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taurus [48]
3 years ago
9

Sometimes a nuclide is referenced by the name of the element followed by the:

Chemistry
1 answer:
pentagon [3]3 years ago
8 0

Answer:

electric charge

Explanation:

A nuclide is represented by the name of the element, preceded by the mass number (A) as a superscript on the left and the atomic number (Z) as subscript.

On the right, as a superscript is the electrical charge (e.g. +2) and is following the element name.

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What reaction might we use to synthesize nickel sulfate,niso4? Write equations for acid base neutralization. Include states in y
il63 [147K]

Answer:

Here's what I get  

Explanation:

1. Nickel sulfate

base + acid ⟶ salt + water

NiSO₄ is a salt of the base Ni(OH)₂ and the acid sulfuric acid.

Hydroxides of transition metals are insoluble; most sulfates are soluble.

\underbrace{\hbox{Ni(OH)$_{2}$(s)}}_{\hbox{base}} + \underbrace{\hbox{H$_{2}$SO$_{4}$(aq)}}_{\hbox{acid}} \longrightarrow \, \underbrace{\hbox{NiSO$_{4}$(aq)}}_{\hbox{salt}} + \underbrace{\hbox{2H$_{2}$O(l)}}_{\hbox{water}}

2. Carbonate + acid

Most carbonates are insoluble.

They react with acids to form carbonic acid (H₂CO₃), which decomposes into water and carbon dioxide.

\rm NiCO_{3}(s) + H_{2}SO_{4}(aq) \longrightarrow \, NiSO_{4}(aq) + H_{2}O(l) + CO_{2}(g)

5 0
3 years ago
What element is a F-18 Atom
VikaD [51]

Answer:

Sodium Fluoride F 18 Injection is a positron emitting radiopharmaceutical, no-carrier added.

Explanation:

Element Name Fluorine

Element Symbol F

Atomic Number 9

4 0
3 years ago
What kind of molecule is table sugar? a. Monosaccharide c. Trisaccharide b. Disaccharide d. Octosaccharide
Aleksandr-060686 [28]
Table sugar is a
B. Disaccharide
3 0
3 years ago
A sample of element X contains 90% X-35 atoms, 8.0% X-37 atoms, and 2.0% X-38 atoms. The average atomic mass will be closest to
Ne4ueva [31]

To find average atomic mass you multiply the mass of each isotope by its percentage, and then add the values up.

35 * 0.90 + 37 * 0.08 + 38 * 0.02 = 35.22

Average atomic mass closest to 35.22 amu.

6 0
3 years ago
Read 2 more answers
A 110.0-mL sample of a solution that is 3.0×10−3 M in AgNO3 is mixed with a 230.0- mL sample of a solution that is 0.10 M in NaC
almond37 [142]

Answer:

[Ag⁺] = 0.0666M

Explanation:

For the addition of Ag⁺ and CN⁻, the (Ag(CN)₂⁻ is produced, thus:

Ag⁺ + 2CN⁻  ⇄  Ag(CN)₂⁻

Kf = 1x10²¹ = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺]

As initial concentrations of Ag⁺ and CN⁻ are:

[Ag⁺] = 0.110L × (3.0x10⁻³mol / L) = 3.3x10⁻⁴mol / (0.110L + 0.230L) = 9.7x10⁻⁴M

[CN⁻] = 0.230L × (0.1mol / L) = 0.023mol / (0.110L + 0.230L) = 0.0676M

The equilibrium concentrations of each compound are:

[CN⁻] = 9.7x10⁻⁴M - x

[Ag⁺] = 0.0676M - x

[Ag(CN)₂⁻] = x

<em>Where x is reaction coordinate</em>

Replacing in Kf formula:

1x10²¹ = [x] / [9.7x10⁻⁴M - x]² [0.0676M - x]

1x10²¹ = [x] / 6.36048×10⁻⁸ - 0.000132085 x + 0.06954 x² - x³

-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = x

-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = 0

Solving for x:

X = 9.614x10⁻⁴M

Thus, equilibrium concentration of Ag⁺ is:

[Ag⁺] = 0.0676M - 9.614x10⁻⁴M = <em>0.0666M</em>

6 0
2 years ago
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