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Ipatiy [6.2K]
4 years ago
5

Which are advantages of using a reflecting telescope instead of a refracting telescope? Check all that apply. Reflecting telesco

pes are less expensive. Reflecting telescopes produce clearer images. Reflecting telescopes require only concave lenses. Reflecting telescopes can gather light from objects farther in space. Reflecting telescopes usually take up less space.
Chemistry
1 answer:
slega [8]4 years ago
7 0
The answers are:
Reflecting telescopes are less expensive, the materials are not costly.

Reflecting telescopes can gather light from objects. This becomes basis the system to see Celestial objects.

Reflecting telescopes uses mirrors and Refracting telescopes use lenses. It is also much bigger in size because of the mirrors placed inside.
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Explain how matter changes from one state to another such as liquid
beks73 [17]

Answer:

When heated or cooled, matter can transform from one state to another. When you heat ice (a solid), it turns into water (a liquid). MELTING is the term for this transformation. When water is heated, it becomes steam (a gas).

Explanation:

i hope thats the answer you want

8 0
2 years ago
if you are using a formula where you need the change in temperature, explain why it is not important whether your temperatures a
alexandr1967 [171]

Answer:

This is because, Kelvins and Celcius degrees both agree at fixed points i.e; the lower fixed point and upper

5 0
3 years ago
What actions show conserving by reusing? Choose the two correct answers
hram777 [196]
A and D

Hope this helps
3 0
3 years ago
N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol
docker41 [41]

Answer:

A

Explanation:

Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:


\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)

Therefore, from the chemical equation, we have that:


\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} +  \Delta H^\circ_f \text{ H$_2$O}  \right]   -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}

In conclusion, our answer is A.

5 0
3 years ago
Is there a difference between a homogeneous mixture of hydrogen and oxygen in a 2:1 ratio and a sample of water vapor?
Feliz [49]

Answer:

Yes

Explanation:

There is a difference between the homogeneous mixture of the hydrogen and the oxygen in a 2:1 ratio and the sample of the water vapor.

In the homogeneous mixture of the hydrogen and the oxygen which are present in the ratio, 2:1 , the elements are not chemically combined. They are explosive also as both shows their specific properties. They can be separated by physical means (Condensation, diffusion).

On the other hand, in water vapor, the two elements are chemically bonded in a specific mixture which cannot be separated via physical means. Water has its unique properties and they can be separated by chemical means only.

5 0
3 years ago
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