What is the total number of valence electrons in the propylene molecule??

How does density change as you go deeper in earth

stepladder [879]

Answer:
The density increases.
Explanation:
As you go deeper in depth, pressure increases. Density = mass/volume. The layers beneath us due to pressure get packed to the point of being very dense.

8 0

4 years ago

What is the basic unit of all matter? O A. neutron O B. atom KDC. electron OD proton NUNE nucleus Reset Next

Salsk061 [2.6K]

Answer:

Atom

Explanation:

atom: The basic unit of matter; the smallest unit of an element, having all the characteristics of that element; consists of negatively-charged electrons and a positively-charged center called a nucleus.

7 0

3 years ago

Read 2 more answers

Example of neutral outcome ?

Katarina [22]

Answer:

drops a glass cup of water outcome mad parents

Explanation:

Mad pagents mean death no surviva.

4 0

3 years ago

Read 2 more answers

The reaction for producing glucose in plants, called photosynthesis, is 6 CO 2 + 6 H 2 O light −−→ C 6 H 12 O 6 + 6 O 2 6CO2+6H2

Nostrana [21]

Answer:

30.36moles

Explanation:

6CO2 + 6H2O → C6H12O6 + 6O2

From the equation above,

6moles of H2O are required to produce 1 mole of C6H12O6

Therefore, Xmol of H2O will be require to 5.06moles of C6H12O6 i.e

Xmol of H2O = 6 x 5.06 = 30.36moles

Therefore, 30.36moles of H2O are needed.

5 0

3 years ago

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0

3 years ago