Answer: The derivative of a constant term is always 0. So the acceleration of the body would be zero. Hence, the acceleration of a body moving with uniform velocity will always be zero.
Hope this helps :) :)
Answer:
9.79740949850 moles
Explanation:
- 1 mole = Avogardo's Number <<6.022 E 23 <<particles, atoms, etc.>>
- This problem can be solved using dimensional analysis by multiplying atoms (5.9E24 atoms) by (1) mole and then dividing the number by Avogardo's number (6.022 E 23 atoms).
- Note: E = * 10
Side Note: Please let me know if you need any clarifications about this!
<h3>
Answer: b) 0.250 mol</h3>
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Work Shown:
Using the periodic table, we see that
- 1 mole of carbon = 12 grams
- 1 mole of oxygen = 16 grams
These are approximations and these values are often found underneath the atomic symbol. For example, the atomic weight listed under carbon is roughly 12.011 grams. I'm rounding to 2 sig figs in those numbers listed above.
So 1 mole of CO2 is approximately 12+2*16 = 44 grams. The 2 is there since we have 2 oxygens attached to the carbon atom.
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Since 1 mole of CO2 is 44 grams, we can use that to convert from grams to moles.
11.0 grams of CO2 = (11.0 grams)*(1 mol/44 g) = (11.0/44) mol = 0.250 mol of CO2
In short,
11.0 grams of CO2 = 0.250 mol of CO2
This is approximate.
We don't need to use any of the information in the table.
Answer:
Explanation:
M(s) → M (g ) + 20.1 kJ --- ( 1 )
X₂ ( g ) → 2X (g ) + 327.3 kJ ---- ( 2 )
M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ----- ( 3 )
( 3 ) - 2 x ( 2 ) - ( 1 )
M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s) → M X₄ (g ) - 98.7 kJ - 2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ
0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ
4 X (g ) + M (g ) = M X₄ (g ) - 773.4kJ
heat of formation of M X₄ (g ) is - 773.4 kJ
Bond energy of one M - X bond = 773.4 / 4 = 193.4 kJ / mole
Both A and B are the answer.
