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puteri [66]
3 years ago
14

(Help) which of these is correct?

Chemistry
1 answer:
PolarNik [594]3 years ago
6 0
The third one I think
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You have 1 m3 of wood and 1 m3 of aluminum. which is heavier? prove your answer with calculation.
Oduvanchick [21]

You have 1 m³ of wood and 1 m³ of aluminum.

Aluminum is the heavier material with a higher density.

What is density?

The mass-to-volume ratio, or mass per unit volume, is known as density. It assesses how much material an object contains in relation to its volume (cubic metre or cubic centimeter). In essence, density is a measurement of how closely together matter is packed.

<u>Density of the wood =1500kg/1m³ = 1500kg/m³ ⟹ 1m³ has a mass of 1500kg Density of Aluminum(Al) =2700kg/m³ = 2700kg/m³⟹ 1m³ has a mass of 2700kg </u>

Therefore, Aluminum is the heavier material with a higher density.

To learn more about density from the given link below,

brainly.com/question/17590787

#SPJ4

5 0
1 year ago
Larissa pumps up a soccer ball
Alex73 [517]

Whats the question ? all you said was she pumps up a soccer ball

3 0
3 years ago
What’s the best answer
In-s [12.5K]
C) is correct, because there are 2 carbons and 4 oxygens on both sides, making the equation balanced.
7 0
2 years ago
In an attempt to maximize the yield of methanol (amount of methanol produced), a chemist would try to shift the equilibrium as f
MakcuM [25]

Answer:

removing the methanol as it is formed

Explanation:

One of the ways to drive the equilibrium position towards the right is to remove one of the products formed.

According to Me Chatelier's principle, the imposition of a constraint on a system in a equilibrium causes the equilibrium position to shift towards a new position that annuls the constraint. Hence, removing the methanol causes the equilibrium position to shift to the far right in order to reestablish equilibrium according to Le Chatelier's principle.

3 0
3 years ago
Assuming all the nh3 dissolves and that the volume of the solution remains at 0.300 l , calculate the ph of the resulting soluti
barxatty [35]
<span>Assume p=735 Torr V= 7.6L R=62.4 T= 295 PV-nRT (735 Torr)(7.60L)= n (62.4Torr-Litres/mole-K)(295K) 0.30346 moles of NH3 Find moles 0.300L solution of 0.300 M HCL = 0.120 moles of HCL 0.30346 moles of NH3 reacts with 0.120 moles of HCL producing 0.120 moles of NH4+ ION, and leaving 0.18346 mole sof NH3 behind Find molarity 0.120 moles of NH4+/0.300L = 0.400 M NH4+ 0.18346 moles of NH3/0.300L = 0.6115 M NH3 NH4OH --> NH4 & OH- Kb = [NH4+][OH]/[NH4OH] 1.8 e-5=[0.300][OH-]/[0.6115] [OH-]=1.6e-5 pOH= 4.79 PH=9.21 .</span>
3 0
3 years ago
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