The following information is given for ether, C2H5OC2H5, at 1atm:

boiling point = 34.6 °C Hvap(34.6 °C) = 26.5 kJ/mol
specific heat liquid = 2.32 J/g°C

At a pressure of 1 atm,__kJ of heat are needed to vaporize a 25.1 g sample of liquid ether at its normal boiling point of 34.6 °C.

q = mols ether x Hvap.

mols = grams/molar mass = ?
You know Hvap and can solve for mols. Substitute and solve for q.

To calculate the heat needed to vaporize a sample of liquid ether at its normal boiling point, we can use the equation:

q = m * Hvap

Where:
q is the heat needed
m is the mass of the sample
Hvap is the molar heat of vaporization

First, we need to calculate the moles of ether in the sample. To do this, we can use the molar mass of ether, which is the sum of the atomic masses of its elements (C = 12.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol):

molar mass = (2 * C) + (5 * H) + (1 * O)
molar mass = (2 * 12.01 g/mol) + (5 * 1.01 g/mol) + (1 * 16.00 g/mol)
molar mass = 74.12 g/mol

moles = mass / molar mass
moles = 25.1 g / 74.12 g/mol

Now, let's calculate the heat needed to vaporize the sample at its normal boiling point:

q = moles * Hvap
q = (25.1 g / 74.12 g/mol) * 26.5 kJ/mol

Finally, we can calculate the heat needed to vaporize the sample:

q = (25.1 g / 74.12 g/mol) * 26.5 kJ/mol
q ≈ 8.98 kJ

Therefore, approximately 8.98 kJ of heat is needed to vaporize a 25.1 g sample of liquid ether at its normal boiling point of 34.6 °C.

To find the heat needed to vaporize the liquid ether, we can use the heat of vaporization (Hvap) and the given mass of the liquid.

First, convert the given mass of the liquid ether from grams to moles using its molar mass. The molar mass of C2H5OC2H5 (ether) is calculated as:

(2 * atomic mass of carbon) + (6 * atomic mass of hydrogen) + (1 * atomic mass of oxygen)
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (1 * 16.00 g/mol)
= 74.12 g/mol

So, the number of moles (n) of liquid ether is:
n = mass / molar mass
n = 25.1 g / 74.12 g/mol

Next, we can use the heat of vaporization to find the heat energy needed to vaporize the liquid ether. The Hvap value is given as 26.5 kJ/mol.

The heat energy (q) needed to vaporize the liquid ether can be calculated using the equation:
q = n * Hvap

Substituting the values, we have:
q = (25.1 g / 74.12 g/mol) * 26.5 kJ/mol

Now, let's calculate the heat energy (q):
q = (25.1 / 74.12) * 26.5 kJ

To find the answer, simply multiply the result with the appropriate units:
q ≈ 8.94 kJ

Therefore, approximately 8.94 kJ of heat are needed to vaporize a 25.1 g sample of liquid ether at its boiling point of 34.6 °C.