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mars1129 [50]
3 years ago
12

The metallic radius of a lithium atom is 152 pm. What is the volume of a lithium atom in cubic meters?

Chemistry
1 answer:
Anuta_ua [19.1K]3 years ago
8 0

Answer:

Volume of lithium atom is found to be 1.47 X 10⁻²⁹ m³

Explanation:

Let us consider the volume of atom as a sphere (but it is little complex than that). This volume is mathematically expressed as,

V=\frac{4}{3}\pi  R^{3}----------------------------------------------------------------------------------------(Eq. 1)

Here, R is the radius of lithium atom. The radius is given in picometers, so firstly let us convert it into meters

R=(152pm )(1X10^{-12}\frac{m}{pm})

R = 1.52 X 10^{-10}m

placing this value in Eq.1 the required result is achieved

V=\frac{4}{3}\pi  {1.52X10^{-10}}^{3}

V= 1.47 X 10⁻²⁹ m³

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In balancing the nuclear reaction 7535Br → E + 01e, the identity of element E is ________. In balancing the nuclear reaction Br
nordsb [41]

<u>Answer:</u> The element E is Krypton (Kr).

<u>Explanation:</u>

In a nuclear reaction, the total mass and total atomic number remains the same.

The given nuclear reaction follows:

_{35}^{75}\textrm{Br}\rightarrow ^{A}_{Z}\textrm{X}+^0_{-1}\textrm{e}

  • <u>To calculate A:</u>

Total mass on reactant side = total mass on product side

75 = A + 0

A = 75

  • <u>To calculate Z:</u>

Total atomic number on reactant side = total atomic number on product side

35 = Z + (-1)

Z = 36

The isotopic symbol of element E is _{36}^{75}\textrm{Kr}

Hence, the element E is Krypton (Kr).

4 0
3 years ago
A solution containing 8.3 g of a nonvolatile, nondissociating substance dissolved in 1.00 mol of chloroform, CHCl3, has a vapor
Margaret [11]

Answer:

a. Xm = 0.0229

b. 0.0234 moles

c. 354.1 g/mol

Explanation:

ΔP = P° . Xm

ΔP = P° - P', where P° is vapor pressure of pure solvent and P', vapor pressure of solution-

This is the formula for lowering vapor pressure.

If we apply the data given: 523 Torr - 511 Torr = 523 . Xm

Xm = ( 523 Torr - 511 Torr) / 523 Torr → 0.0229

Xm = Mole fraction of solute → Moles of solute / Total moles (sv + solute)

We can make this equation to determine moles of solute

0.0229 = Moles of solute / Moles of solute + 1

0.0229 (Moles of solute + 1) = Moles of solute

0.0229 = Moles of solute - 0.0229 moles of solute

0.0229 = 0.9771 moles of solute → 0.0229 / 0.9971 = 0.0234 moles

Molecular mass of solute → g/mol → 8.3 g / 0.0234 mol = 354.1 g/mol

7 0
3 years ago
Based on my scale, I have 35g of aluminum. How many aluminum atoms is this? SHOW YOUR WORK.
Aleksandr-060686 [28]
One mole of a substance contains 6.02×10∧23 particles,
1 mole of a aluminium contains 27 g
35 g of aluminium contains 35/27 =1.296 moles
 Thus, the number of particles will be 1.296 × 6.02 ×10∧23
      = 7.804 × 10∧23 particles,
Hence, 35 g of Aluminium contains 7.804 × 10∧23 atoms
3 0
3 years ago
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