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egoroff_w [7]
3 years ago
14

Estructura de Lewis MnCO3

Chemistry
1 answer:
Nana76 [90]3 years ago
5 0

Answer:

Explanation:Explanation is^{} in a filely/3fcEdSx

bit.^{}

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<span>how many molesof oxygen are produced when 8 mols of Al are produced? it is 6</span>
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What is the isotope notation for an ion of silver-109 with a charge of positive 1
Arisa [49]

Answer:

The isotope notation for an ion of silver-109 with a charge of positive 1 is _{47}^{109}Ag^{1+}.

Explanation:

The elements with same atomic number with different mass numbers are called isotopes.

The isotopes of silver is as follows.

_{47}^{109}Ag,_{47}^{107}Ag

From the above two isotopes have same atomic number and different mass number.

From the given,the isotope notation for an ion of silver-109 with a charge of positive 1

It can be represented is as follows.

_{47}^{109}Ag^{1+}

General representation of isotope is as in attachment.

6 0
3 years ago
Which action would shift this reaction away from solid silver chloride and toward the dissolved ions
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Removing chloride ion is the reaction away
3 0
2 years ago
In the electron cloud model of the atom, an orbital is defined as the most probable(1) charge of an electron
Galina-37 [17]

The correct answer would be 3.) Location of an electron
3 0
3 years ago
Which reactant will be used up first if 78.1g of o2 is reacted with 62.4g of c4h10?
dlinn [17]

Answer:

Reagent O₂ will be consumed first.

Explanation:

The balanced reaction between O₂ and C₄H₁₀ is:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:

  • C₄H₁₀: 2 moles
  • O₂: 13 moles
  • CO₂: 8 moles
  • H₂O: 10 moles

Being:

  • C: 12 g/mole
  • H: 1 g/mole
  • O: 16 g/mole

The molar mass of the compounds that participate in the reaction is:

  • C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
  • O₂: 2*16 g/mole= 32 g/mole
  • CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
  • H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:

  • C₄H₁₀: 2 moles* 58 g/mole= 116 g
  • O₂: 13 moles* 32 g/mole= 416 g
  • CO₂: 8 moles* 44 g/mole= 352 g
  • H₂O: 10 moles* 18 g/mole= 180 g

If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

mass of O_{2} =\frac{416grams of O_{2}*62.4 grams ofC_{4}H_{10}   }{116 grams of C_{4}H_{10}}

mass of O₂= 223.78 grams

But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>

3 0
3 years ago
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