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2 years ago

Is it possible to decrease resistance of the wire without changing the material it's made out of?

1 answer:

3 0

Yes, it is possible to decrease the resistance of a wire without changing the material it made out of. This is because, there are many factors which affect the resistance of a wire. These factors can be manipulated to change the resistance of the wire. The factors include: cross sectional area of the wire, length of the wire, temperature and the material of the wire. The other three factors can be manipulated to change the resistance of the wire without changing the material of the wire.

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what is the new pressure acting on a 2.5 l balloon if its original volume was 5.8 l at 3.7 ATM of pressure

Ivan

Answer : The new pressure acting on a 2.5 L balloon is, 8.6 atm.

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure = 3.7 atm

$P_2$ = final pressure = ?

$V_1$ = initial volume = 5.8 L

$V_2$ = final volume = 2.5 L

Now put all the given values in the above equation, we get:

$3.7atm\times 5.8L=P_2\times 2.5L$

$P_2=8.6atm$

Thus, the new pressure acting on a 2.5 L balloon is, 8.6 atm.

8 0

2 years ago

According to the lab guide, which changes below will you look for in order to test the hypothesis? check all that apply. changes

viktelen [127]

Answer:

All of them are.

Explanation:

8 0

2 years ago

Read 2 more answers

The number of sp2 hybrid orbitals on the carbon atom in CO32– is

Vladimir79 [104]

The number of sp2 hybrid orbitals on the carbon atom in CO32– is 3. Because hybrids = combination of 2 different types of orbitals
sp2 = 1/3 s character + 2/3 p character

7 0

2 years ago

Read 2 more answers

A 8.6 g sample of methane and 15.6 g sample of oxygen react according to the reaction in the video. identify the limiting reacta

GalinKa [24]

Answer:

23.6 g carbon dioxide comes from 8.6 g of CH4 or 10.7 g carbon dioxide comes from 15.6 g O that means the 15.6 g of oxygen is still the limiting reactant because it gets used up and only makes 10.7 g of CO2. 

Explanation:

1) Balanced chemical equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

2) mole ratios:
1 mol CH₄ : 2mol O₂ : 1 mol CO₂ : 2 mol H₂O

3) molar masses
CH₄: 16.04 g/mol
O₂: 32.0 g/mol
CO₂: 44.01 g/mol

4) Convert the reactant masses to number of moles, using the formula

number of moles = mass in grams / molar mass

CH₄: 8.6g / 16.04 g/mol = 0.5362 moles


O₂: 15.6 g / 32.0 g/mol = 0.4875 moles

5) If the whole 0.5632 moles of CH₄ reacted that yields to the same number of moles of CO₂ and that is a mass of:
mass of CO₂ = number of moles x molar mass = 23.60 g of CO₂

Which is what the first part of the answer says.

6) If the whole 0.4875 moles of O₂ reacted that would yield 0.4875 / 2 = 0.24375 moles of CO₂, and that is a mass of:
mass of CO₂ = 0.4875 grams x 44.01 g/mol = 10.7 grams of CO₂.

Which is what the second part of the answer says.

7) From the mole ratio you know infere that 0.5362 moles of CH₄ needs more twice number of moles of O₂, that is 1.0724 moles of O₂, and since there are only 0.4875 moles of O₂, this is the limiting reactant.

Which is what the chosen answer says.

8) From the mole ratios 0.4875 moles of O₂ produce 0.4875 / 2 moles of CO₂, and that is:
0.4875 / 2 mols x 44.01 g/mol = 10.7 g of CO₂, which is the last part of the answer.

3 0

2 years ago

Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 3

ZanzabumX [31]

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)× $\frac{1molZn}{65,38g}$ × $\frac{1molH_{2}(g)}{1molZn}$ = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = 190,1L

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL× $\frac{101,325}{1atmL}$ = 19,26 kJ

I hope it helps!

5 0

2 years ago