A 46.6-mgmg sample of boron reacts with oxygen to form 150 mgmg of the compound boron oxide. Part A What is the empirical formula of boron oxide

B₂O₃

Explanation:

Step 1: Calculate the mass of oxygen in 150 mg of boron oxide

Of 150 mg of boron oxide, 46.6 mg belong to boron. The mass of oxygen is:

150 mg - 46.6 mg = 103.4 mg

Step 2: Calculate the percent by mass of each element

We will use the following expression.

%Element = mElement/mCompound × 100%

%B = 46.6 mg/150 mg × 100% = 31.1%

%O = 103.4 mg/150 mg × 100% = 68.9%

Step 3: Divide each percentage by the atomic mass of the element

B: 31.1/10.81 = 2.88

O: 68.9/16.00 = 4.31

Step 4: Divide both numbers by the smallest one (2.88)

B: 2.88/2.88 = 1

O: 4.31/2.88 ≈ 1.5

Step 5: Multiply both numbers by 2 so that they are integers

B: 1 × 2 = 2

O: 1.5 × 2 = 3

The empirical formula is B₂O₃.

personal, she chose to do it.

option c: 1605.1 kj

the balanced chemical reaction for combustion of methane is as follows:

here, enthalpy formation of is -74.7 kj/mol, 0 kj/mol, -393.5 kj/mol and -241.82 kj/mol respectively.

the heat released or can be calculated as follow:

putting the values,

thus, heat released by combustion of 1 mol of methane is 802.54 kj/mol. thus, for 2 mol, heat released will be:

therefore, heat released by the combustion of 2 mol of methane is 1605.1 kj.