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romanna [79]
3 years ago
15

While doing a lab a student found the density of a piece of pure aluminum to be 2.85 g/cm3 the accepted value for the density of

aluminum is 2.70 g/cm3. What was the percent error
Chemistry
1 answer:
____ [38]3 years ago
3 0

Answer:

THE PERCENT ERROR IS 5.55 %

Explanation:

To calculate the percent error, we use the formula:

Percent error = Found value - accepted value / accepted value * 100

Found value = 2.85 g/cm3

Accepted value = 2.70 g/cm3

Solving for the percent error, we have:

Percent error = 2.85 g/cm3 - 2.70 g/cm3 / 2.70 g/cm3 * 100

Percent error = 0.15 / 2.70 * 100

Percent error = 0.05555 * 100

Percent error = 5.55 %

In conclusion, the percent error is 5.55 %

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Order the various instruments used for measuring water in this experiment (balances, graduated cylinders, beaker and pipette) fr
lara [203]

Answer:

(Most accurate) pippete>graduated cylinder>beaker>balance (Least accurate)

Explanation:

  1. <em>Most accurate. A pipette prived the most accurate method for delivering a known volume of solution, for example, a 10mL transfer pipette has an accuracy of ±0.02mL</em>
  2. A graduated cylinder is specifically used to deliver a known volume, its typical accuracy is ±1%, this means that a 100ml graduated cylinder is accurate to ±1mL.
  3. A beaker is a multipurpose cylindrical glass mainly used to hold liquids. Even though they are graduated, these marks are an estimation, the beaker's accuracy is around 10%.
  4. Least accurate. A balance measures an object's mass, even though water's density is close to 1, a balance is not the ideal equipment to measure volume, its capacity usually goes between 100-200grams and can measure mass to the nearest ±0.01mg to ±1mg.

I hope you find this information useful and interesting! Good luck!

4 0
3 years ago
If the molar heat of combustion of liquid benzene at constant volume and 300k is -3272KJ. Calculate the heat of combustion at co
vladimir2022 [97]

Answer:

The heat at constant pressure is -3,275.7413 kJ

Explanation:

The combustion equation is 2C₆H₆ (l) + 15O₂ (g)  → 12CO₂ (g) + 6H₂O (l)

\Delta n_g = (12 - 15)/2 = -3/2

We have;

\Delta H = \Delta U + \Delta n_g\cdot R\cdot T

Where R and T are constant, and ΔU is given we can write the relationship as follows;

H = U + \Delta n_g\cdot R\cdot T

Where;

H = The heat at constant pressure

U = The heat at constant volume = -3,272 kJ

\Delta n_g = The change in the number of gas molecules per mole

R = The universal gas constant = 8.314 J/(mol·K)

T = The temperature = 300 K

Therefore, we get;

H = -3,272 kJ + (-3/2) mol ×8.314 J/(mol·K) ×300 K) × 1 kJ/(1000 J) = -3,275.7413 kJ

The heat at constant pressure, H = -3,275.7413 kJ.

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