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3 years ago

If you had to choose to live on the moon or mars, explain why you would choose

Chemistry

1 answer:

FinnZ [79.3K]3 years ago

8 0

Answer:

I would choose to live on mars

Explanation:

I would choose to live on mars because not only is it the most habitable planet in our solar system other than Earth, there is enough gravity on Mars to be believed that our bodies can adapt. It is also not too hot or cold for us. There is soil that water can be extracted from, and we can also use solar panels to collect energy. Although we would have to use advanced technology when going there, but I think these are a few reasons why I would prefer living on Mars over the Moon.

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. What is the mass, in grams, of a sample of 7.83 × 1024 atoms of helium (He)?

erica [24]

Moles He = 7.83 x 10^24 / 6.02 x 10^23 =13.0

<span>mass He = 13.0 mol x 4.00 g/mol = 52.0 g</span>

6 0

3 years ago

Fatima is designing an advanced graphics card for a next-generation video game system. She needs to be able to precisely control

Salsk061 [2.6K]

Answer:svbehsbss c

Explanation:

D xxxx

8 0

2 years ago

What is the energy released or absorbed as heat when one mole of a compound is produced by combination of its elements? select o

suter [353]

B and D is out. It cant be A because heat of combustion is substance not compound. So the answer is D.

7 0

2 years ago

Some oxides are given below. (i)Na2O (ii)NO2 (iii) CO2 (iv) MgO a) Which are the basic oxides among these? b) What is the name o

umka2103 [35]

<h3>Further explanation</h3>

Basic oxides ⇒ metal(usually alkali/alkaline earth) +O₂

L + O₂ ⇒ L₂O

L + O₂ ⇒ LO

Dissolve in water becomes = basic solution

L₂O+H₂O⇒ 2LOH

LO + H₂O⇒ L(OH)₂

So the basic oxides : Na₂O and MgO

Na₂O + H₂O⇒NaOH

MgO +H₂O⇒Mg(OH)₂

The aqueous solution of CO₂(dissolve in water)

CO₂ + +H₂O⇒ H₂CO₃(carbonic acid)

5 0

2 years ago

Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c

Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

$m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO} *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2$

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

$Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%$

Best regards!

8 0

2 years ago