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pickupchik [31]
3 years ago
9

A battery is connected at its positive end by a wire to the head of an iron nail. The wire then coils aground the nail to its ti

p and from there to the negative end of the battery.
Predict what would happen to the strength of the electromagnet in each situation.
Using a battery with a higher voltage:
Wrapping the wire around the nail only three times:
Changing the direction of the current by reversing the battery connections:
Using a plastic stick in place of the iron nail:
Holding the nail vertically rather than horizontally: it
Chemistry
2 answers:
cupoosta [38]3 years ago
7 0

Answer:

sing a battery with a higher voltage:  

✔ stronger force of attraction

Wrapping the wire around the nail only three times:  

✔ weaker force of attraction

Changing the direction of the current by reversing the battery connections:  

✔ no change

Using a plastic stick in place of the iron nail:  

✔ weaker force of attraction

Holding the nail vertically rather than horizontally:  

✔ no change

Explanation: You're Welcome

puteri [66]3 years ago
7 0

Answer:

✔ stronger force of attraction

✔ weaker force of attraction

✔ no change

✔ weaker force of attraction

✔ no change

Explanation:

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Draw the other possible resonance structure of each organic ion. In each case, draw the structure that minimizes formal charges.
pshichka [43]

Three resonance structures  can be drawn for the allyl cation while two resonance structures can be drawn for the amidate ion.

Sometimes, we cannot fully describe the bonding in a chemical specie using a single chemical structure. In such cases, we have to use a number of structures which cooperatively represent the actual bonding in the molecule. These structures are called resonance or canonical structures.

The resonance structures of the allyl cation and the amidate ion are shown in the images attached to this answer. These structures show the different bonding extremes in these organic ions.

Learn more: brainly.com/question/4933048

7 0
3 years ago
If 61.5 moles of an ideal gas occupies 97.5 liters at 473 K, what is the pressure of the gas, in mmHg?
kiruha [24]
I’d say for the answer 13.13 mmHg?
5 0
3 years ago
A chocolate bar weighs 4.00 ounces. If the chocolate delivers 2.78 calories per gram, how many
dimulka [17.4K]

Answer:

40.79

Explanation:

an ounce is equal to approximately 28.3 grams.

if you have 4 ounces then it would be equal to about 113.4 grams. then you would divide that by 2.78 which will equal about 40.8

6 0
3 years ago
What two parts are found in the nucleus
cricket20 [7]
Protons and neutrons
5 0
3 years ago
The Haber Process synthesizes ammonia at elevated temperatures and pressures. Suppose you combine 1580 L of nitrogen gas and 351
ikadub [295]

Answer : The volume of reactant measured at STP left over is 409.9 L

Explanation :

First we have to calculate the moles of N_2 and H_2 by using ideal gas equation.

<u>For N_2 :</u>

PV_{N_2}=n_{N_2}RT

where,

P = Pressure of gas at STP = 1 atm

V = Volume of N_2 gas = 1580 L

n = number of moles N_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

1atm\times 1580L=n_{N_2}\times (0.0821L.atm/mol.K)\times 273K

n_{N_2}=70.49mole

<u>For H_2 :</u>

PV_{H_2}=n_{H_2}RT

where,

P = Pressure of gas at STP = 1 atm

V = Volume of H_2 gas = 3510 L

n = number of moles H_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

1atm\times 3510L=n_{H_2}\times (0.0821L.atm/mol.K)\times 273K

n_{H_2}=156.6mole

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

From the balanced reaction we conclude that

As, 3 mole of H_2 react with 1 mole of N_2

So, 156.6 moles of H_2 react with \frac{156.6}{3}\times 1=52.2 moles of N_2

From this we conclude that, N_2 is an excess reagent because the given moles are greater than the required moles and H_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of N_2 reactant (unreacted gas).

Excess moles of N_2 reactant = 70.49 - 52.2 = 18.29 moles

Now we have to calculate the volume of reactant, measured at STP, is left over.

PV=nRT

where,

P = Pressure of gas at STP = 1 atm

V = Volume of gas = ?

n = number of moles of unreacted gas = 18.29 moles

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

1atm\times V=18.29mole\times (0.0821L.atm/mol.K)\times 273K

V=409.9L

Therefore, the volume of reactant measured at STP left over is 409.9 L

8 0
3 years ago
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