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Yakvenalex [24]
3 years ago
15

Consider two liquids, labeled A and B, that are both pure substances. Liquid A has

Chemistry
1 answer:
Scrat [10]3 years ago
4 0
E is the correct answer
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Which statements accurately describe Ernest Rutherford’s experiment? Check all that apply.
Sunny_sXe [5.5K]

Answer:

The positive particles were deflected due the concentrated positive charge of the atom

Explanation:

Rutherford concluded that the positive particles were deflected through an angle greater than 90 due to electrostatic force of repulsion between the particles and positive part of the atom.

3 0
3 years ago
Read 2 more answers
An atom is electrically neutral because
serg [7]

Answer:

c

Explanation:

neutrons have no charge

protons and electrons have opposite charges and balance each other out in equal numbers

7 0
3 years ago
Differentiate between satured and unsatured fats
olga_2 [115]

Answer:

...

Explanation:

in saturated fats there is no double bond between the acids and are tightly packed and unsaturated fats arent tight and loosely packed/put together

saturated- solid at room temperature

unsaturated= liquid at room temperature

two types of unsaturated fats, Polyunsaturated fats and Monounsaturated fats

6 0
3 years ago
Two moles of a monatomic ideal gas are contained at a pressure of 1 atm and a temperature of 300 K; 34,166 J of heat are transfe
anygoal [31]

Answer:

Final temperature is 302 K

Explanation:

You can now initial volume with ideal gas law, thus:

V = \frac{n.R.T}{P}

Where:

n are moles: 2 moles

R is gas constant: 0,082 \frac{atm.L}{mol.K}

T is temperature: 300 K

P is pressure: 1 atm

V is volume, with these values: <em>49,2 L</em>

The work in the expansion of the gas, W, is: 1216 J - 34166 J = <em>-32950 J</em>

This work is:

W = P (Vf- Vi)

Where P is constant pressure, 1 atm

And Vf and Vi are final and initial volume in the expansion

-32950 J = -1 atm (Vf-49,2L) × \frac{101325 J}{1 atm.L}

Solving: <em>Vf = 49,52 L</em>

Thus, final temperature could be obtained from ideal gas law, again:

T = \frac{P.V}{n.R}

Where:

n are moles: 2 moles

R is gas constant: 0,082 \frac{atm.L}{mol.K}

P is pressure: 1 atm

V is volume: 49,52 L

T is final temperature: <em>302 K</em>

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I hope it helps!

4 0
3 years ago
How many atoms are in mercury (ii) biocarbonate?
bija089 [108]

Alias: Mercuric Hydrogen Carbonate; Mercury(II) Bicarbonate

Formula: Hg(HCO3)2

Molar Mass: 322.6237

8 0
3 years ago
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