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soldi70 [24.7K]
4 years ago
11

How does water vapor get into the atmosphere?

Chemistry
1 answer:
Brums [2.3K]4 years ago
6 0

Answer:

through the water cycle

Explanation:

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Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or
Ahat [919]

Answer :

Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

Rules for Oxidation Numbers are :

  • The oxidation number of a free element is always zero.
  • The oxidation number of a monatomic ion equals the charge of the ion.
  • The oxidation number of  Hydrogen (H)  is +1, but it is -1 in when combined with less electronegative elements.
  • The oxidation number of  oxygen (O)  in compounds is usually -2.
  • The oxidation number of a Group 17 element in a binary compound is -1.
  • The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
  • The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

Now we have to determine the oxidation state of the elements in the compound.

(a) H_2SO_4

Let the oxidation state of 'S' be, 'x'

2(+1)+x+4(-2)=0\\\\x=+6

Hence, the oxidation state of 'S' is, (+6)

(b) Ca(OH)_2

Let the oxidation state of 'Ca' be, 'x'

x+2(-2+1)=0\\\\x=+2

Hence, the oxidation state of 'Ca' is, (+2)

(c) BrOH

Let the oxidation state of 'Br' be, 'x'

x+(-2)+1=0\\\\x=+1

Hence, the oxidation state of 'Br' is, (+1)

(d) ClNO_2

Let the oxidation state of 'N' be, 'x'

-1+x+2(-2)=0\\\\x=+5

Hence, the oxidation state of 'N' is, (+5)

(e) TiCl_4

Let the oxidation state of 'Ti' be, 'x'

x+4(-1)=0\\\\x=+4

Hence, the oxidation state of 'Ti' is, (+4)

(f) NaH

Let the oxidation state of 'Na' be, 'x'

x+(-1)=0\\\\x=+1

Hence, the oxidation state of 'Na' is, (+1)

4 0
3 years ago
Which professional most likely uses x-ray technology a financial analyst or a dentist
amid [387]

Answer:

The professionals that most likely use X-ray technology are a dentist and an airport-security scanner.

A dentist would scan your teeth using X-ray technology to see whether there are any cavities within the teeth themselves which he couldn't otherwise see with the naked eye. An airport-security scanner would use this technology to scan your luggage to check whether there are any weapons and other illegal products.

Explanation:

5 0
3 years ago
Read 2 more answers
Can someone help me with that question
Katen [24]

Answer:

Gene - determines the trait (phenotype)

6 0
3 years ago
Silver (ag) has a molar mass of 107. 8682 g, nitrogen has a molar mass of 14. 0067 g, and oxygen has a molar mass of 15. 9994 g.
hjlf

Answer:

M(AgNO3) = 169,8731 g mol

Explanation:

M(Ag) = 107,8682 g mol

M(N) = 14.0067 g mol

M(O) = 15.9994 g mol

M(AgNO3) = M(Ag) + M(N) + 3 x M(O) = 107.8682 + 14.0067 + 3 x 15.9994 =

= 169,8731 g mol

Just sum up all the respective elements' molar mass.

3 0
2 years ago
A galvanic cell at a temperature of 42 degrees Celcius is powered by the following redox reaction:
seropon [69]

<u>Answer:</u> The cell voltage of the given reaction is 1.86 V

<u>Explanation:</u>

The given chemical equation follows:

3Cu^{2+}(aq.)+2Al\rightarrow 2Al^{3+}(aq.)+2Au(s)

<u>Oxidation half reaction:</u> Al(aq.)\rightarrow Al^{3+}(aq.)+3e^-;E^o_{Al^{3+}/Al}=1.66V       ( × 2)

<u>Reduction half reaction:</u> Cu^{2+}(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.16V       ( × 3)

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}=0.16-(-1.66)=1.82V

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

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3}

where,

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

E^o_{cell} = standard electrode potential of the cell = +1.82 V

n = number of electrons exchanged = 2

R = Gas constant = 8.314 J/mol Kl

T = temperature = 42^oC=[42+273]K=315K

F = Faraday's constant = 96500

[Al^{3+}]=1.63M

[Cu^{2+}]=3.43M

Putting values in above equation, we get:

E_{cell}=1.82-\frac{2.303\times 8.314\times 315}{2\times 96500}\times \log(\frac{(1.63)^2}{(3.43)^3})\\\\E_{cell}=1.86V

Hence, the cell voltage of the given reaction is 1.86 V

4 0
3 years ago
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