Answer:
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O=
Answer: 9.04 g of H2O
Explanation:
First set up equation: C4H10 (g)+ O2(g) -> CO2(g) + H2O(g)
Next balance it: 2C4H10 (g)+ 13O2(g) -> 8CO2(g) + 10H2O (g)
Use equation to get moles and plug given
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O
Let A be the 80% solution and B be the 20% solution and P be the produce solution of 70%. Va and Vb and Vp are the volumes of A and B and P respectively.
Va + 60 = Vp
0.7Vp = 0.8Va + 0.2(60)
Substituting the value of Vp from the first equation:
0.7(Va + 60) = 0.8Va + 12
30 = 0.1Va
Va = 300 gallons
It is B nevermind I thought it was c but it wasn't.
Answer:
Temperature at which molybdenum becomes superconducting is-272.25°C
Explanation:
Conductor are those hard substances which allows path of electric current through them. And super conductors are those hard substances which have resistance against the flow of electric current through them.
As given, molybdenum becomes superconducting at temperatures below 0.90 K.
Temperature in Kelvins can be converted in °C by relation:
T(°C)=273.15-T(K)
Molybdenum becomes superconducting in degrees Celsius.
T(°C)=273.15-0.90= -272.25 °C
Temperature at which molybdenum becomes superconducting is -272.25 °C
3.47 x 
atoms of gold have mass of 113.44 grams.
Explanation:
Data given:
number of atoms of gold = 3.47 x
mass of the gold in given number of atoms = ?
atomic mass of gold =196.96 grams/mole
Avagadro's number = 6.022 X
from the relation,
1 mole of element contains 6.022 x atoms.
so no of moles of gold given = 
0.57 moles of gold.
from the relation:
number of moles = 
rearranging the equation,
mass = number of moles x atomic mass
mass = 0.57 x 196.96
mass = 113.44 grams
thus, 3.47 x atoms of gold have mass of 113.44 grams
