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dexar [7]
1 year ago
8

b) Write a few paragraphs describing the chemical reaction and explaining the energy change in the reaction. Your document shoul

d: i. identify the reactants and products. ii. describe the change in energy that occurs as bonds are broken and formed. iii. identify how the potential energy of the reactants compares to the potential energy of the products. iv. state and explain the net change in enthalpy. v. identify whether the reaction is endothermic or exothermic. vi. explain how energy is conserved between the reaction and the surrounding environment.
Chemistry
2 answers:
Lena [83]1 year ago
5 0

Answer:

The energy change in a chemical reaction is due to the difference in the amounts of stored chemical energy between the products and the reactants. This stored chemical energy, or heat content, of the system is known as its enthalpy.

Explanation:

if u want to u can give me the crown btw have a good day

Maksim231197 [3]1 year ago
3 0

Answer:

here is what i put for this project:

The reaction in this project is exothermic. The energy change in a chemical reaction is due to the difference in the amounts of stored chemical energy between the products and the reactants. This stored chemical energy, or heat content, of the system is known as its enthalpy. The potential energy in reactants is higher than the potential energy of products in an exothermic reaction. Since energy is given off, the products are lower in energy than the reactants. The energy that is given off is a result of the formation of new bonds.

The change of the enthalpy happening in the chemical reaction is different from the pathway in between the initial and the final states. The addition of the equation gives the net change of the enthalpy. Enthalpy change is the amount of heat evolved or absorbed in the reaction under constant pressure. This can be calculated by checking the amount of heat evolved during the reaction. This reaction occurs at a constant pressure.

You might be interested in
Just Lemons Lemonade Recipe Equation:
zalisa [80]

Answer:

Explanation:

Hello!

<em>Complete text:</em>

<em>Honors Stoichiometry Activity WorksheetInstructions: </em>

<em>Activity Two: Just Lemons, Inc. Production</em>

<em>Here's a one-batch sample of Just Lemons lemonade production. Determine the percent yield and amount of leftover ingredients for lemonade production and place your answers in the data chart.</em>

<em>Hint: Complete stoichiometry calculations for each ingredient to determine the theoretical yield. Complete a limiting reactant-to-excess reactant calculation for both excess ingredients. </em>

<em>Water 946.36 g </em>

<em>Sugar 196.86 g </em>

<em>Lemon Juice 193.37 g </em>

<em>Lemonade 2050.25g</em>

<em>Leftover Ingredients?</em>

<em>Just Lemons Lemonade Recipe Equation:</em>

<em>2 water + sugar + lemon juice = 4 lemonade</em>

<em>Mole conversion factors:</em>

<em>1 mole of water = 1 cup = 236.59 g</em>

<em>1 mole of sugar = 1 cup = 225 g</em>

<em>1 mole of lemon juice = 1 cup = 257.83 g</em>

<em>1 mole of lemonade = 1 cup = 719.42 g</em>

You have the information on the ingredients used to produce one batch of lemonade and the amount of lemonade produced. To determine which ingredients be leftovers, you have to determine first, which one is the limiting reactant, i.e. the ingredient that will be used up first.

According to the recipe, to make 4 moles of lemonade, you use 2 moles of water, one mole of sugar and one mole of lemon juice, expressed in grams:

2 water  + sugar + lemon juice = 4 lemonade

2*(236.59) + 225g + 257.83g  = 4*(719.42)g

    473.18g + 225g + 257.83g = 2877.68g

So for every 2877.68g of lemonade made, they use 473.18g of water, 225g of sugar, and 257.83g of lemon juice.

You know that they made a batch of 2050.25g, so to detect the limiting reactant, first, you have to calculate, in theory, how much of each ingredient you need to make the given amount of lemonade:

Use cross multiplication

<u>Water:</u>

2877.68g lemonade → 473.18g water

2050.25g lemonade → X= (2050.25*473.18)/2877.68= 337.12g water

Following the recipe, to elaborate 2050.25g of lemonade, you need to use 337.12g of water.

<u>Sugar:</u>

2877.68g lemonade → 225g sugar

2050.25g lemonade → X= (2050.25*225)/2877.68= 160.30g sugar.

To elaborate 2050.25f of lemonade you need to use 160.30g of sugar.

<u>Lemon juice:</u>

2877.68g lemonade → 257.83g lemon juice

2050.25g lemonade → X= (2050.25*257.83)/2877.68= 183.69g lemon juice.

To elaborate 2050.25f of lemonade you need to use 183.69g lemon juice.

Available ingredients vs. theoretical yields for 2050.25g of lemonade:

Water 946.36 g → 337.12g

Sugar 196.86 g → 160.30g

Lemon Juice 193.37 g → 183.69g

The lemon juice will be the first ingredient to be used up, there will be a surplus of water and sugar.

I hope this helps!

7 0
2 years ago
Determine the overall charge on particles with the subatomic makeups given.
Gwar [14]

Answer:

The correct solution will be "+3".

Explanation:

The given values are:

Number of protons

= 13

Number of neutrons

= 14

Number of electrons

= 10

As we know,

⇒  Charge = Number \ of \ protons - Number \ of \ electrons

On putting the estimated values, we get

⇒               =13 - 10

⇒               =+3

5 0
2 years ago
Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.
maria [59]
The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
5 0
1 year ago
At high temperature, 2.00 mol of HBr was placed in a 4.00 L container where it decomposed in the reaction: 2HBr(g) H2(g) Br2(g)
viktelen [127]

Answer: K_c for this reaction at this temperature is 0.029

Explanation:

Moles of  HBr = 2.00 mole

Volume of solution = 4.00 L

Initial concentration of HBr=\frac{moles}{Volume}=\frac{2.00}{4.00L}=0.500M

The given balanced equilibrium reaction is,

                            2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial conc.              0.500 M              0  M        0 M  

At eqm. conc.            (0.500-2x) M   (x) M   (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}

Equilibrium concentration of [Br_2] = x =  0.0955 M

Now put all the given values in this expression, we get :

K_c=\frac{0.0955\times 0.0955}{0.500-2\times 0.0955}

K_c=0.029

Thus K_c for this reaction at this temperature is 0.029

7 0
2 years ago
A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical formula? The second step is to calcul
Free_Kalibri [48]
Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.

2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O

1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O

Empirical formula
H2SO4
8 0
1 year ago
Read 2 more answers
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