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Scorpion4ik [409]
3 years ago
7

Jenna built the electrical circuit seen here. What would happen to the electrical energy if Jenna removed one of the clips on th

e battery?
A) The electrical energy would be transformed into heat.
Eliminate
B) The electrical energy would pulse and the light would flicker.
C) The electrical energy would continue to flow but take another path.
D) The electrical energy would stop because the circuit is in
Chemistry
1 answer:
otez555 [7]3 years ago
6 0

Answer: Option (D) is the correct answer.

Explanation:

An electric circuit works well when all the connections are complete but if any of the connections in the circuit is loose or disconnected then it is possible that current will not flow from the circuit.

Therefore, when Jenna removed one of the clips on the battery then circuit becomes incomplete and as a result there will no flow of current.

Thus, we can conclude that the electrical energy would stop because the circuit is incomplete.

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Be sure to answer all parts. Write an unbalanced equation to represent each of the following reactions: Do not include phase abb
Eva8 [605]

<u>Answer:</u> The unbalanced chemical equations are written below.

<u>Explanation:</u>

An unbalanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side is not equal to the total number of individual atoms on the product side. These equations does not follow law of conservation of mass.

  • <u>For a:</u>

The chemical equation for the reaction of nitrogen gas and oxygen gas follows:

N_2+O_2\rightarrow NO_2

The product formed is nitrogen dioxide.

  • <u>For b:</u>

The chemical equation for the decomposition of dinitrogen pentaoxide follows:

N_2O_5\rightarrow N_2O_4+O_2

The product formed is dinitrogen tetroxide and oxygen gas.

  • <u>For c:</u>

The chemical equation for the reaction of ozone to oxygen gas follows:

O_3\rightarrow O_2

The product formed is oxygen gas.

  • <u>For d:</u>

The chemical equation for the reaction of chlorine and sodium iodide follows:

Cl_2+NaI\rightarrow NaCl+I_2

The product formed is sodium chloride and iodine gas

  • <u>For e:</u>

The chemical equation for the reaction of magnesium and oxygen gas follows:

Mg+O_2\rightarrow MgO

The product formed is magnesium oxide

3 0
3 years ago
How many moles of oxygen gas (O2) are required to completely react with 10 moles of hydrogen gas (H2)?
Brums [2.3K]

Answer:

5 moles of oxygen are required.

Explanation:

Given data:

Moles of O₂ required = ?

Moles of H₂ present = 10 mol

Solution:

Chemical equation:

O₂ + 2H₂       →     2H₂O

Now we will compare the moles of oxygen and hydrogen.

                      H₂        :        O₂

                        2         :         1

                       10         :     1/2×10 = 5 mol

5 moles of oxygen are required.

3 0
3 years ago
In two or more complete sentences, explain how to balance the combustion equation: ___C5H12 + ___O2 ⟶ ___CO2 + ___H2O
egoroff_w [7]

Answer:

First, place no. 5 in front of the CO2 in order to balance the carbon atoms. Next, place no. 6 in front of H2O to balance the hydrogen atoms. Lastly place no. 8 in front of the O2 so that there are 16 oxygen atoms on both sides of the reaction.

7 0
3 years ago
Universal indicator solution
Lynna [10]
Shred red cabbage ~ (3/4 of a very small head)

Put the cabbage pieces in a small container ~ ( you can use a Pyrex-4-cup measure, a bowl or even a plastic zipper bag)

Cover the cabbage with very hot water. Let it sleep until the water has cooled. (somewhere between lukewarm and room-temperature) 

The purple liquid you've made is your indicator. 
Pour it into a container and compost the cabbage.

Now look for substances that may be acids or bases. 
Liquids are good, like fruits. 

You can also use solids around for baking are good too. (such as baking soda, salt, sugar, cream of tartar...)

Get containers for mixing (such as tea cups, because they are small, shallow and white inside)

Pour the indicator into the tea cups and add an acid or base.

Lemon juice, rice wine vinegar, and apple cider vinegar, turn the cabbage-water indicator into a pink.

Orange juice or fresh oranges (same thing) turn the cabbage-water indicator into an orangish-pinkish color. 

Baking soda turns the cabbage-water indicator blue.

Milk (non-fat) turns the cabbage-water indicator turn opaque and milky, yet purple.

An egg white (which won't get into the solution immediately until after a lot of stirring) turns the cabbage-water indicator blue.

Hint:
Bases mostly turn the indicator towards blue-ish colors such as purple, light blue, dark blue, opaque blue...

Acids mostly turn the indicator towards pink-ish colours such as orange-ish pink, floral pink...

(You'll have to keep on testing the cabbage-water indicator in after a day or two to see if the indicator quality persists or degrades.
6 0
3 years ago
Read 2 more answers
Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgC
sesenic [268]

Answer:

a. 63.2%

b. 11.7%

c. 73.3%

d. 0.995%

e. 55.5%

Explanation:

An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).

The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:

%IC = (1 - e^{-0.25*(x_A - x_B)^2}) *100%

Where x_A - x_B is the electronegativity difference of the elements. Thus, consulting an electronegativity table:

a. x_{Ti} = 1.5

x_{O} = 3.5

%IC = (1 - e^{-0.25*(3.5 - 1.5)^2})*100%

%IC = 63.2%

b. x_{Zn} = 1.6

x_{Te} = 2.1

%IC = (1 - e^{-0.25*(2.1 - 1.6)^2})*100%

%IC = 11.7%

c. x_{Cs} = 0.7

x_{Cl} = 3.0

%IC = (1 - e^{-0.25*(3.0 - 0.7)^2})*100%

%IC = 73.3%

d. x_{In} = 1.7

x_{Sb} = 1.9

%IC = (1 - e^{-0.25*(1.9 - 1.7)^2})*100%

%IC = 0.995 %

e. x_{Mg} = 1.2

x_{Cl} = 3.0

%IC = (1 - e^{-0.25*(3.0 - 1.2)^2})*100%

%IC = 55.5%

4 0
3 years ago
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