Answer:
CH₂
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 1 g
Mass of CO₂ = 3.14 g
Mass of H₂O = 1.29 g
Empirical formula =?
Next, we shall determine the mass of Carbon and hydrogen present in the compound. This can be obtained as follow:
For Carbon, C:
Mass of CO₂ = 3.14 g
Molar mass of CO₂ = 12 + (2×16)
= 12 + 32
= 44 g/mol
Molar mass of C = 12 g/mol
Mass of C =?
Mass of C = molar mass of C/ Molar mass of CO₂ × Mass of CO₂
Mass of C = 12/44 × 3.14
Mass of C = 0.86 g
For hydrogen, H:
Mass of C = 0.86 g
Mass of compound = 1 g
Mass of H =?
Mass of H = (Mass of compound) – (mass of C)
Mass of H = 1 – 0.86
Mass of H = 0.14 g
Finally, we shall determine the empirical formula of the cyclopropane. This can be obtained as follow:
Mass of C = 0.86 g
Mass of H = 0.14 g
Divide by their molar mass
C = 0.86 / 12 = 0.07
H = 0.14 / 1 = 0.14
Divide by the smallest
C = 0.07 / 0.07 = 1
H = 0.14 / 0.07 = 2
Thus, the empirical formula of cyclopropane is CH₂
Answer:
10.
Explanation:
Hello,
In this case, for the given reaction:
![Cr + Fe(NO_3)_2 \rightarrow Fe + Cr(NO_3)_3](https://tex.z-dn.net/?f=Cr%20%2B%20Fe%28NO_3%29_2%20%5Crightarrow%20Fe%20%2B%20Cr%28NO_3%29_3)
By knowing the law of conservation of mass must be respected, the following coefficients do balance the chemical equation:
![2Cr + 3Fe(NO_3)_2 \rightarrow 3Fe + 2Cr(NO_3)_3](https://tex.z-dn.net/?f=2Cr%20%2B%203Fe%28NO_3%29_2%20%5Crightarrow%203Fe%20%2B%202Cr%28NO_3%29_3)
So the sum of all the coefficients is 10.
Regards.
1.08 atm is the pressure for a certain tire in atmosphere.
<u>Explanation:</u>
One kilo pascal (1 kPa) corresponds to 1000 pascal. Another common unit used for pressure is atmosphere (symbolised as ‘atm’). 1 atm refers the standard atmospheric pressures and corresponds to 760 mm Hg and 101.3 kPa. Atmospheric pressures are commonly referred as square inches (psi)/ pounds.
![1 \mathrm{atm}=101.3 \mathrm{kPa}=101,325 \mathrm{Pa}=760 \mathrm{mm} \mathrm{Hg}=760 \text { torr }=14.7 \mathrm{lb} / \mathrm{in}^{2}(\mathrm{psi})](https://tex.z-dn.net/?f=1%20%5Cmathrm%7Batm%7D%3D101.3%20%5Cmathrm%7BkPa%7D%3D101%2C325%20%5Cmathrm%7BPa%7D%3D760%20%5Cmathrm%7Bmm%7D%20%5Cmathrm%7BHg%7D%3D760%20%5Ctext%20%7B%20torr%20%7D%3D14.7%20%5Cmathrm%7Blb%7D%20%2F%20%5Cmathrm%7Bin%7D%5E%7B2%7D%28%5Cmathrm%7Bpsi%7D%29)
Given:
The air pressure for a certain tire = 109 kPa
We need to find pressure in atmospheres
So, we know,
1 atm = 101.3 kPa
Hence,
![\frac{109 \mathrm{kPa}}{1} \times \frac{1 \mathrm{atm}}{101.3 \mathrm{kPa}}=1.076=1.08 \mathrm{atm}](https://tex.z-dn.net/?f=%5Cfrac%7B109%20%5Cmathrm%7BkPa%7D%7D%7B1%7D%20%5Ctimes%20%5Cfrac%7B1%20%5Cmathrm%7Batm%7D%7D%7B101.3%20%5Cmathrm%7BkPa%7D%7D%3D1.076%3D1.08%20%5Cmathrm%7Batm%7D)
1.08 atm is the pressure for a certain tire in atmosphere.
Answer:
18 moles
Explanation:
Here the combustion of one mole of glucose ----> carbon dioxide + water, releases 2870 kilojoules / moles.
_______________________________________________________
With one contraction cycle requiring 55 kilojoules,
2870 / 55 ≈ 52.18
And with the efficiency being 35 percent,
52.1818..... * 0.35 = ( About ) 18 moles
<u><em>Hope that helps!</em></u>