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Airida [17]
4 years ago
8

The men's world record for running a mile outdoors (as of 1999) is 3 min 43.13 s. At this rate, how long would it take to run a

2 km race (1 mi =1609 m)?
Chemistry
1 answer:
Leno4ka [110]4 years ago
8 0

Answer:

4 min 37.35 seconds

Explanation:

Given:

Time taken to run 1 mile = 3 min 43.13 seconds

1 min = 60 seconds

thus,

3 min = 180 seconds

therefore total time taken to run 1 mile = 180 + 43.13 = 223.13 seconds

Now,

1 mi = 1609 m

1 km = 1000 m

thus,

1 mi = 1.609 Km

or

1 km = 0.6215 mi

therefore,

2 km = 0.6215 × 2 = 1.243 mi

hence, the time taken to run 2 km (1.243 mi) = 223.13 × 1.243 = 277.35 seconds = 4 min 37.35 seconds

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Answer:

hope it helps you....

Explanation:

natural and social environment.

5 0
2 years ago
A voltaic cell made of a Cr electrode in a solution of 1.0 M in Cr3+ and a gold electrode in a solution that is 1.0 M in Au3+.
Vikki [24]

Answer: a) Anode: Cr\rightarrow Cr^{3+}+3e^-

Cathode: Au{3+}+3e^-\rightarrow Au

b) Anode : Cr

Cathode : Au

c) Au^{3+}+Cr\rightarrow Au+Cr^{3+}

d) E_{cell}=2.14V

Explanation: - 

a) The element Cr with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element Au with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.

At cathode: Au{3+}+3e^-\rightarrow Au

At anode: Cr\rightarrow Cr^{3+}+3e^-

b) At cathode which is a positive terminal, reduction occurs which is gain of electrons.

At anode which is a negative terminal, oxidation occurs which is loss of electrons.

Gold acts as cathode ad Chromium acts as anode.

c) Overall balanced equation:

At cathode: Au{3+}+3e^-\rightarrow Au     (1)

At anode: Cr\rightarrow Cr^{3+}+3e^-        (2)

Adding (1) and (2)

Au^{3+}+Cr\rightarrow Au+Cr^{3+}

d)E^0_(Cr^{3+}/Cr)= -0.74 V

E^0_(Au^{3+}/Au)= 1.40 V  

E^0{cell}=E^0{cathode}-E^0{anode}=1.40-(-0.74)=2.14V

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Au^{3+}]}{[Cr^{3+}]^}

where,

n = number of electrons in oxidation-reduction reaction = 3

E^o_{cell} = standard electrode potential = 2.14 V

E_{cell}=2.14-\frac{0.0592}{3}\log \frac{[1.0}{[1.0]}

E_{cell}=2.14

Thus the standard potential for an electrochemical cell with the cell reaction is 2.14 V.

6 0
4 years ago
Where do meteorologists obtain the data they use to make a <br> weather forecast?
andre [41]

Answer:

doppler radar, radiosondes, weather satellites, buoys

8 0
3 years ago
In which orbitals would the valence electrons for carbon (C) be placed?
Alexxandr [17]

Answer: orbitals supernumerary

Explanation:

8 0
4 years ago
If you had a 100 mL of a solution of 0.01 M NaF, how many moles would that solution contain?
Daniel [21]

Answer:

0.001 mole of NaF.

Explanation:

From the question given above, the following data were obtained:

Volume of solution = 100 mL

Molarity = 0.01 M

Mole of NaF =?

Next, we shall convert 100 mL to litre (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

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100 mL = 0.1 L

Thus, 100 mL is equivalent to 0.1 L.

Finally, we shall determine the number of mole of NaF present in the solution. This can be obtained as follow:

Volume of solution = 0.1 L

Molarity = 0.01 M

Mole of NaF =?

Molarity =mole /Volume

0.01 = mole of NaF / 0.1

Cross multiply

Mole of NaF = 0.01 × 0.1

Mole of NaF = 0.001 mole.

Thus, 0.001 mole of NaF is present in 100 mL of the solution.

5 0
3 years ago
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