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3 years ago

Match the following scientist with his experiment or contribution

Chemistry

1 answer:

Verdich [7]3 years ago

3 0

J. J. Thomson - oil drop

Albert Einstein - atomic theory

Niels Bohr - model of the atom

Ernest Rutherford - gold foil experiment

John Dalton - cathode ray tube

Robert Millikan - photoelectric effect

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The energy levels of the positive ions that have just a single electron, like He , Li2 , or Be3 , etc., follow the formula Bohr

Sidana [21]

The answer is false. Hope that helps

3 0

2 years ago

Cu + 2 AgNO3 + Cu(NO3)2 + 2Ag

Allushta [10]

Answer:

7.28 moles Ag°

Explanation:

Cu° + 2 AgNO₃ => Cu(NO₃)₂ + 2Ag°

Given 7.28 moles 7.28 moles

To determine limiting reactant, divide the mole values by the respective coefficient of balanced equation. The resulting smallest value is the limiting reactant. Note: this is a short cut method for determining limiting reactant only. Once the limiting reactant is determined one must use the given mole values of the limiting reactant to solve problem. That is ...

Limiting reactant determination:

Cu° + 2 AgNO₃ => Cu(NO₃)₂ + 2Ag°

Cu: 7.28 / 1 = 7.28

AgNO₃ : 7.28 / 2 = 3.64 => Limiting Reactant is AgNO₃

Solving Problem depends on AgNO₃; Cu will be in excess.

Since coefficients of AgNO₃ & Ag° are equal, then the moles AgNO₃ used equals moles Ag° produced and is therefore 7.28 moles Ag°.

5 0

3 years ago

HELLPPP PLZZ ASAPP!

Sphinxa [80]

Answer:

Condensation and Depositon

Explanation:

Condensation is from gas to liquid

Deposition is from gas to solid

7 0

3 years ago

The Henry's law constant for helium gas in water at 30 °C is 3.70 × 10-4 M/atm. When the partial pressure of helium above a samp

igor_vitrenko [27]

Answer:

The concentration of helium in the water is 2.405×10^-4 M

Explanation:

Concentration = Henry's law constant × partial pressure of helium

Henry's law constant = 3.7×10^-4 M/atm

Partial pressure of helium = 0.65 atm

Concentration = 3.7×10^-4 × 0.65 = 2.405×10^-4 M

4 0

3 years ago

Se realiza una mezcla de minerales de Cu y Fe: 20 kg FeS2 (pirita), 70 kg de Fe2O3 (hemetita) 15 kg de CuFe2 (calcopirita) y 90

Artemon [7]

Answer:

34.78% Fe

39.66% Cu

5.48% S

20.07% O

Explanation:

Para resolver esta pregunta debemos hallar la masa de cada átomo en cada mineral. Así, podremos hallar el porcentaje de cada átomo:

Pirita (Fe: 55.845g/mol; S: 32.065g/mol; FeS2: 119.975g/mol)

Masa Fe:

20kg FeS2 * (1*55.845g/mol / 119.975g/mol) = 9.31kg Fe

Masa S:

20kg FeS2 * (2*32.065g/mol / 119.975g/mol) = 10.69kg S

Hemetita (Fe: 55.845g/mol; O: 16g/mol; Fe2O3: 159.688g/mol)

Masa Fe:

70kg Fe2O3 * (2*55.845g/mol / 159.688g/mol) = 48.96kg Fe

Masa O:

70kg Fe2O3 * (3*16g/mol / 159.688g/mol) = 21.04kg O

Calcopirita (Fe: 55.845g/mol; Cu: 63.546g/mol; CuFe2: 175.236 g/mol)

Masa Fe:

15kg CuFe2 * (2*55.845g/mol / 175.236 g/mol) = 9.56kg Fe

Masa Cu:

15kg CuFe2 * (1*63.546g/mol / 175.236 g/mol) = 5.44kg Cu

Tenorita (O: 16g/mol; Cu: 63.546g/mol; CuO: 79.545 g/mol)

Masa O:

90kg CuO * (1*16g/mol / 79.545 g/mol) = 18.10kg O

Masa Cu:

90kg CuO * (1*63.546g/mol / 79.545 g/mol) = 71.90kg Cu

Masa Total: 20kg + 70kg + 15kg + 90kg = 195kg

Porcentaje Hierro:

9.31kg Fe + 48.96kg Fe + 9.56kg Fe / 195kg * 100 =

34.78% Fe

Porcentaje Cobre:

5.44kg Cu + 71.90kg Cu / 195kg * 100 =

39.66% Cu

Porcentaje Azufre:

10.69kg S / 195kg * 100 =

5.48% S

Porcentaje Oxígeno:

21.04kg O + 18.10kg O/ 195kg * 100 =

20.07% O

8 0

3 years ago