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scZoUnD [109]
3 years ago
9

he electrical leads from the Constant Current System are damaged slightly, leading to a small "short-circuit" that results in 10

% of the measured current passing through the short circuit without going through the metal strips. The current displayed (and hence the value of the integral) are based on the total current, but only 90% of this actually leads to dissolution of metal. What effect would this have on your experimental value of Avogadro's Number? (Note that you complete your calculations assuming that 100% of the current is applied to dissolving the metal strip.)
Chemistry
1 answer:
VikaD [51]3 years ago
8 0

Answer:

The Avogadro's Number will reduce in value

Explanation:

From the question we are told that

The current pass through the short circuit is 0.1 *I

Here I is the total current

Generally Avogadro's Number is mathematically represented as

N_A =  \frac{Q}{ 2 *  n * e}

Here Q is the charge which is mathematically represented as

Q =  It

        n  is number of moles

        e is the charge on an electron

 So

       N_A =  \frac{I * t }{ 2 *  n * e}

Now when the current decreases from 100% to  90% due to the short circuit . this will also reduce  the no of moles of metal  dissolved hence from the equation we see that the Avogadro's Number will reduce

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WILL MARK BRAINIEST <br><br> Can someone please help me understand this?
Mila [183]

To convert from moles to grams you divide by the molar mass of the element. To convert from grams to moles you X by the molar mass element

5 0
3 years ago
An ethylene gas torch requires 300 L of gas at 0.8 atm. What will be the pressure of the gas if ethylene is supplied by a 200.0
Vlad [161]

Answer:

1.2 atm

Explanation:

Given data

  • Volume of the gas in the tank (V₁): 200.0 L
  • Pressure of ethylene gas in the tank (P₁): ?
  • Volume of the gas in the torch (V₂): 300 L
  • Pressure of the gas in the torch (P₂): 0.8 atm

If we consider ethylene gas to be an ideal gas, we can find the pressure of ethylene gas in the tank using Boyle's law.

P_1 \times V_1 = P_2 \times V_2\\P_1 = \frac{P_2 \times V_2}{V_1} = \frac{0.8atm \times 300L}{200.0L} = 1.2 atm

3 0
3 years ago
The compound methyl butanoate smells like apples. Its percent composition is 58.8% C, 9.9% H, and 31.4% O. What’s the empirical
blsea [12.9K]
To find the empirical formula you would first need to find the moles of each element:

58.8g/ 12.0g = 4.9 mol C

9.9g/ 1.0g = 9.9 mol H

31.4g/ 16.0g = 1.96 O

Then you divide by the smallest number of moles of each:

4.9/1.96 = 2.5

9.9/1.96 = 6

1.96/1.96 = 1

Since there is 2.5, you find the least number that makes each moles a whole number which is 2.

So the empirical formula is C5H12O2.
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4 years ago
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Answer:

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Explanation:

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