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telo118 [61]
3 years ago
14

What is the concentration of a solution if 0.450 mol of NaCl dissolved in 750 mL of water?

Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
3 0

Answer:

.6mol/L

Explanation:

molarity = number of moles / volume of solvent (in L)

750mL / 1000mL/L = .75L

M = .450mol / .75L

M = .6mol/L

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The enthalpy of vaporization of water is 2,257,000 J/kg. If I have a 1 kg sample, how much energy is needed to boil all of it
worty [1.4K]

Answer:

2257000 J

Explanation:

Applying,

Q = Cₓm.................. Equation 1

Where Q = amount of energy need to boil the water, Cₓ = Enthalpy of vaporization of water, m = mass of water.

From the question,

Given: Cₓ = 2257000 J/kg, m = 1 kg

Substitute these values into equation 1

Q = 2257000×1

Q = 2257000 J

Hence the energy needed to boil all of the water is 2257000 J

8 0
3 years ago
The properties of protons, neutrons and electrons.
Mashutka [201]
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I hope this helps! :)


7 0
3 years ago
Why is Potassium not used in school laboratory
densk [106]
<h3>Because it is harmful for school environment.</h3>

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<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em>

6 0
3 years ago
Read 2 more answers
The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.
mihalych1998 [28]

<u>Answer:</u> The potential of the given cell is 1.551 V

<u>Explanation:</u>

The given chemical cell follows:

Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)

<u>Oxidation half reaction:</u> Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V

<u>Reduction half reaction:</u> Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V       ( × 2)

<u>Net cell reaction:</u> Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.799-(-0.762)=1.561V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}

where,

E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

[Zn^{2+}]=0.125M

[Ag^{+}]=0.240M

Putting values in above equation, we get:

E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})

E_{cell}=1.551V

Hence, the potential of the given cell is 1.551 V

6 0
4 years ago
Bisect tension can only be determined for a corner pole that has conductor spans of equal length on either side.
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A square mile = 27,878,400 square feet

A square mile also = 640 acres.

31,580/.
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2 years ago
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