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The exact molecular mass for butane (C4H10) is
12.0096*4+1.0079*10=58.1174 which is 58.1 to 3 significant figures.
Proportion of carbon in the compound
12.0096*4: 58.1174
=>
48.0384 : 58.1174
The mass of carbon in 2.50 grams of butane can be obtained by proportion, namely
Mass of carbon
= 2.50 * (48.0384/58.1174)
= 2.0664
= 2.07 g (approximated to 3 significant figures)
I am sorry bro I don't know
Answer:
0.36 M
Explanation:
There is some info missing. I think this is the complete question.
<em>Suppose a 250 mL flask is filled with 0.30 mol of N₂ and 0.70 mol of NO. The following reaction becomes possible:
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<em>N₂(g) +O₂(g) ⇄ 2 NO(g)
</em>
<em>The equilibrium constant K for this reaction is 7.70 at the temperature of the flask. Calculate the equilibrium molarity of O₂. Round your answer to two decimal places.</em>
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Initially, there is no O₂, so the reaction can only proceed to the left to attain equilibrium. The initial concentrations of the other substances are:
[N₂] = 0.30 mol / 0.250 L = 1.2 M
[NO] = 0.70 mol / 0.250 L = 2.8 M
We can find the concentrations at equilibrium using an ICE Chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with the concentration or change in the concentration.
N₂(g) +O₂(g) ⇄ 2 NO(g)
I 1.2 0 2.8
C +x +x -2x
E 1.2+x x 2.8 - 2x
The equilibrium constant (K) is:
![K=7.70=\frac{[NO]^{2}}{[N_{2}][O_{2}]} =\frac{(2.8-2x)^{2} }{(1.2+x).x}](https://tex.z-dn.net/?f=K%3D7.70%3D%5Cfrac%7B%5BNO%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%20%3D%5Cfrac%7B%282.8-2x%29%5E%7B2%7D%20%7D%7B%281.2%2Bx%29.x%7D)
Solving for x, the positive one is x = 0.3601 M
[O₂] = 0.3601 M ≈ 0.36 M
Answer:
1.51 x 10²⁴molecules
Explanation:
Given parameters:
Number of moles of NaCl = 2.5moles
Unknown:
Number of molecules = ?
Solution:
The number of molecules that can be found in a given mole is given as the Avogadro's constant;
1 mole of substance = 6.02 x 10²³ molecules
2.5 mole of NaCl = 2.5 x 6.02 x 10²³ molecules
= 1.51 x 10²⁴molecules