Answer:
Oobleck is a mixture of cornstarch & water, a substance that behaves like the Mantle.
Explanation:
the chemical equation will be XY2
Answer: The correct option is Current W flows at a higher rate than Current Z.
Explanation: To answer this question, we will require Ohm's law.
Ohm's Law states that the current flowing through a conductor across two points is directly proportional to the voltage difference across that two points.
Mathematically,
where, V = voltage
I = Current
R = resistance
For the given question, assuming that the resistance is constant. So, the current is directly proportional to the voltage.
Hence, as the current W is greater of all the given currents so, it will flow at a higher rate.
Therefore, the correct answer is Current W flows at a higher rate than Current Z.
Answer:
7.38 g/cm³ is the density of the metal
Explanation:
In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).
To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.
- <em>Volume of the unit cell</em>
Volume = a³
a = √8×r
(r = 198x10⁻¹²m)
a = 5.6x10⁻¹⁰ m
Volume = 1.756x10⁻²⁸ m³
1m = 100cm → 1m³ = (100cm)³:
1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =
<h3> 1.756x10⁻²² cm³ → Volume of the unit cell in cm³</h3><h3 />
- <em>Mass of the unit cell:</em>
<em>There are 4 atoms of gold:</em>
4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold
As 1 mole weighs 195.08g:
6.64x10⁻²⁴ moles of gold × (195.08g / mol) =
<h3>1.296x10⁻²¹g is the mass of the unit cell</h3><h3 />
- <em>Density of the metal:</em>
1.296x10⁻²¹g / 1.756x10⁻²² cm³ =
<h3>7.38 g/cm³ is the density of the metal</h3>
The
atomic mass of an element is a result of the weighted average of the
masses of its corresponding isotopes. <span>
We are given:
Ag-107: mass of 106.905 amu, 52 percent abundance
Ag-109: mass of 108.905 amu,
48 percent abundance
To be able to calculate the atomic mass of
silver, we multiply the percent abundances of the isotopes by their
respective atomic masses. Then, we add,
Atomic mass = (Atomic Mass of Ag-107) (%
Abundance) + (Atomic Mass of Ag-109) (% Abundance )
Substituting the given values, we have
Atomic mass = 106.905 (.52) + 108.905 (.48)
Atomic mass = 107.865 amu
Therefore, the atomic mass of silver would be 107.865 amu.</span>
