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shusha [124]
2 years ago
9

Which type of energy is stored within an object and is partially determined by the object’s position?

Chemistry
2 answers:
guapka [62]2 years ago
5 0

Answer:

D. Potential energy

Explanation:

Vsevolod [243]2 years ago
3 0
It is A. Electrical energy
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Will mark as brainiest <br> Please help
Levart [38]

Answer:

  1. metal sulfate
  2. metal sulfate
  3. copper sulfate
  4. copper nitrate
  5. copper chloride
  6. copper phosphate
  7. hydrochloric acid, water
  8. Potassium, sulfuric acid, water

(Correct me if I am wrong)

4 0
3 years ago
Read 2 more answers
Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39
nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1)   Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)    ΔH = −23 kJ

2)   3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g)   ΔH = −39 kJ

3)   Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3:  Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

We have to multiply the third equation by 2.

2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ

<u>This equation we add to the second equation</u>

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

ΔH°  = 2*18 + (-39) = 36 - 39 =  -3 kJ

<u>This new equation we will divide by 3 </u>

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) →  CO2(g) + 2FeO(s)

ΔH°  =-3/3 = -1 kJ

<u>Now we will substract this new equation from the first equation</u>

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

<u>The next equation we will divide by 2</u>

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

CO(g)  + FeO(s)  → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

7 0
3 years ago
1. Carson completed conductivity testing on the solutions shown in the data table below and recorded his findings in the conduct
pishuonlain [190]

Answer:

See below :)

Explanation:

There is an evident reason why some of the solutions Carson's has listed and observed, does conduct electricity and some that do.

A flow of electrical charge is called an electric current. Ions are atoms, or sets of atoms, that contain an electrical charge. There are two types of ions, cation or a positively charged ion containing a deficiency of electrons, and anion or a negatively charged ion which contains a surplus of electrons. When a solution conducts electricity the charge is carried within by ions that move through the solution. The larger the number of ions in the solution, the better the conductivity of the solution is. Pure water does not conduct very well because it contains very few ions, but when table salt (NaCl) is dissolved in the water, this solution does conduct well because the solution contains a more abundance of ions. The majority of the ions come from the table salt, chemically names sodium chloride. Because Sodium contains its sodium ions, and these are positive charge and chloride ions which is a negative charge, it is called an ionic substance. Not every substance is made up of ions, one such example is sugar (C12H22O11). Sugar is made up of uncharged particles also called molecules. Although sugar is a substance its molecules do not hold a charge, thus when sugar is dissolved in water, the solution does not conduct electricity, due to the lack of ions in the solution.

Therefore, depending on the ions that make up the compound, the substance would or would not conduct electricity.

4 0
2 years ago
Help me<br><br> Extra points
aleksley [76]
G: maintain homeostasis
3 0
3 years ago
Read 2 more answers
t 745 K, the reaction below has an equilibrium constant (Kc) of 5.00 × 102. H2 (g) + I2 (g) ⇌ 2 HI (g) If a mixture of 0.10 mol
spayn [35]

Answer : The concentration of HI (g) at equilibrium is, 0.643 M

Explanation :

The given chemical reaction is:

                        H_2(g)+I_2(g)\rightarrow 2HI(g)

Initial conc.    0.10        0.10      0.50

At eqm.        (0.10-x)  (0.10-x)   (0.50+2x)

As we are given:

K_c=5.00\times 10^2

The expression for equilibrium constant is:

K_c=\frac{[HI]^2}{[H_2][I_2]}

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

5.00\times 10^2=\frac{(0.50+2x)^2}{(0.10-x)\times (0.10-x)}

x = 0.0713  and x = 0.134

We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.0713

The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M

Thus, the concentration of HI (g) at equilibrium is, 0.643 M

8 0
3 years ago
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