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3 years ago

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Describe what does happen to the electrons when metal atoms are bonded together.

1 answer:

tankabanditka [31]3 years ago

4 0

Answer: In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “sea” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions. Metals are shiny.

Explanation: Hope this helped!

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A substance that will change shape to fit its container but has a definite volume is in a _____ phase of matter

larisa86 [58]

It’s a liquid because it can form any shape to fit container but the amount of liquid always stayed the same unless dropped some.

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4 years ago

Which compound has both ionic and covalent bonding? A) CaCO3 B) CH2Cl2 C) CH3OH D) C6H12O6

Ber [7]

The answer is letter A.

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4 years ago

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Calculate the mass percent composition of sulfur in Al2(SO4)3. Calculate the mass percent composition of sulfur in Al2(SO4)3. 28

ludmilkaskok [199]

Answer:

The mass percent of sulfur is 28.12% (option 1)

Explanation:

Step 1: Data given

Atomic mass of Al = 26.98 g/mol

Atomic mass of S = 32.065 g/mol

Atomic mass of O = 16.0 g/mol

Step 2: Calculate the molar mass of Al2(SO4)3

In 1 molecule Al2(SO4)3 we have 2 Aluminium atom, 3 Sulfur atoms and 12 oxygen atoms

Molar mass of Al2(SO4)3 = 2*26.98 + 3*32.065 + 12*16 .0

Molar mass of Al2(SO4)3 = 342.155 g/mol

Step 3: Calculate the mass percent of sulfur

In 1 molecule Al2(SO4)3 we have 2 Aluminium atom, 3 Sulfur atoms and 12 oxygen atoms

Mass % sulfur = (3*atomic mass of S/ molar mass of Al2(SO4)3) * 100 %

Mass % sulfur = (3*32.065 / 342.155) * 100%

Mass % sulfur = 28.12 %

The mass percent of sulfur is 28.12% (option 1)

3 0

3 years ago

What is the mass (in grams) of 9.42 × 1024 molecules of methanol (CH3OH)?

Furkat [3]

(9.42x10^24)/(6.02x10^23)=15.6478 mol methanol
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3 years ago

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The half-life of gold-198 is 2.7 days. After

Pie

Answer: 8.1 days

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant = x

a - x = amount left after decay process= $\frac{x}{4}$

a) to find rate constant

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$k=\frac{0.693}{2.7days}=0.257days^{-1}$

b) for completion of one fourth of reaction

$t=\frac{2.303}{k}\log\frac{x}{\frac{x}{4}}$

$t=\frac{2.303}{0.257}\log{4}$

$t=8.1days$

Thus after 8.1 days , one fourth of original amount will remain.

8 0

4 years ago