CH4 + 2O2→CO2 + 2H2O
Explanation:
CH4 + O2→CO2 + H2O
First look at the C atoms. At first glance they are balanced with 1 C on each side.
Now look at the H atoms. They are not balanced. There are 4 H atoms on the left side and 2 H atoms on the right. Place a coefficient of 2 in front of the H2O. We now have 4 H atoms on both sides.
CH4 + O2→CO2 + 2H2O
Now look at the O atoms. They are not balanced. There are 2 O atoms on the left side and 4 on the right. Place a coefficient of 2 in front of the O2. We now have 4 O atoms on both sides.
CH4 + 2O2→CO2 + 2H2O
The equation is now balanced. Each side has 1 C atom, 4 H atoms, and 4 O atoms.
For machines such as a rake the output force is less than the input force. If the input force of the rake is 10 newtons and the output force is 5 Newtons mechanical advantage is .5
Answer:
hhh
Explanation:
Pyranose is collective term for the saccharides which have chemical structure which includes six-membered ring that consists of one oxygen atom and five carbon atoms.
Furanose is collective term for carbohydrates which have chemical structure which includes five-membered ring system that consists of one oxygen atom and four carbon atoms.
Glucose exists both in pyranose and furanose form. It's structure is shown in image.
Answer:
130.4 grams of sucrose, would be needed to dissolve in 500 g of water.
Explanation:
Colligative property of boiling point elevation:
ΔT = Kb . m . i
In this case, i = 1 (sucrose is non electrolytic)
ΔT = Kb . m
0.39°C = 0.512°C/m . m
0.39°C /0.512 m/°C = m
0.762 m (molality means that this moles, are in 1kg of solvent)
If in 1kg of solvent, we have 0.712 moles of sucrose, in 500 g, which is the half, we should have, the hallf of moles, 0.381 moles
Molar mass sucrose = 342.30 g/m
Molar mass . moles = mass
342.30 g/m . 0.381 m = 130.4 g
Answer:
1.16atm
Explanation:
We are going to derive the mass of ether from density
mass=density *volume
Also moles=mass/molecular mass
molar mass C2H5OC2H5 =74.12 g/mole
the density of ether is 0.7134 g/ml
mass C2H5OC2H5 = 5.30 ml x 0.7134 g/ml = 3.78 g
moles C2H5OC2H5 =3.78 g x 1 mole/74.12 g = 0.0509 moles
PV = nRT where P=?; n=0.0509 moles; V=6.50L; R=0.0821 Latm/Kmol; T=35ºC +273 = 308K
P = nRT/V = 0.0509)(0.0821)(308)/6.50
P = 0.198 atm (to 3 significant figures (this is the partial pressure of diethyl ether).
TOTAL PRESSURE
P1+p2+p3
= 0.198 atm + 0.750 atm + 0.207 atm =1.1550atm
1.16atm(3 significant figures)
