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azamat
3 years ago
6

Write the symbol for each of the following ions:

Chemistry
1 answer:
kap26 [50]3 years ago
3 0

Answer:

a) _{31}^{71}\textrm {P^{3+}}

b) _{35}^{80}\textrm {Br^{-}}

c) _{90}^{232}\textrm {Th^{4+}}

d) _{38}^{87}\textrm {Sr^{2+}}

Explanation:

For neutral atoms:

Atomic Number (Z)= number of protons = number of electrons

Mass number (A) = number of protons + number of neutrons

For ions with positive net charge:

Number of protons = Z + net charge

For ions with negative net charge:

Number of protons= Z - net charge

a) A = 71, Charge = +3

Number of electrons = 28

Number of protons = 28 +3 =31

_{31}^{71}\textrm {P^{3+}}

b)  A = 35, Z = 45+35=80, Charge = -1

Number of protons =35

Number of neutrons = 45

Number of electrons = 36

Charge = Number of protons- Number of electrons =35-36 = -1

_{35}^{80}\textrm {Br^{-}}

c)  Charge = +4

Number of electrons = 86

Number of protons = Z = 86+4 = 90

mass number = A = 90+142 = 232

_{90}^{232}\textrm {Th^{4+}}

d)  Charge = +2

Atomic number = Number of protons = Z = 38

mass number = A = 87

_{38}^{87}\textrm {Sr^{2+}}

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The answer would be c as the cart is not in motion therefor ruling out kinetic and it is completely at rest making all of it energy potential

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3 years ago
How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles
sattari [20]

1) Balance the chemical equation.

3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

<em>3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4

<em>We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. </em>This is the excess reactant.

<em>3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2

<em>We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. </em>This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2

<em>The mass of Ca3(PO4)2 produced is</em> 351 g Ca3(PO4)2.

Option D.

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1 year ago
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Answer:

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2 years ago
How many atoms are in a 1.8 mol sample of Magnesium (Mg)?
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Answer:

1.1 × 10²⁴ atoms Mg

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify</em>

[Given] 1.8 mol Mg

[Solve] atoms Mg

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 1.8 \ mol \ Mg(\frac{6.022 \cdot 10^{23} \ atoms \ Mg}{1 \ mol \ Mg})
  2. [DA] Multiply [Cancel out units]:                                                                      \displaystyle 1.08396 \cdot 10^{24} \ atoms \ Mg

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.08396 × 10²⁴ atoms Mg ≈ 1.1 × 10²⁴ atoms Mg

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