Answer: n = 3.0 moles
V = 60.0 L
T = 400 K
From PV = nRT, you can find P
P = nRT/V = (3.0 mol)(0.0821 L-atm/K-mol)(400 K)/60.0L
P = 1.642 atm = 1.6 atm (to 2 significant figures)
Answer:
47.9 g of ethanol
Explanation:
Combustion is a chemical reaction in which a substance reacts with oxygen to produce heat and light. Combustion reactions have been very useful as a source of energy. Ethanol is now burnt for energy purposes as a fuel. Ethanol has even been proposed as a possible alternative to fossil fuels.
Since 1 mole of ethanol when combusted releases 1367 kJ/mol of energy
x moles of ethanol releases 1418 kJ/mol.
x= 1 × 1418 kJ/mol/ 1367 kJ/mol
x= 1.04 moles of ethanol.
Mass of ethanol = number of moles × molar mass
Molar mass of ethanol = 46.07 g/mol
Mass of ethanol = 1.04 moles × 46.07 g/mol
Mass of ethanol= 47.9 g of ethanol
Answer:
long range order
Explanation:
A crystal consists of atoms, ions or molecules having both short range and long range order. The atoms, ions or molecules are arranged in a regular pattern throughout the lattice both at immediate vicinities and across the entire crystal structure.
This order accounts for the definite shape and unique properties of crystals which include their sharp melting and boiling points which distinguishes them from amorphous substances.
You would need exactly 50 molecules of glucose.
Answer: The enthalpy change for formation of butane is -125 kJ/mol
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f%7Breactants%7D%5D)
Putting the values we get :
![\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B4%5Ctimes%20H_f_%7BCO_2%7D%2B5%5Ctimes%20H_f_%7BH_2O%7D%5D-%5B1%5Ctimes%20H_f_%7BC_4H_%7B10%7D%7D%2B%5Cfrac%7B13%7D%7B2%7D%5Ctimes%20H_f_%7BO_2%7D%5D)
![-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]](https://tex.z-dn.net/?f=-2877%3D%5B%284%5Ctimes%20-393%29%2B%285%5Ctimes%20-286%29%5D-%5B1%5Ctimes%20H_f_%7BC_4H_%7B10%7D%7D%2B%5Cfrac%7B13%7D%7B2%7D%5Ctimes%200%5D)
Thus enthalpy change for formation of butane is -125 kJ/mol
