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son4ous [18]
3 years ago
7

What is Keystone XL pipeline?

Chemistry
1 answer:
Leno4ka [110]3 years ago
3 0
Is a oil pipeline in Canada

hope it helped<span />
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Nguyên tử nguyên tố sắt có 26e<br> a. Viết cấu hình e của sắt<br> b. Sắt thuộc nguyên tố s,p,d,f
vichka [17]

Answer:

1s2 2s2 2p6 3s2 3p6 3d6 4s2

Explanation:

3 0
3 years ago
Why copper should not be disposed of in landfill sites?
Dafna1 [17]
Copper is a heavy metal that can harm many people. This process of heavy metals building in your body is called bioaccumulation and the process of this building in the food chain is called biomagnification. Its hard to rid of this heavy metal because it is very dense, therefore, cannot breakdown in our bodies. Heavy metals get consumed by producers and then reach to the top of the pyramid (us) so we have the highest risk of getting diseases and so on. Make sure you explain that heavy metals get passed from industrial purposes where companies do not dispose of waste properly.
Source: Exam as well
Good luck!  
4 0
3 years ago
Read 2 more answers
What is this plzzzzzzzzzzzzzzzzzzz help!!!!
motikmotik

Answer:

convex

converges

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3 0
3 years ago
We might think of a porous material as being a composite wherein one of the phases is a pore phase. Estimate upper and lower lim
eduard

Answer:

The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are

Ku = 38.252 W/mK

K lower = 0.199 W/mK

Explanation:

As we know  

Ku = Vp * Kair + Vmagnesium * K metal  

Ku = 0.10 *0.02 + (1-0.25) * 51

Ku = 38.252 W/mK

The lower limit  

K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)

K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)

K lower = 0.199 W/mK

8 0
2 years ago
How many grams of CO₂ can be produced from the combustion of 2.76 moles of butane according to this equation: 2 C₄H₁₀(g) + 13 O₂
Vilka [71]

Answer:

485.76 g of CO₂ can be made by this combustion

Explanation:

Combustion reaction:

2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)

If we only have the amount of butane, we assume the oxygen is the excess reagent.

Ratio is 2:8. Let's make a rule of three:

2 moles of butane can produce 8 moles of dioxide

Therefore, 2.76 moles of butane must produce (2.76 . 8)/ 2 = 11.04 moles of CO₂

We convert the moles to mass → 11.04 mol . 44g / 1 mol = 485.76 g

5 0
3 years ago
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