The atomic mass of the isotope copper-_

cluponka [151]

Answer:

A partial lunar eclipse occurs when the Earth moves between the Sun and Moon but the three celestial bodies do not form a straight line in space. When that happens, a small part of the Moon's surface is covered by the darkest, central part of the Earth's shadow, called the umbra.

Explanation:

5 0

3 years ago

Read 2 more answers

By what factor will the kinetic energy change if the speed of the baseball is decreased to 54.8 mi/h? Express your answer as an

Serggg [28]

Answer:

The Kinetic Energy is approximately 3 times decreased

Explanation:

A baseball weighs 5.13 oz.

a)What is the kinetic energy, in joules, of this baseball when it is thrown by a major league pitcher at 95.o mi/h?

b) By what factor will the kinetic energy change if the speed of the baseball is decreased to 54.8 mi/h? Express your answer as an integer.

Kinetic Energy (KE)=0.5×mass×velocity ^ 2

Kinetic Energy (KE)=0.5×mass × velocity ^ 2

Joules = kg×m^2/s^2

1 mile = 1609.344 meters

1 hour = 3600 sec

1 Oz = 28.34952 g = 0.02834952 kg

a) KE=0.5×m×v^2

=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(95 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2

=130.761 kg×m^2/s^2 = 130.761 Joules

b) KE=0.5×m×v^2

=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(54.8 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2

=43.51028 kg×m^2/s^2 = 43.51028 Joules

= 130.761 / 43.51028 = 3.00528,

As such the Kinetic Energy is approximately 3 times decreased

4 0

4 years ago

How many elements are found in the following chemical equation? K + Cl → KCl A. 1 B. 2 C. 3 D. 0

Gnesinka [82]

K is Potassium
Cl is Chlorine
There are 2 elements.

5 0

3 years ago

Read 2 more answers

Which of the following atoms has the greatest atomic radius? Oxygen (o) fluorine (F) Chlorine (CL) Carbon (C)

Andrei [34K]

Periodic Trend:

The Atomic radius of atoms generally decreases from left to right across a period

Group Trend:

The atomic radius of atoms generally increases from top to bottom within a group. As atomic number increases down a group, there is a increase in the positive nuclear charge, however the co-occurring increase in the number of orbitals wins out, increasing the atomic radius down a group in the periodic table

Answer :

The Atom with the greatest atomic radius is chlorine. Fluorine can be ruled out because it is in the same period as oxygen and further to the right down the period. Chlorine has the largest atomic size because it is farthest down the group of any of the above elements listed.

5 0

3 years ago

You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=1.60 \times {300.0\times 10^{-3}}\ moles$

Moles of potassium iodide = 0.48 moles

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=1.20\times {285\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.342 moles

According to the given reaction:

$2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.

Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.102/0.585 M = 1.174 M

Statement A is correct.

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g

Statement B is correct.

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.48/0.585 M = 0.821 M

Statement C is correct.

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

So, Concentration = 0.684/0.585 M = 1.169 M

Statement D is incorrect.

4 0

3 years ago

The atomic mass of the isotope copper-__ is 62.930 amu