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ser-zykov [4K]
3 years ago
12

Three elements are represented by the letters A, B, and C. the three elements have the same number of valence electrons. Element

C has electrons in three energy levels and element A is the lightest of the three. Elements B has a total of twenty electrons. Identify the elements.
Chemistry
1 answer:
AfilCa [17]3 years ago
3 0
Answer:
A- beryllium
B- calcium
C- magnesium

Explanation
NOTE: all element in group 2 have 2 balance electrons

First let’s start with B- number of electrons= number of protons which is equal to the atomic number. therefore, the answer is calcium as it’s atomic number is 20

C- magnesium will have three energy levels considering it has 12 electrons (2,8,2).

A- beryllium is the lightest one in group 2 as it has the atomic mass of 9.0122.
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A 1.00 liter container holds a mixture of 0.52 mg of He and 2.05 mg of Ne at 25oC. Determine the partial pressures of He and Ne
Ymorist [56]

Answer:

pHe = 3.2 × 10⁻³ atm

pNe = 2.5 × 10⁻³ atm

P = 5.7 × 10⁻³ atm

Explanation:

Given data

Volume = 1.00 L

Temperature = 25°C + 273 = 298 K

mHe = 0.52 mg = 0.52 × 10⁻³ g

mNe = 2.05 mg = 2.05 × 10⁻³ g

The molar mass of He is 4.00 g/mol. The moles of He are:

0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol

We can find the partial pressure of He using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 3.2 × 10⁻³ atm

The molar mass of Ne is 20.18 g/mol. The moles of Ne are:

2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol

We can find the partial pressure of Ne using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 2.5 × 10⁻³ atm

The total pressure is the sum of the partial pressures.

P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm

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The reaction is first order in cyclopropane and has a measured rate constant of k=3.36×10−5 s−1 at 720 k. if the initial cyclopr
sergey [27]

Answer:

0.0277 M.

Explanation:

The integral rate law of a first order reaction:

<em>Kt = ln ([A₀]/[A]),</em>

where, k is the rate constant of the reaction <em>(k = 3.36 × 10⁻⁵ s⁻¹)</em>,

t is the time of the reaction <em>(t = 235.0 min = 14100 s)</em>,

[A₀] is the initial concentration of cyclopropane <em>([A₀] = 0.0445 M)</em>

<em>∵ Kt = ln ([A₀]/[A]),</em>

∴ (3.36 × 10⁻⁵ s⁻¹)(14100 s) = ln (0.0445 M)/[A]

Taking the exponential of both sides:

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<em>∴ [A] = (0.0445 M)/1.6 = 0.0277 M.</em>

<em />

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