Ba2+(aq)+SO42−(aq)→BaSO4(s)

Mashutka [201]

A two carbon chain attached to a benzene ring: ethylbenzene

6 0

3 years ago

Explain where you would find the three different types of muscles (cardiac, smooth and skeletal) and how these different types o

yawa3891 [41]

Explanation: Skeletal muscle, attached to bones, is responsible for skeletal movements. The peripheral portion of the central nervous system (CNS) controls the skeletal muscles. Thus, these muscles are under conscious, or voluntary, control. The basic unit is the muscle fiber with many nuclei. These muscle fibers are striated (having transverse streaks) and each acts independently of neighboring muscle fibers.

4 0

3 years ago

How many moles are in 100.0 g of gold?

Savatey [412]

Answer:

196.96655

Explanation:

The answer is 196.96655. I assume you are converting between grams Gold and mole. The molecular formula for Gold is Au. The SI base unit for the amount of substance is the mole.

7 0

3 years ago

Indicate whether each of the statements below is true or false. 1. CBr4 is more volatile than CCl4. 2. CBr4 has a higher vapor p

Rashid [163]

Answer:

1. False

2. False

3. True

4. False

Explanation:

1. CBr4 is more volatile than CCl4 False

The molecular weight of CBr4 is is greater than the CCl4, therefore it has less tendency to escape to the gas phase. Also, the CBr4 has greater London dispersion forces compared to CCl4 since bromine is a larger atom than chlorine.

2. CBr4 has a higher vapor pressure at the same temperature than CCl4 False

For the same reasons as above, the vapor pressure of CBr4 is smaller than the vapor pressure of CCl4

3. CBr4 has a higher boling point than CCl4 True

Again, CBr4 having a molecular weight greater than CCl4 ( 331 g/mol vs 158.2 g/mol) is heavier and less volatile with a higher boiling point than CCl4.

4. CBr4 has weaker intermolecular forces than CCl4 False

Both molecules are non-polar because the dipole moments in C-Cl and C-Br bonds cancel in the tetrahedron. The only possible molecular forces are of the London dispersion type which are temporary and greater for larger atoms.

8 0

3 years ago

How many grams of C5H12 must be burned to heat 1.39 kg of water from 21.2 °C to 97.0 °C? Assume that all the heat released durin

faust18 [17]

Answer:

m = 8.9856 g

Explanation:

In order to do this, we need to write the expressions that are to be used. First, to calculate heat:

Q = m*C*ΔT (1)

Where C would be heat capacity of the substance.

The heat can also be relationed with the moles and enthalpy of a compound using the following expression:

Q = n*ΔH (2)

Finally for the mass of any compound, we use the following expression:

m = n*MM (3)

So, in order to calculate the grams of pentane (C5H12), we need to calculate the moles of the compound, and to do that, we need the heat exerted.

So, as we are using water, let's calculate the heat that is been exerted with the water. The C of the water is 4.186 J/g °C so:

Q = (1.39 * 1000) * 4.186 * (21.2 - 97)

Q = -441,045.33 J

This is the heat neccesary to burn pentane and heat water. Now, with this value, let's calculate the moles used of pentane with expression (2). The ΔH of the pentane is -3,535 045.kJ/mol or -3.535x10⁶ J/mol. Solving for n we have:

n = -441,045.3 / -3.535x10⁶

n = 0.1248 moles

Finally, we can calculate the grams needed with expression (3). The molar mass of pentane is 72 g/mol

m = 0.1248 * 72

m = 8.9856 g

This is the mass needed to heat 1.39 kg of water

6 0

4 years ago