This element has 80 electrons and is considered a transitional element. What am I?

Balance the equation xcl2(aq)+agno3(aq)=x(no3)2(aq)+agcl(s)

lisabon 2012 [21]

Explanation:

$XCl _{2(aq)} + 2AgNO _{3(aq)}→X(NO _{3}) _{2(aq)} +2 AgCl _{(s)}$

6 0

3 years ago

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We have a sample of water vapor which cools from 150 °C to 125 °C. What happens to the motion of the molecules during this time

pentagon [3]

The motion of the molecules decreases.

Explanation:

Gases are formed when the energy in a system overcomes the attractive forces between the molecules. The gases expand to fill the space they occupy. In this way, the gas molecules interact little. In the gaseous state, the molecules move very quickly. As the temperature decreases, the amount of movement of the individual molecules also decreases.

The fast-moving particle slows down. When a particle speeds up, it has more kinetic energy. When a particle slows down, it has less kinetic energy. The particles in solid form are commonly connected through electrostatic powers. They don't get enough space to move around, therefore, their speed diminishes, they can't keep their standard speed like in the vaporous or fluid state.

5 0

3 years ago

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

4 0

3 years ago

How do tidal bulges form?

Troyanec [42]

Answer: The pull of the moon's gravity on Earth's water causes tidal bulges to form on the side closest to the moon and farthest from the moon. In the place where there are tidal bulges, high tide occurs along coastline.

Explanation:

4 0

1 year ago

Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What

coldgirl [10]

Answer:

$\rm_{90}^{231}\text{Th}$

Explanation:

The unbalanced nuclear equation is

$\rm _{92}^{235}\text{U} \longrightarrow \, _{2}^{4}\text{He} + X$

Let's write X as a nuclear symbol.

$\rm _{92}^{235}\text{U} \longrightarrow \, _{2}^{4}\text{He} + _{Z}^{A}\text{X}$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and of the subscripts must be the same on each side of the reaction arrow.

Then

235 = 4 + A , so A = 235 - 4 = 231, and

92 = 2 + Z , so Z = 92 - 2 = 90

And your nuclear equation becomes

$\rm _{92}^{235}\text{U} \longrightarrow \, _{2}^{4}\text{He} +\, _{90}^{231}\text{X}$

Element 90 is thorium, so

$\rm X = _{90}^{231}\text{Th}$

7 0

3 years ago