What is a shortened definition for chemistry?

Please help with this question.

marysya [2.9K]

Answer:

It is C.

Explanation:

4 0

3 years ago

when u walk your dog ur using energy from the sunlight to power this activity explain (ANSWER PLESSSSSSSSSSSSSSZZZ)

wariber [46]

Answer:

When you walk your dog, you are using energy from the sunlight to power the activity. Explain. The energy you use is transferred from glucose, which got it directly from sunlight.

Explanation:

4 0

3 years ago

The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi

avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

$Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)$

The balanced two-half reactions will be,

Oxidation half reaction : $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction : $Ni^{2+}+2e^-\rightarrow Ni$

The expression for reaction quotient will be :

$Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(0.0141)}{(0.00104)}=13.6$

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

$E^o_{cell}$ = standard electrode potential of the cell = 0.51 V

$E_{cell}$ = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

$E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)$

$E_{cell}=0.47V$

Therefore, the actual cell potential of the cell is 0.47 V

4 0

3 years ago

Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co

amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as $p_{d}$ and the proportion of air that is by-passed as $p_{bp}$ , being $p_{d}+p_{bp}=1$ .

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

$0.004 = 0.016*p_{bp}+ 0.002*p_{d}$

Replacing the first equation in the second one we have

$0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143$

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

$100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW$

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

7 0

4 years ago

What is the name of the acid whose formula is hi? spell out the full name of the acid?

serious [3.7K]

The name of the acid whose formula is Hi is Hydrogen iodide

6 0

3 years ago