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TiliK225 [7]
4 years ago
15

How did early chemists determine which substances were elements?

Chemistry
1 answer:
padilas [110]4 years ago
8 0

。☆✼★ ━━━━━━━━━━━━━━  ☾  

The correct answer would be option D

Have A Nice Day ❤    

Stay Brainly! ヅ    

- Ally ✧    

。☆✼★ ━━━━━━━━━━━━━━  ☾

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Work is being done in which of these situations? All motions are at a constant velocity (including velocity = zero).
tigry1 [53]

Answer:

None of the above

Explanation:

Work is defined as the force that moves an object through a particular distance:

    Workdone = force x displacement

At constant velocity, the force on a body is zero because such a body is not accelerating.  

    Force is a product of mass and acceleration. Acceleration occurs only when velocity changes with time. If no change in velocity, acceleration does not occur. Therefore, no force is being applied.

 From all the situations given, the condition is set at constant velocity. At constant velocity, there is no acceleration. This implies that no work is done.

5 0
3 years ago
Read 2 more answers
An instrument which measures relative humidity is a ________ (n)
Misha Larkins [42]
An instrument which measure relative humidity is a hygrometer
3 0
4 years ago
Assuming dopant atoms are uniformly distributed in a silicon crystal, how far apart are these atoms when the doping concentratio
enyata [817]

Answer:

d =~ 5.8μm

d =~ 0.13 μm

Explanation:

when the doping concentrations are 5 × 10^15 cm^-3

d = v^-1/3  ; where d represent the distance between the atoms , and v  represent the volume

d =1/ ∛v

d = 1/ ∛5 × 10^15

d = 1/ 170997.5

d = 5.85 × 10 ^ -6

d =~ 5.8μm

when the doping concentrations are 5 × 10^20 cm^-3

d = v^-1/3  ; where d represent the distance between the atoms , and v  represent the volume

d =1/ ∛v

d = 1/ ∛5 × 10^20

using the principle of surds and standard forms, we have

d = 1/ ∛0.5 × 10^21  

d = 1/7937005.26

d = 1.26 × 10 ^ -7

d = 0.126 × 10 ^ -6

d =~ 0.13 μm

8 0
3 years ago
A mixture of 80.0 g of oxygen and 80.0 g of helium has a total
iren [92.7K]

Answer:

Partial pressure of O₂ =  0.0198 atm

Partial pressure of He = 0.1584 atm

Explanation:

Given data:

Mass of oxygen = 80.0 g

Mass of helium = 80.0 g

Total pressure = 0.1800 atm

Partial pressure of each gas = ?

Solution:

First of all we will calculate the number of moles.

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 80.0 g/ 32 g/mol

Number of moles = 2.5 mol

Number of moles of helium:

Number of moles = mass/molar mass

Number of moles = 80.0 g/ 4 g/mol

Number of moles = 20 mol

Total number of moles = 20 mol + 2.5 mol = 22.5 mol

Mole fraction of O₂ = 2.5 mol/ 22.5 mol = 0.11

Mole fraction of He = 20 mol / 22.5 mol = 0.88

Partial pressure:

Partial pressure of O₂ =  0.11× 0.1800 atm = 0.0198 atm

Partial pressure of He = 0.88 × 0.1800 atm = 0.1584 atm

3 0
3 years ago
For the reaction So3 + H2O -> H2SO4, how many grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid
zalisa [80]

Answer:

320 grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid.

Explanation:

The balanced reaction is:

SO₃ + H₂O → H₂SO₄

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

  • SO₃: 1 mole
  • H₂O: 1 mole
  • H₂SO₄: 1 mole

Being the molar mass of each compound:

  • SO₃: 80 g/mole
  • H₂O: 18 g/mole
  • H₂SO₄: 98 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • SO₃: 1 mole* 80 g/mole= 80 grams
  • H₂O: 1 mole* 18 g/mole= 18 grams
  • H₂SO₄: 1 mole* 98 g/mole= 98 grams

Then you can apply the following rule of three: if 1 mole of sulfuric acid is produced by the reaction of 80 grams of sulfur trioxide, 4 moles of sulfuric acid is produced from how much mass of sulfur trioxide?

mass of sulfur trioxide= \frac{4 moles of sulfuric acid*  80 grams of sulfur trioxide}{1 mole of sulfuric acid }

mass of sulfur trioxide= 320 grams

<u><em>320 grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid.</em></u>

3 0
3 years ago
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