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3 years ago

1. Which type of telescope is most useful for detecting a star that emits very long

1 answer:

5 0

Which type of telescope is most useful for detecting a star that emits very long electromagnetic waves?

A. a radio telescope B. an X-ray telescope C. an optical telescope D. a gamma ray telescope
A.)a radio telescope

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HELP ASAP! GIVING BRAINLIEST!! 15 POINTS

Rufina [12.5K]

I think Kinetic energy forms Motion energy and Potential energy forms Gravitational Potential energy.

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3 years ago

4.41 g of propane gas (C3H8) is injected into a bomb calorimeter and ignited with excess oxygen, according to the reaction below

gayaneshka [121]

Answer : The heat of the reaction is -221.6 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter

$q_{rxn}=-q_{cal}$

$q_{cal}=c_{cal}\times \Delta T$

where,

$q_{rxn}$ = heat released by the reaction = ?

$q_{cal}$ = heat absorbed by the calorimeter

$c_{cal}$ = specific heat of calorimeter = $97.1kJ/^oC=97100J/^oC$

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=(27.282-25.000)=2.282^oC$

Now put all the given values in the above formula, we get:

$q_{cal}=(97100J/^oC)\times (2.282^oC)$

$q_{cal}=221582.2J=221.6kJ$

As, $q_{rxn}=-q_{cal}$

So, $q_{rxn}=-221.6kJ$

Thus, the heat of the reaction is -221.6 kJ

6 0

3 years ago

What is the correct answer please

Anestetic [448]

Answer:

I think it's the second answer --If you increase the acidity..

I hope answer I can answer your question!

7 0

3 years ago

Number of molecules of water in 6 moles

Anit [1.1K]

I mol of anything is 6.02 * 10^23 (in this case molecules of water)
6 mols of water = x

1/6 = 6.02 * 10^23 / x Cross multiply
x = 6 * 6.02* 10^23
x = 3.612 * 10^24 molecules in 6 mols of water

5 0

3 years ago

2.0L of hydrogen gas is mixed with 3.0L of nitrogen gas at STP in a rigid 5.0L vessel. A reaction occurs, producing ammonia gas

IrinaK [193]

Answer:

Moles NH₃: 0.0593

0.104 moles of N₂ remain

Final pressure: 0.163atm

Explanation:

The reaction of nitrogen with hydrogen to produce ammonia is:

N₂ + 3 H₂ → 2 NH₃

Using PV = nRT, moles of N₂ and H₂ are:

N₂: 1atmₓ3.0L / 0.082atmL/molKₓ273K = 0.134 moles of N₂

H₂: 1atmₓ2.0L / 0.082atmL/molKₓ273K = 0.089 moles of H₂

The complete reaction of N₂ requires:

0.134 moles of N₂ × (3 moles H₂ / 1 mole N₂) = 0.402 moles H₂

That means limiting reactant is H₂. And moles of NH₃ produced are:

0.089 moles of H₂ × (2 moles NH₃ / 3 mole H₂) = 0.0593 moles NH₃

Moles of N₂ remain are:

0.134 moles of N₂ - (0.089 moles of H₂ × (1 moles N₂ / 3 mole H₂)) = 0.104 moles of N₂

And final pressure is:

P = nRT / V

P = (0.104mol + 0.0593mol)×0.082atmL/molK×273K / 5.0L

P = 0.163atm

7 0

3 years ago