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Virty [35]
3 years ago
6

How did the development of the earliest idea about atoms differ from the later work of scientists?

Chemistry
2 answers:
igor_vitrenko [27]3 years ago
7 0

Answer:

C. It was based on the thoughts of an early philosopher.

Explanation:

gtnhenbr [62]3 years ago
4 0

The correct answer is C. It was based on the thoughts of an early philosopher.

Explanation:

An atom is a basic and smallest unit that composes matter and that determine the properties of elements. Regarding the development of ideas related to atoms these did not begin in science but in philosophy; indeed the first person that proposed matter or elements were composed of certain smaller units was the philosophers Leucippus and his pupil Democritus in Ancient Greece, who stated atoms were eternal, infinite and defined the qualities of an object, idea that was supported by other Greek philosophers. But it was not until 16th and 17th centuries after the Middle Ages that the term re-emerged and until the 19th century it was officially proposed and there were experiments by scientists that later became a theory. Therefore, the development of the earliest idea about attoms differs from later work of scientists is that the earliest idea was based on the thoughts of an early philosopher.

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ILL GIVE BRAINLY, 5 STARS , FOLLOW AND HEARTS
Jlenok [28]

Answer:

Second one

Explanation:

I think it's the second cause there's no energy w/o heat. So it's the second.

6 0
3 years ago
Read 2 more answers
24. The air in a small 250 cm birthday balloon is at a pressure of 760 torr. A boy sits on it at
kvv77 [185]

Answer:

1.14atm

Explanation:

Given parameters:

V1 = 250cm³ ;

              1000cm³ = 1dm³;  so this is 0.25dm³

P1  = 760torr

            760torr  = 1atm

       

V2  = 220cm³ ; 0.22dm³

Unknown:

New pressure = ?

Solution:

To solve this problem, we apply Boyle's law and we use the expression below:

       P1 V1 = P2V2

The unknown is P2;

            1 x 0.25  = P2 x 0.22

                 P2  = 1.14atm

7 0
3 years ago
A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce
spin [16.1K]

Explanation:

The given data is as follows.

       Volume of lake = 15 \times 10^{6} m^{3} = 15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}

        Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = 5.6 \times 15 \times 10^{9} mg

                                                                    = 84 \times 10^{9} mg

                                                                    = 84 \times 10^{3} kg

Flow rate of river is 50 m^{3} sec^{-1}

Volume of water in 1 day = 50 \times 10^{3} \times 86400 liter

                                          = 432 \times 10^{7} liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are 2.9792 \times 10^{10} mg or 2.9792 \times 10^{4} kg

Flow rate of sewage = 0.7 m^{3} sec^{-1}

Volume of sewage water in 1 day = 6048 \times 10^{4} liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = 1.8144 \times 10^{10} mg or 1.8144 \times 10^{4}kg

Therefore, total concentration of lake after 1 day = \frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l

                                        = 6.8078 mg/l

                 k_{D} = 0.2 per day

       L_{o} = 6.8078

Hence, L_{liquid} = L_{o}(1 - e^{-k_{D}t}

             L_{liquid} = 6.8078 (1 - e^{-0.2 \times 1})  

                             = 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

                                                             = 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0
4 years ago
In the periodic table , the elements are organized in order of ________atomic number .
jok3333 [9.3K]
The correct answer is increasing
4 0
3 years ago
Read 2 more answers
A chemist prepares a solution of copper(II) sulfate CuSO4 by measuring out 31.μmol of copper(II) sulfate into a 150.mL volumetri
baherus [9]

Answer:

The concentration of the copper (II) sulfate solution is 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM

Explanation:

The concentration of a solution is the amount of solute dissolved in a given volume of solution. In this case, the concentration of the copper(II) sulfate solution in micromoles per liter (symbol ) is the number of micromoles of copper(II) sulfate dissolved in each liter of solution. To calculate the micromoles of copper(II) sulfate dissolved in each liter of solution you must divide the total micromoles of solute by the number of liters of solution.

Here's that idea written as a formula: c= n/V

where c stands for concentration, n stands for the total micromoles of copper (II) sulfate and V stands for the total volume of the solution.

You're not given the volume of the solution in liters, but rather in milliliters. You can convert milliliters to liters with a unit ratio: V= 150. mL * 10^-3 L/ 1 mL = 0.150 L

Next, plug in μmol and liters into the formula to divide the total micromoles of solute by the number of liters of solution: c= 31 μmol/0.150 L = 206.66 μmol/L

Convert this number into scientific notation: 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM

3 0
2 years ago
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