The answer is D. Use the equation PV=nRT
P=(.567mol)(.0821)(300K)/4.5L
Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
Regards.
Hey there!
Values Ka1 and Ka2 :
Ka1 => 8.0*10⁻⁵
Ka2 => 1.6*10⁻¹²
H2A + H2O -------> H3O⁺ + HA⁻
Ka2 is very less so I am not considering that dissociation.
Now Ka = 8.0*10⁻⁵ = [H3O⁺] [HA⁻] / [H2A]
lets concentration of H3O⁺ = X then above equation will be
8.0*10−5 = [x] [x] / [0.28 -x
8.0*10−5 = x² / [0.28 -x ]
x² + 8.0*10⁻⁵x - 2.24 * 10⁻⁵
solve the quardratic equation
X =0.004693 M
pH = -log[H⁺]
pH = - log [ 0.004693 ]
pH = 2.3285
Hope that helps!
To get a result with the best degree of precision, the
number of significant figures should be equal to the smallest number of
significant figures of the given numbers. In this case, the smallest is 3 as
given by the number 9.03 mL.
Therefore density is:
<span>11.50 g / 9.03 mL = 1.27 g/mL</span>
Answer:
1. E. mitochondrion
2. C. chloroplasts
3. G. vacuole
4. A. cell membrane
5. B. cell wall
6. D.cytoplasm
7. F. Nucleus
8. cytoplasm
9. cell membrane
10. chloroplast
11. mitochondrion
12. nucleus
13. cell wall
14. vacuole
Explanation:
- Be familiar which each term.
- Look up diagram to understand image.
- Hope that helped! Please let me know if you need further explanation on each word.
