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2 years ago

In what way is the planet Uranus unique

1 answer:

6 0

It is unique for many many ways:
1.)Unlike the other planets of the solar system, Uranus is tilted so far that it essentially orbits the sun on its side, with the axis of its spin nearly pointing at the star. This unusual orientation might be due to a collision with a planet-size body, or several small bodies, soon after it was formed

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What is potential energy? Give an example please.

FromTheMoon [43]

Answer:

Potential energy can be defined as the energy in a body due to its position

In simple terms potential energy is the energy at rest

Explanation: Examples ;

A spring has more potential energy when it is compressed or stretched.

A steel ball has more potential energy raised above the ground than it has after falling to Earth.

6 0

3 years ago

Suppose you held two magnets a short distance apart,then let go. what would happen?

Murljashka [212]

The two magnets ought to join together given that at least, if not both, have a strong enough magnetic force to do so.

4 0

3 years ago

Read 2 more answers

All atoms of uranium have the same

Aleksandr [31]

When uranium comes in contact with oxygen in the air, it rusts, just like iron does, but uranium rust is black and not red. Like other heavy atoms such as iron, uranium atoms have more neutrons than they do protons. Not all uranium atoms have the same number of neutrons.

8 0

3 years ago

Read 2 more answers

Consider the reaction:

stich3 [128]

Answer:

$\large \boxed{\text{-851.4 kJ/mol}}$

Explanation:

2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s); ΔᵣH = ?

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

$\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})$

2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s)

ΔfH°/kJ·mol⁻¹: 0 -824.3 -1675.7 0

$\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [1(-1675.7) + 2(0)] - [2(0) - 1(-824.3)]\\& = & -1675.7 + 824.3\\& = & \textbf{-851.4 kJ/mol}\\\end{array}\\\text{The enthalpy change is } \large \boxed{\textbf{-851.4 kJ/mol}}$

7 0

3 years ago

Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0

3 years ago