What is a three-dimensional structure that represents the alternating pattern of particles in a crystal? a polyatomic ion
a crystal lattice
an ionic bond
a formula unit

A three-dimensional structure representing alternating pattern of particles in a crystal is : ( B ) A crystal lattice

Crystal lattice is the structural arrangement of ions, atoms and molecules as points inside a solid ( crystalline ). Crystal lattice represents the patterns of particles ( atoms, ions or molecules ) in a crystal solid.

An ionic bond is a type of bonding that involves the transfer of electrons from an atom to another atom due to the presence of electrostatic force of attraction between oppositely charged ions.

Therefore we can conclude that A three-dimensional structure representing alternating pattern of particles in a crystal is A crystal lattice.

Learn more about crystal lattice : link

The three dimensional structure that has been representing the pattern of the particles in the crystal has been a crystal lattice. Thus option B is correct.

The crystal has been the solid structure that has been consisted of the alternating arrangement of the atoms and the ions in the lattice.

The poly atomic ion has been the ionic state of the element, ionic bond has been the sharing of the electrons in the molecule,while formula unit has been the sharing of electrons in the formula unit of the compound.

Thus, the crystal lattice has been the 3-D structure that has been the representation of the alternating particles in the crystals. Thus, option B is correct.

For more information about the crystal, refer to the link:

link

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error