This is a D. combustion reaction
That is because you add O2 which is necessary for a combustion, while the results are CO2 and water. What you are missing is a ---> after the 2O2 (g)
Answer:
C₇H₁₄O₂
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 1.152 g
Mass of CO₂ = 2.726 g
Mass of H₂O = 1.116 g
Empirical formula =?
Next, we shall determine the mass of carbon, hydrogen and oxygen present in the compound. This can be obtained as follow:
For carbon (C):
Mass of CO₂ = 2.726 g
Molar mass of CO₂ = 12 + (2×16)
= 12 + 32
= 44 g/mol
Mass of C = 12/44 × 2.726
Mass of C = 0.743 g
For hydrogen (H):
Mass of H₂O = 1.116 g
Molar mass of H₂O = (2×1) + 16
= 2 + 16
= 18 g/mol
Mass of H = 2/18 × 1.116
Mass of H = 0.124 g
For oxygen (O):
Mass of compound = 1.152 g
Mass of C = 0.743 g
Mass of H = 0.124 g
Mass of O =?
Mass of O = (Mass of compound) – (Mass of C) + (Mass of H)
Mass of O = 1.152 – (0.743 + 0.124)
Mass of O = 1.152 – 0.867
Mass of O = 0.285 g
Finally, we shall determine the empirical formula for the compound as follow:
C = 0.743 g
H = 0.124 g
O = 0.285 g
Divide by their molar mass
C = 0.743 / 12 = 0.062
H = 0.124 / 1 = 0.124
O = 0.285 / 16 = 0.018
Divide by the smallest
C = 0.062 / 0.018 = 3.44
H = 0.124 / 0.018 = 7
O = 0.018 / 0.018 = 1
Multiply by 2 to express in whole number.
C = 3.44 × 2 = 7
H = 7 × 2 = 14
O = 1 × 2 = 2
Empirical formula => C₇H₁₄O₂
That ball could be heavier or rounder
Answer:
The new concentration of CO₂ dissolved is 0.0085 M
Explanation:
Applying the equation below;
![n = \frac{PV}{K}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7BPV%7D%7BK%7D)
![K= \frac{P_1}{n_1} = \frac{P_2}{n_2}](https://tex.z-dn.net/?f=K%3D%20%5Cfrac%7BP_1%7D%7Bn_1%7D%20%3D%20%5Cfrac%7BP_2%7D%7Bn_2%7D)
where;
P₁ is the initial pressure = 4 x 10⁻⁴ atm
P₂ is the final pressure = 0.25 atm
n₁ is the initial concentration = 1.36 x 10⁻⁵ M
n₂ is the final concentration = ?
n₂ = (n₁*P₂)/(P₁)
n₂ = (1.36 x 10⁻⁵ X 0.25)/(4 x 10⁻⁴)
n₂ = 0.0085 M
Therefore, the new concentration of CO₂ dissolved is 0.0085 M
<span>One way to get more energy out of fuels is to increase _____, the percentage of energy used for work.
</span><span>C.efficiency
Hope this helps :)</span>