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Dmitry [639]
3 years ago
8

⚠️⚠️ ⚠️⚠️

Chemistry
2 answers:
kolbaska11 [484]3 years ago
8 0

<em />

_____________________________________________________

おもりを垂直に持ち上げて、重力ポテンシャルエネルギーを取得します。おもりを垂直に持ち上げて、重力ポテンシャルエネルギーを取得します。力が取り除かれると、物体は地面に落下し、重力ポテンシャルエネルギーが落下する物体の運動エネルギーに伝達されます。私が助けて、良い一日を過ごせることを願っています!

Anastaziya [24]3 years ago
3 0

Answer/Explanation: A weight lifted vertically to acquire gravitational potential energy. A weight lifted vertically to acquire gravitational potential energy. If the force were to be removed, the object would fall back down to the ground and the gravitational potential energy would be transferred to kinetic energy of the falling object.

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Lyrx [107]
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5 0
2 years ago
Between which two pints does the object have the greatest speed please help
Akimi4 [234]
The answer is A to B because the distance is rising rapidly as seen by the steep slope segment A to B had
5 0
3 years ago
Calculate the standard free-energy change at 25 ∘C for the following reaction:
lianna [129]

Answer:

Standard free-energy change at 25^{0}\textrm{C} is -3.80\times 10^{2}kJ/mol

Explanation:

Oxidation: Mg(s)-2e^{-}\rightarrow Mg^{2+}(aq.)

Reduction: Fe^{2+}(aq.)+2e^{-}\rightarrow Fe(s)

--------------------------------------------------------------------------------------

Overall: Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)

Standard cell potential, E_{cell}^{0}=E_{Fe^{2+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}

So, E_{cell}^{0}=(-0.41V)-(-2.38V)=1.97V

We know, standard free energy change at 25^{0}\textrm{C}(\Delta G^{0}): \Delta G^{0}=-nFE_{cell}^{0}

where, n is number of electron exchanged during cell reaction, 1F equal to 96500 C/mol

Here n = 2

So, \Delta G^{0}=-(2)\times (96500C/mol)\times (1.97V)=-380210J/mol=-380.21kJ/mol=-3.80\times 10^{2}kJ/mol

8 0
3 years ago
Read 2 more answers
In which reaction does the oxidation number of hydrogen change? In which reaction does the oxidation number of hydrogen change?
dedylja [7]

<u>Answer:</u> The correct answer is 2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)

<u>Explanation:</u>

Oxidation number is defined as the number which is given to an atom when it looses or gains electron. When an atom looses electron, it attains a positive oxidation state. When an atom gains electron, it attains a negative oxidation state.

Oxidation state of the atoms in their elemental state is considered as 0. Hydrogen is present as gaseous state.

For the given chemical reactions:

  • <u>Reaction 1:</u>  2HClO_4(aq.)+CaCO_3(s)\rightarrow Ca(ClO_4)_2(aq.)+H_2O(l)+CO_2 (g)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

  • <u>Reaction 2:</u>  CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(s)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

  • <u>Reaction 3:</u>  HCl(aq.)+NaOH(aq.)\rightarrow NaCl(aq.)+H_2O(l)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

  • <u>Reaction 4:</u>  2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: 0

Thus, the oxidation state of hydrogen is changing.

  • <u>Reaction 5:</u>  SO_2(g)+H_2O(l)\rightarrow H_2SO_3(aq.)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

Hence, the correct answer is 2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)

6 0
3 years ago
Calculate the relative density of carbon monoxide and carbon dioxide with respect to air.
JulsSmile [24]
Density is proportional to molar mass, assuming pressure and temperature remain constant. Therefore, since CO has a molar mass of 28 and CO2 has a molar mass of 44:
The relative density of CO vs air is 28/29 = 0.9655.
The relative density of CO2 vs air is 44/29 = 1.517.

3 0
3 years ago
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