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Lapatulllka [165]
3 years ago
5

Decide which element probably has a density most and least similar to the density of strontium.

Chemistry
1 answer:
gayaneshka [121]3 years ago
3 0

Answer:

Explanation:

If we compare the density of strontium with all periodic table elements then the elements with very similar to the strontium density is Aluminium.

The density of aluminium is 2.698 g/cm³ while the density of strontium is 2.64 g/cm³.

While the least similar is Hassium. The estimated value of density of Hassium is 41 g/cm³. There is big difference between between both values.

If the given options are lead, indium, calcium and nitrogen than the elements with the density similar to the density of strontium is calcium.

The density of calcium is 1.54 g/cm³.

The elements with the density least similar to the density of strontium is  lead.

The density of lead is 11.342 g/cm³.

While the density of indium and nitrogen is 7.310 and 0.0012506  g/cm³ respectively.

It can be written as:

Most similar:  calcium, nitrogen, indium, lead.

Least similar: lead, indium, nitrogen, calcium.

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Write the chemial formula for H3PO4​
rewona [7]

Phosphoric acid, also known as orthophosphoric acid or phosphoric(V) acid, is a weak acid with the chemical formula H3PO4.

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3 years ago
The concentration of ag+ in a solution saturated with ag2c2o4(s) is 1.7 × 10-4 m. calculate ksp for ag2c2o4.
Gelneren [198K]
<span>Silver oxalate dissociation equation is following: 
</span><span>
Ag</span>₂C₂O₄(s) ⇄ 2Ag⁺(aq) + C₂O₄²⁻(aq)

According to reaction follows next stoichiometric ratio:

[Ag⁺] : [C₂O₄²⁻] = 2 : 1

[C₂O₄²⁻] = [Ag⁺] / 2

[C₂O₄²⁻] = (1.7×10⁻⁴)/2 = 8.5×10⁻⁵ M

So, solubility product constants for silver oxalate is:

Ksp = [Ag⁺]² x [C₂O₄²⁻]

Ksp = [1.7×10⁻⁴]² x [8.5×10⁻⁵]

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4 0
4 years ago
How many electrons will the lithium atom give up to become stable?
Angelina_Jolie [31]
Recall that most atoms are stable when their outermost ring has eight electrons. (Some atoms, such as lithium and beryllium, are stable when their outermost ring has two electrons.)
8 0
3 years ago
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In a coffee cup calorimeter, 1.60 g of NH4NO3 is mixed with 75.0 g of water at an initial temperature of 25.00 degrees C. After
igomit [66]

Answer:

+26.6kJ/mol

Explanation:

The enthalpy of dissolution of NH₄NO₃ is:

NH₄NO₃(aq) + ΔH  → NH₄⁺ + NO₃⁻

Where ΔH is the heat of reaction that is absorbed per mole of NH₄NO₃,

The moles that reacts in 1.60g are (Molar mass NH₄NO₃:80g/mol):

1.60g * * (1mol / 80g) = 0.02 moles reacts

To find the heat released in the coffee cup calorimeter, we must use the equation:

Q = m×ΔT×C

Where Q is heat released,

m is mass of the solution

ΔT is change in temperature (Final temperature - Initial temperature)

C is specific heat of the solution (4.18J/g°C)

Mass of the solution is:

1.60g + 75g = 76.60g

Change in temperature is:

25.00°C - 23.34°C = 1.66°C

Replacing:

Q = m×ΔT×C

Q = 76.60g×1.66°C×4.18J/g°C

Q = 531.5J

This is the heat released per 0.02mol. The heat released per mole (Enthalpy change for the dissolution of NH₄NO₃) is:

531.5J / 0.02mol = 26576J/ mol =

+26.6kJ/mol

<em>+ because the heat is absorbed, the reaction is endothermic-</em>

7 0
3 years ago
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