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Zielflug [23.3K]
3 years ago
8

A student has a mass of 90 kg what would his weight be on mars if the acceleration due to gravity is 3.7 ?

Chemistry
1 answer:
Mumz [18]3 years ago
3 0

Answer:

333N

Explanation:

Once we have the object's mass, we can find the weight by multiplying it by the gravitational force, which it is subject to. Being that Mars has a gravitational force of 3.7m/s2, we multiply the object's mass by this quantity to calculate an object's weight on mars.

So an object or person on Mars would weigh 37.83% its weight on earth. Therefore, a person would be much lighter on mars. Conversely, a person is 62.17% heavier on earth than on Mars.

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The right to refuse work that could affect their health and safety and that of others.

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For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia
Fittoniya [83]

Answer : The value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

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

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

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o_{(Cu^{2+}/Cu)} of the reaction, we use the equation:

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

E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)

E^o_{(Cu^{2+}/Cu)}=0.34V

Hence, the value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

8 0
3 years ago
What is the molecular formula of a compound with the empirical formula
ipn [44]

Answer:  C_2H_2O_2

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

The empirical formula is CHO

The empirical weight of CHO = 1(12)+1(1)+1(16)= 29 g.

The molecular weight = 60 g/mole

Now we have to calculate the molecular formula:

n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{60}{29}=2

The molecular formula will be=2\times CHO=C_2H_2O_2

Thus molecular formula will be C_2H_2O_2

5 0
3 years ago
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