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Stels [109]
3 years ago
11

Write the empirical formula for given elements

Chemistry
1 answer:
rusak2 [61]3 years ago
3 0

Answer:

Fe(CN)_{2}\\Fe(C_{2}H_{3}O_{2})_{2}\\\\Fe(CN)_{3}\\\\Fe(C_{2}H_{3}O_{2})_{3}

Explanation:

Fe^{2+}(CN^{-})_{2}--->Fe(CN)_{2}\\Fe^{2+}(C_{2}H_{3}O_{2}^{-})_{2}--->Fe(C_{2}H_{3}O_{2})_{2}\\\\Fe^{3+}(CN^{-})_{3}--->Fe(CN)_{3}\\\\Fe^{3+}(C_{2}H_{3}O_{2}^{-})_{3}--->Fe(C_{2}H_{3}O_{2})_{3}

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The amount of 217 mg of an isotope is given by A(t) = 217 € -0.0171, where t is time in years since the initial amount of 217 mg
Maru [420]

The amount left after 20 years = 154.15 mg

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.  

Usually radioactive elements have an unstable atomic nucleus.  

The main particles are emitted by radioactive elements so that they generally decay are alpha (α), beta (β) and gamma (γ) particles  

The decay formula for isotope :

\tt \large{\boxed{\bold{A(t)=217e^{-0.0171t}}}

Then for t=20 years, the amount left :

\tt A(t)=217e^{-0.0171\times 20}\\\\A(t)=154.15~mg

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3 years ago
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If the force is 2,500 N and the acceleration is 500 m/s/s, what is the objects mass
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25 g of NH, is mixed with 4 moles of O, in the given reaction:
FromTheMoon [43]

Answer:

a. NH3 is limiting reactant.

b. 44g of NO

c. 40g of H2O

Explanation:

Based on the reaction:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l)

4 moles of ammonia reacts with 5 moles of oxygen to produces 4 moles of NO and 6 moles of water.

To find limiting reactant we need to find the moles of each reactant and using the balanced equation find which reactant will be ended first. Then, with limiting reactant we can find the moles of each reactant and its mass:

<em>a. </em><em>Moles NH3 -Molar mass. 17.031g/mol-</em>

25g NH3*(1mol/17.031g) = 1.47moles NH3

Moles O2 = 4 moles

For a complete reaction of 4 moles of O2 are required:

4mol O2 * (4mol NH3 / 5mol O2) = 3.2 moles of NH3.

As there are just 1.47 moles, NH3 is limiting reactant

b. Moles NO:

1.47moles NH3 * (4mol NO/4mol NH3) = 1.47mol NO

Mass NO -Molar mass: 30.01g/mol-

1.47mol NO * (30.01g/mol) = 44g of NO

c. Moles H2O:

1.47moles NH3 * (6mol H2O/4mol NH3) = 2.205mol H2O

Mass H2O -Molar mass: 18.01g/mol-

2.205mol H2O * (18.01g/mol) = 40g of H2O

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How do you do empirical formula
mihalych1998 [28]

Answer:

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Explanation:

Step 1: Obtain the mass of each element present in grams. Element % = mass in g = m.

Step 2: Determine the number of moles of each type of atom present. ...

Step 3: Divide the number of moles of each element by the smallest number of moles. ...

Step 4: Convert numbers to whole numbers.

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