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mars1129 [50]
3 years ago
9

HELP CHEM IM GIVING BRAINLIST

Chemistry
1 answer:
levacccp [35]3 years ago
3 0

Answer:

mol = M x L

mol = 0.25 M x 0.5 L

mol = 0.125 mol

0.125 mol x molar mass = ? g

molar mass for Cu(NO3)2 = 63.546 + 2(14.007) + 6(15.999) = 187.554

0.125 mol x 187.554 = ~23.44425 g

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12. How are secondary sources of energy different from other energy resources?
Mkey [24]

Answer:

A primary source of energy is a fossil fuel , nuclear fuel , wind or sunlight in an unconverted state . A secondary source of energy is created when a primary source of energy is burned or otherwise converted into a form , like electricity, that can be used for useful work.

Explanation:

5 0
3 years ago
The four lines observed in the visible emission spectrum of hydrogen tell us that: a. We could observe more lines if we had a st
Aliun [14]

Answer:

c. Only certain energies are allowed for the electron in a hydrogen atom

Explanation:

Emission spectrum are produced when the excited electron in a atom release the energy in the form of photons to come to ground state. These photons are of different wavelengths depending on the excitation state of emitting electron or transition of electron. These electromagnetic radiation are observed through prism to produce the spectrum.

As the name indicates this spectrum is produced by emission of energy. Although the electron can be excited by different methods such as by heating but the key point is that electrons in hydrogen atom will emit the photons of same energy which they absorb and each electron can absorb only certain type of energy. So four lines were observed in the visible spectrum of hydrogen because only certain energies are observed for hydrogen atom.

6 0
3 years ago
What is the empirical formula of a compound with 35.94% aluminum and 64.06% sulfur?
Marat540 [252]

<u>Given:</u>

% Al = 35.94

% S = 64.06

<u>To determine:</u>

Empirical formula of a compound with the above composition

<u>Explanation:</u>

Atomic wt of Al = 27 g/mol

Atomic wt of S = 32 g/mol

Based on the given data, for 100 g of the compound: Mass of Al = 35.94 g and mass of S = 64.06 g

# moles of Al = 35.94/27 = 1.331

# moles of S = 64.06/32 = 2.002

Divide by the smallest # moles:

Al = 1.331/1.331 = 1

S = 2.002/1,331 = 1.5 ≅ 2

Empirical formula = AlS₂

6 0
3 years ago
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Helga [31]

Answer:

Yeild of CO2 is approximately 54g

Explanation:

Using reaction stoichiomety and coeeficients, and knowing O2 is limiting reactant, 54 g of CO2 is produced.

7 0
4 years ago
Al2(SO4)3 + CaCl2 → AlCl3 + CaSO4
wolverine [178]
Al2(SO4)3 + 3CaCl2 > 2AlCl3 + 3CaSO4
3 0
3 years ago
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