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Trava [24]
3 years ago
15

A student experimentally determines the density of a metal cube. The edge length of the cube is measured using calipers. The cub

e's true density is 12.30 g/cm. Use the student's collected data below and calculate the percent error for the measurement.
Student's Collected Data:
Cube's Mass 551786
Edge Length 1.72 cm
Percent error= _______%
Chemistry
1 answer:
IgorC [24]3 years ago
8 0

Answer:

10.9%.

Explanation:

The first thing to do in order to solve this question is to Determine the value for the volume of the the cube. This can be done by taking the cube root of the length of the cube;

The volume of the cube = (length of the cube)^3 = length × length × length = 1.72 × 1.72 × 1.72 =( 1.72)^3 = 5.09cm^3.

The next thing you do is to Determine the exponential density, the can be done by using the formula below;

The exponential density = mass/ volume = 55. 786/ 5.09 = 10.96 g/cm^3.

Therefore, the percent error = (true density of the cube - exponential density of the cube)÷ true density of the cube × 100.

Hence, the percent error = 12.30 - 10.96/12.30 × 100 = 10.9%.

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Explanation: for 75 ml, take 45*75 mg of BSA and make up to 75 mL with distilled water.

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3 years ago
C + O2 = CO2
geniusboy [140]

Answer:

44 grams of CO₂ will be formed.

Explanation:

The balanced reaction is:

C + O₂ → CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • C: 1 mole
  • O₂: 1 mole
  • CO₂: 1 mole

Being the molar mass of each compound:

  • C: 12 g/mole
  • O₂: 32 g/mole
  • CO₂: 44 g/mole

By stoichiometry the following mass quantities participate in the reaction:

  • C: 1 mole* 12 g/mole= 12 g
  • O₂: 1 mole* 32 g/mole= 32 g
  • CO₂: 1 mole* 44 g/mole= 44 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

If 12 grams of C react, by stoichiometry 32 grams of O₂ react. But you have 40 grams of O₂. Since more mass of O₂ is available than is necessary to react with 12 grams of C, carbon C is the limiting reagent.

Then by stoichiometry of the reaction, you can see that 12 grams of C form 44 grams of CO₂.

<u><em>44 grams of CO₂ will be formed.</em></u>

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3 years ago
The change in velocity over a specific amount of time Includes speeding up. slowing down or changing direction. This is known as
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Answer:

Acceleration

Explanation:

Acceleration = \frac{velocity}{time}

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3 years ago
Please<br>list uses uses aluminium?​
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Answer:

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4 0
3 years ago
Read 2 more answers
Hello, everyone!
marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the <em>law of ideal gases,</em>  

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x \frac{1 g}{1000 mg} x \frac{1 mol}{28.01 g}

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V =  \frac{0.00125 mol x 0.082057 \frac{atm L}{mol K}  x 243 K}{0.92 atm}

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x \frac{1000 cm^{3} }{1 L} = 27.09 cm³

<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>

c = 27 cm³ / m³ = 27 ppm

<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 \frac{cm^{3} }{m^{3} } x \frac{1 m^{3} }{1 000 000 cm^{3} } x 100%

c = 0.003 %

So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.

5 0
3 years ago
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