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Nutka1998 [239]
3 years ago
7

Please help me with this. Thanks

Chemistry
1 answer:
ozzi3 years ago
5 0

The third one because

Most minerals can be characterized and classified by their unique physical properties: hardness, luster, color, streak, specific gravity, cleavage, fracture, and tenacity.

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Write the structure of the ionized form of phenylalanine
Licemer1 [7]
Amino Acids undergo a kind of acid base reaction within a molecule. The Amino group acts as a base and accepts proton from Carboxylic functional group resulting in formation of Ammonium with positive charge and a negative charge on Carboxylate. Such molecules containing both positive and negative charges are called as Zwittor Ions. Phenylalanine both in Un-ionized and Ionized form is shown below,

7 0
3 years ago
What is the formula weight (amu) of the molecule H2O? Use atomic masses of H and O as 1.008 amu and 16.00 amu respectively. Repo
vaieri [72.5K]

Answer:

Formula weight of H₂O molecule is  18.02 amu.

Explanation:

Given data:

Formula weight of H₂O = ?

Atomic mass of H = 1.008 amu

Atomic mass of O = 16.00 amu

Solution:

Formula weight:

"It is the sum of all the atomic weight of atoms present in given formula"

Formula weight of H₂O = 2×1.008 amu + 1×16.00 amu

Formula weight of H₂O = 18.02 amu

Thus, formula weight of H₂O molecule is  18.02 amu.

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3 years ago
Gas a (L^2. atm/mol^2) b(L/mol)
Amiraneli [1.4K]

Answer:

Explanation:

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7 0
3 years ago
300×175<br> Express your answer using the correct number of significant figures.
icang [17]

Answer:

300×175=52500

Explanation:

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3 0
3 years ago
A metal, M, forms an oxide having the formula MO2 containing 59.93% metal by mass. Determine the atomic weight in g/mole of the
Damm [24]

Answer:

See solution.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:

\%  M=\frac{m_M}{m_M+2*m_O}*100 \%\\\\59.93\%  =\frac{m_M}{m_M+32.00}*100 \%

Thus, we solve for the molar mass of the metal to obtain:

59.93\% (m_M+32.00) =m_M*100 \%\\\\m_M*59.93\% +1917.76\% =m_M*100 \%\\\\m_M=47.86g/mol

For the subsequent problems, we proceed as follows:

a.

4.00gO_2*\frac{1molO_2}{32.00gO_2}=0.125molO_2

b.

0.400molH_2S*\frac{2molH}{1molH_2S}*\frac{6.022x10^{23}atomsH}{1molH}=4.82x10^{23}atomsH

c.

0.235gNH_3*\frac{1molNH_3}{17.03gNH_3} *\frac{3molH}{1molNH_3}*\frac{6.022x10^{23}atomsH}{1molH}=2.49x10^{22}atomsH

Regards!

7 0
3 years ago
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