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Luda [366]
3 years ago
11

What are the Group 1A and Group 7A elements examples of? my last wuestion

Chemistry
2 answers:
avanturin [10]3 years ago
8 0

Group 1A  and Group 7A  elements are examples  of  representative  elements

    <u><em>Explanation</em></u>

  • Representative element  - are element in the first two families that is group 1 and 11 element on the far left  of periodic table.
  • In addition  representative element include  the last 6 families  on  the right  side of periodic table.
  •   The representative element include s  and p  blocks element of periodic table.

alisha [4.7K]3 years ago
4 0
Representative elements
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a length of #8 copper wire (radius = 1.63 mm) has a mass of 24.0 Kg and a resistance of 2.061 Ohm per Km. What is the over all r
lubasha [3.4K]
V = m / ρ 
<span>V = (23,200/8.96) × 1000 </span>
<span>V = 2,589,285.71 mm³ </span>

<span>Now we have volume and given the radius we can solve for its length. </span>
<span>L = V/πr² </span>
<span>L = (2,589,285.71/(π×1.63²))/1×10^6 </span>
<span>L = 0.31 km </span>

<span>So </span>
<span>R = rL </span>
<span>R = 2.061(0.31) </span>
<span>R = .6389 Ω </span>

<span>End.</span>
6 0
3 years ago
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How Difficult or easy to obtain geothermal
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6 0
3 years ago
Ella has a mass of 56 kg, and Tyrone has a mass of 68 kg. Ella is standing at the top of a skateboard ramp that is 1.5 meters ta
Ket [755]

Answer:

The correct answer is the last statement that if Tyrone stands at the top of a two meter high ramp, the potential energy of Tyrone will be greater in comparison to Ella.

Explanation:

The energy possessed by an object due to its position is termed as its potential energy. The formula for calculating potential energy is,  

Potential energy = mgh, where m is mass, g is gravity, and h is height.  

The respective masses of Ella and Tyrone as mentioned in the question is 56 Kg and 68 Kg. By looking at all the conditions as mentioned in the given options we get,  

1) When both Ella and Tyrone are standing at the same ramp height:  

Potential energy of Ella = 56*9.8*1.5 = 823.2J (The value of g is 9.8 m.s^-2)

Potential energy of Tyrone = 68*9.8*1.5 = 999.6J

Thus, when both are standing at the same ramp height, the potential energy of Tyrone will be more than Ella.  

2) When Tyrone stands at the top of 1 m high ramp, and Ella at 1.5 m:  

Potential energy of Ella = 56*9.8*1.5 = 823.2J

Potential energy of Tyrone = 68*9.8*1 = 666.4J

Thus, the potential energy of Tyrone will be less in comparison to Ella.  

3) If Tyrone stands at the top of 2 m high ramp and Ella at 1.5 m:  

Potential energy of Ella = 56*9.8*1.5 = 823.2J

Potential energy of Tyrone = 68*9.8*2 = 1332.8J

Thus, the potential energy of Tyrone will become more than Ella. Hence, the condition mentioned in the last statement is correct.  

5 0
3 years ago
Read 2 more answers
Methane (ch4) is a greenhouse gas. how many hydrogen atoms are in 1.4 moles of methane?
kherson [118]

The answer is:

There are 3.6 X 10^24 hydrogen atoms.

The explanation:

1.4 moles of methane will contain 1.4*Avogadro's number

                                       = 1.4*(6.02 X 10^23) =  8.4 X 10^23 methane molecules. But, since the formula for methane is CH4, there are 4 hydrogen atoms for each methane molecule. Thus, 1.4 moles of methane contain

4*(8.4 X 10^23) = 3.6 X 10^24 hydrogen atoms.

8 0
4 years ago
Read 2 more answers
3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH− (aq) → 3 Ni(s) + 2 CrO42−(aq) + 8 H2O(l) ΔG∘ = +87 kJ/mol Given the standard reduction potent
Tom [10]

Answer : The standard reduction potential, E^o_{(Cr^{+6}/Cr^{+3})} is -0.13 V.

Solution : Given,

E^o_{(Ni^{2+}/Ni)}=-0.28V

\Delta G^o=+87KJ/mole=+87000J/mole       (1 KJ = 1000 J)

The net reaction is,

3Ni^{2+}(aq)+2Cr(OH)_3(s)+10OH^-(aq)\rightarrow 3Ni(s)+2CrO^{2-}_4(aq)+8H_2O(l)

The half cell reactions are :

At cathode : Ni^{2+}(aq)+2e^-\rightarrow Ni(s)       E^o_{(Ni^{2+}/Ni)}=-0.28V

At anode : CrO^{2-}_4(aq)+4H_2O(l)+3e^-\rightarrow Cr(OH)_3(s)+5OH^-(aq)  E^o_{(Cr^{+6}/Cr^{+3})}=?

First we have to calculate the E^o_{cell} by using formula,

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = Gibbs's free energy

n = number of electrons in a net chemical reaction = 6 electrons

F = Faraday constant = 96485 C

E^o_{cell} = standard cell potential

Now put all the given values in this formula, we get

+87000KJ/mole=-6\times (96485)\times E^o_{cell}\\E^o_{cell}=-0.15V

Now we have to calculate the E^o_{(Cr^{+6}/Cr^{+3})} by using formula,

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

E^o_{cell}=E^o_{(Ni^{2+}/Ni)}-E^o_{(Cr^{+6}/Cr^{+3})}

Now put all the given values in this formula, we get

-0.15V=-0.28V-E^o_{(Cr^{+6}/Cr^{+3})}

E^o_{(Cr^{+6}/Cr^{+3})}=-0.13V

Therefore, the standard reduction potential, E^o_{(Cr^{+6}/Cr^{+3})} is -0.13 V.

4 0
4 years ago
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