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jeka94
3 years ago
8

Why are collisions between Pluto and Neptune impossible

Chemistry
2 answers:
Pavel [41]3 years ago
7 0
Because plutos orbit takes it much higher above the suns orbital plane when Pluto is at the same point as neptunes orbit it actually MUCH higher up than Neptune so the two planets will never be at the same place at the same time
Brrunno [24]3 years ago
4 0
Even though the orbits of pluto and neptune intersect,they are locked in a gravitational resonance condition which seems to prevent them from coming closer than a few 100 million miles from one another. that is why it is not possible for both planets to collide
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g Phosphorus -32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The h
Setler79 [48]

Answer:

6.88 mg

Explanation:

Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄

The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.

175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P

Step 2: Calculate the rate constant for the decay of ³²P

The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.

k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹

Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days

For first-order kinetics, we will use the following expression.

ln P = ln P₀ - k × t

ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d

P = 6.88 mg

3 0
3 years ago
Ano-ano ang mensahe ng grupo ng el gamma penumbra sa kanilang palabas​
Alexxandr [17]

Answer:

Pag-alaga ng kalikasan, pagpapakita ng mga tanawin o kagandahan ng Pilipinas

6 0
3 years ago
Is the maximum population that a given area can support
Anastaziya [24]

Answer:

Carry capacity

Explanation:

I took the test

Hope this helps

pls mark brainliest :3

8 0
3 years ago
Read 2 more answers
The combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110 kg of carbon dioxide. What is the limiti
nadezda [96]

Answer:

The limiting reactant is the propane gas, C₃H₈ while the percentage yield is 83.77%

Explanation:

Here we have

Propane gas with molecular formula C₃H₈, molar mass  = 44.1 g/mol combining with O₂ as follows

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Therefore, 1 mole of C₃H₈  combines with 5 moles of O₂ to produce 3 moles CO₂ and 4 moles of H₂O

Mass of propane = 0.1240 kg = 124.0 g

Number of moles of propane = mass of propane/(molar mass of propane)

The number of moles of propane = 124/44.1 = 2.812 moles

The molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.3110 kg = 311.0 g

Therefore, number of moles of CO₂ = mass of CO₂/(molar mass of CO₂)

The number of moles of CO₂ = 311.0 kg/ 44.01 g/mol = 7.067 moles

Therefore, since 1 mole of propane produces 3 moles of CO₂, 2.812 moles of propane will produce 3 × 2.812 moles or 8.44 moles of CO₂

Therefore;

The limiting reactant is the propane gas, C₃H₈, since the oxygen is in excess

Hence

The \ percentage \ yield = \frac{Actual \, yield}{Theoretical \, yield} \times 100 = \frac{7.067}{8.44} \times 100 = 83.77 \%

The percentage yield = 83.77%.

7 0
3 years ago
0.023ml= blank cl <br> PLEASE HELP
Rus_ich [418]
0.23cL because 0.023mL= 23.0L so 23.0L= 0.23cL
6 0
3 years ago
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