The formula to convert from Celsius to Fahrenheit is F = 9/5C + 32 or F = (C*1,8)+32
F = 9/5*39 + 32 = 70,2 + 32 = 102,2° F
Answer:
Mass of glucose = 515.34 g
Explanation:
We are given;
Mass; m = 60 kg
Elevation; h = 1550 m
Acceleration due to gravity; 9.8 m/s²
Now, work performed to lift 60kg by 1550m is given by the formula;
W = mgh
W = 60 × 9.8 × 1550
W = 911400 J
We are told the actual work is 4 times the one above.
Thus;
Actual work = 4W = 4 × 911400 = 3,645,600 J
Now,
Molar mass of Glucose(C6H12O6) = 180 g/mol
We are given standard enthalpy of combustion = -1273.3 KJ/mol = -1273300
Moles of glucose = 3645600/1273300 = 2.863mol
Mass of glucose = 2.863 mol × 180 g/mol
Mass of glucose = 515.34 g
Answer:
Difussion
Explanation:
Diffusion is the result of a totally random phenomenon in which the molecules of a fluid come and go between two vessels that can be connected by a pipe. These molecules travel in a single direction, where the solute is more concentrated to where it is more diluted.
This movement of particles will be modified according to the length or area of the pipe and the concentration of solute. The greater the difference in solute concentration along the tube, the greater the diffusion
The percentages are given in weight. You need to transform them into moles.
Take a basis of 100 grams of compound. Then there are 69.94 grams of iron and 30.06 grams of oxygen.
Use the atomic masses of iron and oxygen to obtain each number of moles.
Molar mass of iron: 55.85 g/mol
# moles of iron = 69.94 g / 55.85 g/mol = 1.252
Molar mass of oxygen: 16 g/mol
# moles of oxygen = 30.06 g / 16 g/mol =1.87875
Proportions
1.87875 moles of Oxigen/ 1.252moles of iron = 1.50 /1
3moles of Oxygen: 1 mol of Iron
Empirical formula: FeO3
Molar mass of the empirical formula: 55.85 g/mol + 3(16g/mol) = 103.85 g/mol
Molar mass of the compound / molar mass of the empirical formula = 199.55 g/mol / 103.85 g/mol = 1.92
Round to 2.
Then the molecular formula is the empirical formula times 2.
Fe2O6
