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Marat540 [252]
2 years ago
5

If you reacted 93.7 grams of iron with excess copper sulfate solution, what mass of copper would you expect to make? you may nee

d to balance the equation first. fe cuso4 -> cu feso4
Chemistry
1 answer:
Mariana [72]2 years ago
3 0

Answer is although the question should have 3 significant figures.

<h3>What is significant figures?</h3>
  • Chemical equation

Fe + CuSO4 --> Cu + FeSO4

It is already balanced

  • Molar ratios

1 mol Fe : 1 mol CuSO4 : 1 mol Cu : 1 mol FeSO4

  • atomic masses of Fe and Cu

Fe: 55.8 g/mol

Cu: 63.5 g/mol

  • number of moles of Fe

number of moles = mass in grams / atomic mass = 93.7 g / 55.8 g/mol = 1.679 mol

  • number of moles of Cu:

1 mol Fe / 1 mol Cu => 1.679 mol Fe / 1.679 mole Cu

  • mass of Cu in 1.679 moles Cu

mass = atomic mass * number of atoms = 63.5 g/mol * 1.679 moles = 106.6 g = 107 rounded to 3 significant figures.

To learn more about significant figures, refer to:

brainly.com/question/24491627

#SPJ4

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the volume of liquid can be measured with? A. graduated cylinder B. pan balance C. scale D. metric ruler
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a: is the correct anwser

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In order to change a solid to a liquid or a liquid to a gas, what is needed?
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Latent heat, also called the heat of vaporization, is the amount of energy necessary to change a liquid to a vapour at constant temperature and pressure. The energy required to melt a solid to a liquid is called the heat of fusion
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Olympic cyclists fill their tires with helium to make them lighter. Assume that the volume of the tire is 860 mL , that it is fi
nata0808 [166]

Answer:

a)  the mass of air = 8.24 grams

b) the mass of helium = 1.14 grams

c) the mass difference = 7.10 grams

Explanation:

Step 1: Data given

Volume of the tire = 860 mL

Total pressure = 120 psi

Temperature = 26°C

molar mass of air = 28.8 g/mol

Step 2:  Convert psi to atm

(

120 psi) (1 atm / 14.7 psi) = 8.163

Step 4: Calculate moles

PV = nRT

 ⇒ with P = the pressure = 8.163 atm

⇒ with V = the volume = 860 mL = 0.860 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the universal gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 26 °C = 299 Kelvin

n = (8163*0.860)/(0.08206*299)

n = 0.2861 moles of gas

Step 5: Calculate the mass of air in an air-filled tire.

Mass = moles * molar mass

Mass = (0.2861 moles of gas) (28.8 g/mol)  

Mass = 8.24 grams

Step 6: Calculate the mass of helium in a helium- filled tire.

mass of helium = 0.2861 moles of gas * 4 g/mol)  

mass of helium  = 1.14 grams

Step 7: What is the mass difference between the two?

Δmass=  8.24 grams -  1.14 grams

Δmass= 7.10 grams

8 0
3 years ago
Design a test to determine whether thorium-234 also emits particles. First, explain how Rutherford’s experiment measured positiv
liubo4ka [24]

The characteristics of the α and β particles allow to find  the design of an experiment to measure the ²³⁴Th particles is:

  • On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the beta particle emission from ²³⁴Th.
  • The neutrons cannot be detected in this experiment because they have no electrical charge.

In Rutherford's experiment, the positive particles directed to the gold film were measured on a phosphorescent screen that with each arriving particle a luminous point is seen.

The particles in this experiment are α particles that have two positive charge and two no charged is a helium nucleus.

The test that can be carried out is to place a small ours of Thorium in front of a phosphorescent screen and see if it has flashes, with the amount of them we can determine the amount of particle emitted per unit of time.

Thorium has several isotopes, with different rates and types of emission:

  • ²³²Th emits α particles, it is the most abundant 99.9%
  • ²³⁴Th emits β particles, exists in small traces.

In this case they indicate that the material used is ²³⁴Th, which emits β particles that are electrons, the detection of these particles is more difficult since it has one negative charge, it has much lower mass, but they can travel further than the particles α, therefore, for what type of isotope we have, we can start measuring at a small distance and increase the distance until the reading is constant. At this point all the particles that arrive are β, which correspond to ²³⁴Th.

Neutron detection is much more difficult since these particles have no charge and therefore do not interact with electrons and no flashing on the screen is varied.

In conclusion with the characteristics of the α and β particles we can find the design of an experiment to measure the ²³⁴Th particles is:

  • On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the β particle emission from ²³⁴Th.
  • The neutrons cannot be detected in this experiment because they have no electrical charge.

Learn more about radioactive emission here: brainly.com/question/15176980

7 0
3 years ago
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