

1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.

Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles

Hence, the moles of hydroxide ions present in the sample is 0.0008 moles

8 0

3 years ago

Two objects are brought into contact Object 1 has mass 0.76 kg, specific heat capacity 0.87) g'c and initial temperature 52.2 'C

taurus [48]

Answer:

$T_F=77.4\°C$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

$Q_1=-Q_2$

In terms of mass, specific heat and temperature change is:

$m_1C_1(T_F-T_1)=-m_2C_2(T_F-T_2)$

Now, solve for the final temperature, as follows:

$T_F=\frac{m_1C_1T_1+m_2C_2T_2}{m_1C_1+m_2C_2}$

Then, plug in the masses, specific heat and temperatures to obtain:

$T_F=\frac{760g*0.87\frac{J}{g\°C} *52.2\°C+70.7g*3.071\frac{J}{g\°C}*154\°C}{760g*0.87\frac{J}{g\°C} +70.7g*3.071\frac{J}{g\°C}} \\\\T_F=77.4\°C$

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.

Regards!

4 0

3 years ago