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Lina20 [59]
4 years ago
6

The melting point of chromium is 1890. The melting point of argon is -189. Suggest a reason for this difference.

Chemistry
1 answer:
DanielleElmas [232]4 years ago
4 0

Answer:

Different forces between atoms and difference in structure.

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Factors that affect the acceleration of an object include the..?
iren [92.7K]
As per the Equation,
Acceleration = Force/Mass

We Can Conclude That,
Acceleration of an object is affected by A). Net Force Acting On the Object and C). Object's Mass.

!! Hope It Helps !!
7 0
3 years ago
How many moles of plutonium
bixtya [17]

0.139 moles of plutonium are present in a sample containing 8.4 x 10²² atoms of Pt.

<h3>How to calculate number of moles?</h3>

The number of moles of a substance can be calculated using the following expression:

no of moles = no of molecules ÷ 6.02 × 10²³

According to this question, 8.4 × 10²² atoms of plutonium are present in the molecule of plutonium.

no of moles = 8.4 × 10²² = 6.02 × 10²³

no of moles = 1.39 × 10-¹

no of moles = 0.139moles

Therefore, 0.139 moles of plutonium are present in a sample containing 8.4 x 10²² atoms of Pt.

Learn more about no of moles at: brainly.com/question/14919968

#SPJ1

4 0
2 years ago
Would it still be a "mixture"?
pashok25 [27]
Conglomerate rock, water and oil, a portion salad, trail mix, and concrete (not cement).
7 0
3 years ago
What is carbon fixation? carbon atoms are joined with oxygen molecules to make carbon dioxide carbon atoms are added to molecule
Sidana [21]

Carbon fixation is the process of conversion in which carbon-dioxide (inorganic carbon) converts into organic compounds with the help of living organisms and example of this process is photosynthesis.

From the definition, it is confirmed that there is no formation of carbon dioxide takes place i.e. carbon doesn't joined with oxygen molecules.

Thus, carbon fixation is the process in which carbon atoms are added to the molecules of hydrogen and oxygen to make carbohydrates.


3 0
3 years ago
Suppose that 10.0 mol C2H6(g) is confined to 4.860 dm3 at 27 °C. Predict the pressure exerted by the ethane from (i) the perfect
Afina-wow [57]

Answer:

P=35.16

Z=4.6

Explanation:

Hello,

In this case, since the VdW equation is:

P=\frac{nRT}{V-n*b}-a(\frac{n}{V} )^2

Since the moles are 10.0 moles, the temperature in K is 300.15 K and the volume is liters is also 4.860 L (1 dm³= 1L), the pressure exerted by the ethane is:

P=\frac{10.0mol*0.082\frac{atm*L}{mol*K}*300.15K}{4.860mol-10.0mol*0.0651\frac{L}{mol} }-5.507\frac{atm*L^2}{mol^2}(\frac{10.0mol}{4.86L} )^2\\\\P=58.48atm-23.3atm\\\\P=35.16

Thus the compression factor turns out:

Z=\frac{PV}{RT}=\frac{23.3atm*4.86L}{  0.082\frac{atm*L}{mol*K}*300.15K}\\\\Z=4.6

Regards.

8 0
3 years ago
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