What are some of the different substances that make up a pizza?

The question is in the attached word document.

Alekssandra [29.7K]

Answer:i read it Explanation:

3 0

1 year ago

Explain why we dissected the eggs that didn't hatch. What was the outcome of our dissection? PLEASE HELP DUE TONIGHT

stiks02 [169]

Answer:

It helps us develop our obervation skills and we learn what the egg looks like. We can learn the nutritional values of the egg.

Explanation:

Hope that helped

5 0

2 years ago

The change of matter, solid, liquid and gas is called a_________.

valkas [14]

Answer:

This would be a phase change

Explanation:

Since the diffrent phases solid,liquid,gas are changing to other phases.

Sorry for the bad explanation.

7 0

2 years ago

Read 2 more answers

CO(g)+2H2(g)⇌CH3OH(g)CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial concentrations o

Katarina [22]

Answer:

9.4

Explanation:

The equation for the reaction can be represented as:

$CO_{(g)}$ + $2H_2O_{(g)}$ ⇄ $CH_3OH_{(g)}$

The ICE table can be represented as:

$CO_{(g)}$ + $2H_2_{(g)}$ ⇄ $CH_3OH_{(g)}$

Initial 0.27 0.49 0.0

Change -x -2x x

Equilibrium 0.27 - x 0.49 -2x x

We can now say that the concentration of $CH_3OH_{(g)}$ at equilibrium is x;

Let's not forget that at equilibrium $CH_3OH_{(g)}$ = 0.11 M

So:

x = [ $CH_3OH_{(g)}$ ] = 0.11 M

[ $CO_{(g)}$ ] = 0.27 - x

[ $CO_{(g)}$ ] = 0.27 - 0.11

[ $CO_{(g)}$ ] = 0.16 M

[ $2H_2_{(g)}$ ] = (0.49 - 2x)

[ $2H_2_{(g)}$ ] = (0.49 - 2(0.11))

[ $2H_2_{(g)}$ ] = 0.49 - 0.22

[ $2H_2_{(g)}$ ] = 0.27 M

$K_C = \frac{[CH_3OH]}{[CO][H_2]^2}$

$K_C = \frac{(0.11)}{(0.16)[(0.27)^2}$

$K_C = 9.4307$

$K_C$ = 9.4

∴ The equilibrium constant at that temperature = 9.4

8 0

2 years ago

Calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxid

kiruha [24]

Answer:

-471 Kj/mole acrylic acid

Explanation:

THIS IS THE COMPLETE QUESTION BELOW;

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In the first step, calcium carbide and water react to form acetylene and calcium hydroxide: CaC (s) + 2 H2O(g) - CH (9) + Ca(OH),(s) AH -414. kJ In the second step, acetylene, carbon dioxide and water react to form acrylic acid: 6 C H (9) + 3 CO2(9) + 4H2O(g) - SCH,CHCO,H) AH-132. kJ Calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ?

The two equations from the reaction can be written as;

a)CaC₂(s) + 2H₂O(l) ------->C₂H₂(g) + CaOH₂(s)

Δ H= -414Kj ........................ equation (a)

b)6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj ...................... equation (b)

In equation (b)acrylic acid was produced by the reaction between Acetylene carbon dioxide and water

Then we can multiply equation(a) by factor of 6 and the ΔH Then we have (6× -414Kj)= ΔH= -2484Kj.

6CaC₂(s) + 12H₂O(l) ------->6C₂H₂(g) + 6CaOH₂(s)

Δ H= -2484Kj.................. equation (c)

6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj

Then add equation (c) and equation(b) then we have

6CaC₂(s) + 16H₂O(l)+3CO₂(g)------> 5CH₂CHCO₂H(g) + 6CaOH₂(s) ΔH= -2352Kj

ΔH(net)= -2352Kj/5moles

=-471Kj/mole

therefore, net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ? is -471Kj/mole acrylic acid

6 0

2 years ago