CH4+2O2====>2H2O+CO2.calculate the mass of CO2 and H2O formed by combustion of 64g of CH4.

144g of 2H2O

ELASTICITY, LIKE YOUR SOCKS (10/10 points)

The crystal structure of graphite is shown below. Use the figure to answer the following questions.

Compare the Young's moduli for the directions indicated in the figure. Fill in the blank.

The modulus along a is _________ that along b

equal to - correct
The modulus along a is _________ that along c

greater than - correct
The modulus along b is _________ that along c

greater than - correct
Which of the following reasons best explain your reasoning when comparing the moduli of b and c?

The Poisson ratio is greater than 0.3 in graphite.
The two directions are crystallographically identical
Fundamental bending and stretching of bonds is different in the two directions
Van der Waals bonding has a lower bond strength than colavent bonding - correct <--
Stretching of covalent bonds requires greater force than bending such bonds
The opportunity to stetch rings in one direction gives one direction a lower modulus

Which of the following reasons best explain your reasoning when comparing the moduli of a and b?

Stretching of covalent bonds requires greater force than bending such bonds
The Poisson ratio is greater than 0.3 in graphite.
The two directions are crystallographically identical - correct <---
Fundamental bending and stretching of bonds is different in the two directions
The opportunity to stetch rings in one direction gives one direction a lower modulus
Van der Waals bonding has a lower bond strength than colavent bonding

To calculate the mass of CO2 and H2O formed by the combustion of 64g of CH4, you need to use the stoichiometry of the balanced equation.

The balanced equation is:
CH4 + 2O2 → 2H2O + CO2

First, we will calculate the moles of CH4 using its molar mass:
Molar mass of CH4 = 12.01 g/mol (C) + 4*1.01 g/mol (H) = 16.04 g/mol

moles of CH4 = mass of CH4 / molar mass of CH4
moles of CH4 = 64g / 16.04 g/mol = 4 moles

Using the stoichiometry of the balanced equation, we can see that every 1 mole of CH4 produces 1 mole of CO2. Therefore, 4 moles of CH4 will produce 4 moles of CO2.

Molar mass of CO2 = 12.01 g/mol (C) + 2*16.00 g/mol (O) = 44.01 g/mol

mass of CO2 = moles of CO2 * molar mass of CO2
mass of CO2 = 4 moles * 44.01 g/mol = 176.04 g

Therefore, the mass of CO2 formed by the combustion of 64g of CH4 is 176.04g.

Similarly, from the stoichiometry of the balanced equation, every 1 mole of CH4 produces 2 moles of H2O. Therefore, 4 moles of CH4 will produce 8 moles of H2O.

Molar mass of H2O = 2*1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol

mass of H2O = moles of H2O * molar mass of H2O
mass of H2O = 8 moles * 18.02 g/mol = 144.16 g

Therefore, the mass of H2O formed by the combustion of 64g of CH4 is 144.16g.