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pochemuha
3 years ago
12

A student collected a total of 36.04 g of water from a reaction. What number of moles of water does this represent

Chemistry
1 answer:
umka21 [38]3 years ago
4 0
To find the number of moles from a mass given, simply look to the formula n (moles) = m (mass, g) / MM (molar mass).
Mass was given, 36.04
Molar mass is the total atomic mass of all the atoms present. Water is H20, so that means 2 hydrogen and 1 oxygen. The atomic mass of hydrogen is 1 and atomic mass of oxygen is 16. Therefore MM= 1 + 1 + 16= 18.
Plug that value in and the full equation is
n = 36.04/18
n = 2.002 moles
= 2 moles
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6.201cm+7.4cm+0.68 cm+12.0cm
zimovet [89]
If you want the answer is centimeters is it going to be:
26.281 cm
If it make it easier for you to solve add 6.201 and 7.4 which will equal 13.601. Then add .68 to 13.601 which equals 14.281. Last add 12 to 14.281 which equals 26.281.

Hope this helps.

8 0
3 years ago
Which explains the information needed to calculate speed and velocity?
USPshnik [31]

Answer:

Both require time, but velocity requires displacement and speed requires distance

Explanation:

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But for calculating the velocity we require time as well as displacement because velocity is defined as the displacement per unit time and as velocity is a vector quantity it has direction

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8 0
3 years ago
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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
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