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Molodets [167]
3 years ago
15

HELP NOW FIRST=BRANILYEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Which statement describes the difference between an atom of silver and an atom of gold?
A.
They have the same number of electrons.


B.
The atomic numbers are the same.


C.
Their atomic masses are different.


D.
They have exactly the same isotopes.
Chemistry
2 answers:
Oduvanchick [21]3 years ago
8 0

Answer:C

it's c easy..........

cestrela7 [59]3 years ago
3 0

Answer:

the answer is c. their atomic masses are different clearly because an atom of gold has 79 protons and the atom can be divided multiple times. An atom of silver has an atomic number of 47. 47 electrons. Clearly different. Hope it helps :)

Explanation:

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Use the periodic table to determine the electron configuration for dysprosium (Dy) and americium (Am) in noble-gas notation
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Answer:

Dysprosium [Dy]=[Xe]4f^{10}6s^2

Americium [Am]=[Rn]5f^77s^2

Dysprosium is a chemical element with symbol Dy and atomic number of 66. It is a rare earth metal and as it contains partially filled f sub shells, it belongs to f block. Xe is the nearest noble gas and has atomic number of 54.

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What is a DNA fingerprint and why is it considered individual evidence?
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Convert 779 gallons to cubic yards
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A chemist prepares a solution of calcium bromide CaBr2 by measuring out 4.81μmol of calcium bromide into a 50.mL volumetric flas
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Answer:

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Explanation:

<u>Step 1:</u> Data given

Moles of Calciumbromide (CaBr2) = 4.81 µmol

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<u>Step 2:</u> Calculate the concentration of Calciumbromide

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The normal freezing point of a certain liquid
slavikrds [6]

Answer : The molal freezing point depression constant of liquid X is, 4.12^oC/m

Explanation :  Given,

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Mass of liquid X (solvent) = 450 g  = 0.450 kg

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Formula used :  

\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of liquid X Kg}}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution = -0.5^oC

\Delta T^o = freezing point of liquid X = 0.4^oC

i = Van't Hoff factor = 1 (for non-electrolyte)

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Now put all the given values in this formula, we get

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K_f=4.12^oC/m

Therefore, the molal freezing point depression constant of liquid X is, 4.12^oC/m

3 0
3 years ago
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