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padilas [110]
3 years ago
6

Which words describe the image below? CHECK ALL THAT APPLY.

Chemistry
2 answers:
3241004551 [841]3 years ago
3 0

Answer:

A) liquid

Explanation:

hope this helps you!!

QveST [7]3 years ago
3 0
Liquid

Explanation : In liquids, the particles are near each other, but not tightly packed. The particles in a liquid are able to slide past each other.

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What is the isolated system?
Mnenie [13.5K]

Answer:

in physical science, an isolated system is either of the following:

1. a physical system so far removed from other systems that it does not interact with them.

2. a thermodynamic system enclosed by rigid immovable walls through which neither mass nor energy can pass.

Explanation:

4 0
3 years ago
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A buffer was prepared by mixing 1.00 mole of ammonia and 1.00 mole of ammonium chloride to form an aqueous solution with a total
lidiya [134]

Answer:

Idek sorry

Explanation:

So sorry i think its 1.00

3 0
3 years ago
10 points) Which is larger? Ba²+ or Ba<br> explain
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Answer:

<u>Ba is larger than Ba2 + B a 2 + because it has two more electrons.</u>

Explanation:

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1 year ago
How much heat do you need to raise the temperature of 100 g of aluminum from 30°C to 150°C?
BaLLatris [955]

Heat energy can be calculated by using the specific heat of a substance multiplying it to the mass of the sample and the change in temperature. It is expressed as:

 <span>Energy = mCΔT</span>

<span>Energy = 100 (0.90) (150 -  30)</span>

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3 0
4 years ago
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You are trying to determine the volume of the balloon needed to match the density of the air in the lab. You know that if you ca
vova2212 [387]
The equation to be used are:

PM = ρRT
PV = nRT
where
P is pressure, M is molar mass, ρ is density, R is universal gas constant (8.314 J/mol·K), T is absolute temperature, V is volume and n is number of moles

The density of air at 23.5°C, from literature, is 1.19035 kg/m³. Its molar mass is 0.029 kg/mol.

PM = ρRT
P(0.029 kg/mol) = (1.19035 kg/m³)(8.314 J/mol·K)(23.5+273 K)
P = 101,183.9 Pa

n = 0.587 g * 1 kg/1000 g * 1 mol/0.029 kg = 0.02024 mol
(101,183.9 Pa)V = (0.02024 mol)(8.314 J/mol·K)(23.5+273 K)
Solving for V,
V = 4.931×10⁻⁴ m³
Since 1 m³ = 1000 L
V = 4.931×10⁻⁴ m³ * 1000
V = 0.493 L

7 0
3 years ago
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