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statuscvo [17]
2 years ago
10

Lavar el arroz y escurrirle el agua es un ejemplo de mezcla....

Chemistry
1 answer:
Tresset [83]2 years ago
4 0

Answer:

me no oblo espenul

Explanation:

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A sample of nitrogen gas occupies a volume of 1.40L at 297K with a pressure of 776mmHg. The volume increases to 2.70L and the te
konstantin123 [22]

Hey it isnt letting me submit my answer to your question on the Japanese chart, so imma just submit it here

6 0
3 years ago
rutherford's experiment indicated that matter was not as uniform as it appears what part of his experimental results implied thi
MaRussiya [10]

Rutherford theorized that atoms have their charge concentrated in a very small nucleus.

This was famous Rutherford's Gold Foil Experiment: he bombarded thin foil of gold with positive alpha particles (helium atom particles, consist of two protons and two neutrons).  

Rutherford observed the deflection of alpha particles on the photographic film and notice that most of alpha particles passed straight through foil.

That is different from Plum Pudding model, because it shows that most of the atom is empty space.

According to Rutherford model of the atom:

1) Atoms have their charge concentrated in a very small nucleus.

2) Major space in an atom is empty.

3) Atoms nucleus is surrounded by negatively charged particles called electrons.  

4) An atom is electrically neutral.

6 0
3 years ago
Which of these gases will have the greatest density at the same specified temperature and pressure?
ch4aika [34]

Answer:

Neon

The highest density among the inert gases is of Neon (Ne). This a factual data. Thus the highest density among given options is of Ne, as all the options are of inert gases.

Explanation:

hope it helps u

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5 0
2 years ago
A sample of chlorine gas is held at a pressure of 1023.6 kPa. When the pressure is decreased to 8114 kPa the
eduard

Answer:

T₁  = 39 K

Explanation:

Given data:

Initial pressure = 1023.6 kpa

Final pressure = 8114 kpa

Final temperature = 36°C (36+ 273= 309K)

Initial temperature = ?

Solution:

P₁/T₁ = P₂/T₂

T₁  = P₁×T₂ /P₂

T₁  = 1023.6 kpa × 309 K /8114 kpa

T₁  = 316292.4 K. Kpa /8114 kpa

T₁  = 39 K

Thus original pressure was 39 k.

3 0
3 years ago
Determine the temperature of 1.42 mol of a gas contained in a 3.00-L container at a pressure of 123 kPa
Rashid [163]

Answer:

The temperature is 30,92K

Explanation:

We use the formula PV=nRT. We convert the unit of pressure in kPa into atm.

101,325kPa----1atm

121kPa-------x=(121,3kPax 1 atm)/101,325kPa=1, 2 atm

PV=nRT---->T= (PV)/(RT)

T=(1,2 atm x 3L)/(1,42 mol x 0,082 l atm/K mol )= 30, 91721058 K

6 0
3 years ago
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