1. The mass of 1.33×10²² mole of Sb is 1.62×10²⁴ g
2. The mass of 4.75×10¹⁴ mole of Pt is 9.26×10¹⁶ g
3. The mass of 1.22×10²³ mole of Ag is 1.32×10²⁵ g
4. The mass of 9.85×10²⁴ mole of Cr is 5.12×10²⁶ g
<h3>1. Determination of the mass of 1.33×10²² mole of Sb</h3>
- Mole of Sb = 1.33×10²² mole
- Molar mass of Sb = 122 g/mol
Mass = mole × molar mass
Mass of Sb = 1.33×10²² × 122
Mass of Sb = 1.62×10²⁴ g
<h3>2. Determination of the mass of 4.75×10¹⁴ mole of Pt</h3>
- Mole of Pt = 4.75×10¹⁴ mole
- Molar mass of Pt = 122 g/mol
Mass = mole × molar mass
Mass of Pt = 4.75×10¹⁴ × 195
Mass of Pt = 9.26×10¹⁶ g
<h3>3. Determination of the mass of 1.22×10²³ mole of Ag</h3>
- Mole of Ag = 1.22×10²³ mole
- Molar mass of Ag = 108 g/mol
Mass = mole × molar mass
Mass of Ag = 1.22×10²³ × 108
Mass of Ag = 1.32×10²⁵ g
<h3>4. Determination of the mass of 9.85×10²⁴ mole of Cr</h3>
- Mole of Cr = 9.85×10²⁴ mole
- Molar mass of Cr = 52 g/mol
Mass = mole × molar mass
Mass of Cr = 9.85×10²⁴ × 52
Mass of Cr = 5.12×10²⁶ g
Learn more about mole:
brainly.com/question/13314627
Answer:
newton's third law of motion
Explanation:
it states about action and reaction when you see an action you take reaction like taking off your foot from the gas pedal
Answer:
The much higher power density offered by lithium ion batteries is a distinct advantage. Electric vehicles also need a battery technology that has a high energy density. ... Lithium ion cells is that their rate of self-discharge is much lower than that of other rechargeable cells such as Ni-Cad and NiMH forms.
Put this into your own words or teachers will make you redo it
Answer:
Explanation:
The crabs cannot see the plankton they eat near the ocean floor. For the crabs to see the plankton, some color of visible light would need to reach the plankton so that it can be reflected into the crabs' eyes.
Answer:
ΔHreaction = 263.15 kJ/mol
Explanation:
The reaction is as follow:
OH + CF₂Cl₂ → HOF + CFCl₂
You need to calculate the enthalpy of reaction and for this it is necessary to know the standard enthalpies for each of the compounds. These enthalpies are as follows and can be found in your textbook or on the Internet.
ΔHreaction = ∑ΔHproducts - ∑ΔHreactants
