To determine the number of cups of milk, we first calculate for the volume of the milk needed. Then, we use a conversion factor for the volume from cubic centimeter to cups. From literature, 1 cubic centimeter is equal to 0.0042 cup. We do as follows:
Volume of milk = ( 2.50 kg ) ( 1000 g / 1 kg ) / 1.03 g /cm^3 = 2427.18 cm^3
cups of milk = 2427.18 cm^3 ( 0.0042 cup / 1 cm^3 ) = 10.19 cups
When an atom emits a beta
particle from the nucleus, the nucleus only has one more proton and one less
neutron and this will make the atomic mass number remains unchanged while the
atomic number increases by 1.
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Two things plants need to have in order to stay in orbit and rotate is speed and less velocity.
Answer:
Elemental gold to have a Face-centered cubic structure.
Explanation:
From the information given:
Radius of gold = 144 pm
Its density = 19.32 g/cm³
Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:


a = 407 pm
In a unit cell, Volume (V) = a³
V = (407 pm)³
V = 6.74 × 10⁷ pm³
V = 6.74 × 10⁻²³ cm³
Recall that:
Net no. of an atom in an FCC unit cell = 4
Thus;


density d = 19.41 g/cm³
Similarly; For a body-centered cubic structure

where;
r = 144


a = 332.56 pm
In a unit cell, Volume V = a³
V = (332.56 pm)³
V = 3.68 × 10⁷ pm³
V 3.68 × 10⁻²³ cm³
Recall that:
Net no. of atoms in BCC cell = 2
∴


density =17.78 g/cm³
From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.
This makes the elemental gold to have a Face-centered cubic structure.
[CO] = 1 mol / 2L = 0.5 M
[
According to the equation:
and by using the ICE table:
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
initial 0.5 0.5 0 0
change -X -X +X +X
Equ (0.5-X) (0.5-X) X X
when Kc = X^2 * (0.5-X)^2
by substitution:
1.845 = X^2 * (0.5-X)^2 by solving for X
∴X = 0.26
∴ [CO2] = X = 0.26