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Nesterboy [21]
3 years ago
7

Shelly was surprised to learn the dwarf planet Pluto used to be considered a planet. What would have had to be true for Pluto to

remain categorized as a planet?
Chemistry
2 answers:
meriva3 years ago
8 0
Pluto would need to clear the neighborhood of its orbit would be your answer, hope this helps! :D
AlladinOne [14]3 years ago
6 0

The answer is: Pluto would need to clear the neighborhood of its orbit.

Pluto is a dwarf planet in the Kuiper belt.

According to the International Astronomical Union a planet must satisfy the following three criteria:

1) Planet must have enough mass that its own gravity pulls it into a roughly spheroidal shape.

2) Planet must be an object which independently orbits the Sun.

3) Planet must be large enough to dominate its orbit, its mass must be much larger than anything else which crosses its orbit.

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In an oxidation-reduction reaction, the total number of electrons lost is
netineya [11]
In redox reactions, there is no net loss or gain of electrons, so the answer is (1) equal to the total number of electrons gained
7 0
4 years ago
Read 2 more answers
The enthalpy of reaction changes somewhat with temperature. Suppose we wish to calculate ΔH for a reaction at a temperature T th
Contact [7]

Answer:

-99.8 kJ

Explanation:

We are given the methodology to answer this question, which is basically  Kirchhoff law . We just need to find the heats of formation for the reactants and products and perform the calculations.

The standard heat of reaction is

ΔrHº = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

where ν are the stoichiometric coefficients in the balanced equation, and ΔfHº are the heats of formation at their  standard states.

  Compound                 ΔfHº (kJmol⁻¹)

        SO₂                             -296.8

         O₂                                    0

         SO₃                            -395.8

The balanced chemical equation is

SO₂(g) + ½O₂(g) → SO₃(g)

Thus

Δr, 298K Hº( kJmol⁻¹ ) =  1 x (-395.8) - 1 x (-296.8) = -99.0 kJmol⁻¹

Now the heat capacity of reaction  will be be given in a similar fashion:

Cp rxn = ∑ ν x Cp of products - ∑ ν x Cp of reactants

where ν is as above the stoichiometric coefficient in the balanced chemical equation.

Cprxn ( JK⁻¹mol⁻¹) = 50.7 - ( 39.9 + 1/2 x 29.4 ) = - 3.90

                         = -3.90 JK⁻¹mol⁻¹

Finally Δr,500 K Hº = Δr, 298K Hº +  CprxnΔT

Δr,500 K Hº = - 99 x 10³ J + (-3.90) JK⁻¹ ( 500 - 298 ) K = -99,787.8

                     = -99,787.8 J x 1 kJ/1000 J  = -99.8 kJ

Notice thie difference is relatively small that is why in some problems it is o.k to assume the change in enthalpy is constant over a temperature range, especially if it is a small range of temperatures.

3 0
3 years ago
Carbon, oxygen And nitrogen all types of
posledela

molecules or protein molecules

8 0
4 years ago
Read 2 more answers
What is the mass present in a 10.0L container of oxygen at a pressure of 105kPa and 20 degrees Celsius
omeli [17]

1.31 × 10⁴ grams.

<h3>Explanation</h3>

Assume that oxygen acts like an ideal gas. In other words, assume that the oxygen here satisfies the ideal gas law:

P \cdot V = n \cdot R\cdot T,

where

  • P the pressure on the gas, \bf P = 10^{5}\;\textbf{kPa}=10^{8}\;\textbf{Pa};
  • V the volume of the gas, V = 10.0 \;\text{L} = 10.0\times 10^{-3}\;\text{m}^{3}=10^{-2}\;\text{m}^{3};
  • n the number of moles of the gas, which needs to be found;
  • T the absolute temperature of the gas, T=20\;\textdegree{}\text{C} = (20 + 273.15)\;\text{K} = 293.15\;\text{K}.
  • R the ideal gas constant, R = 8.314 if P, V, and T are in their corresponding SI units: Pa, m³, and K.

Apply the ideal gas law to find n:

n = \dfrac{P\cdot V}{R\cdot T} = \dfrac{{\bf 10^{8}\;\textbf{Pa}}\times 10^{-2}\;\text{m}^{3}}{8.314 \;\text{Pa}\cdot\text{m}^{3}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\times 293.15\;\text{K}} = 410.3\;\text{mol}.

In other words, there are 410.3 moles of O₂ molecules in that container.

There are two oxygen atoms in each O₂ molecules. The mass of mole of O₂ molecules will be {\bf 2}\times 16.00 = 32.00\;\text{g}. The mass of 410.3 moles of O₂ will be:

410.3 \times 32.00 = 1.31\times10^{4}\;\text{g}.

What would be the mass of oxygen in the container if the pressure is approximately the same as STP at 10^{5}\;\textbf{Pa} or  10^{2}\;\text{kPa} instead?

6 0
4 years ago
What are the ways humans use, collect, and distribute renewable and nonrenewable natural resources?
guajiro [1.7K]

Answer:

Air and water pollution, damage to public health, wildlife and habitat loss, water use, land use, and global warming emissions.

<h2>Explanation: <u><em>HOPE THIS HELPS!!!!</em></u></h2>

8 0
2 years ago
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