What charge would you expect from a Group 13 atom when it becomes an ion?

almond37 [142]

Air has mass and density, and. cold air is denser than warm air.
The same amount of water as oil air

7 0

3 years ago

The gravitational pull of the Moon has the greatest influence on the water levels of Earth’s ocean tides. If the distance betwee

kherson [118]

The water levels would lower, because as the moon gets farther away, it's gravitational pull decreases. Hope this helps! :)

6 0

3 years ago

What terms refers to the process by which astronomers study distant objects by dissecting examining the spectrum of light an obj

mart [117]

Spectroscopy because it talks about the study of spectrum of light.

3 0

3 years ago

The mass of an electron is 9.11× 10–31 kg. What is the uncertainty in the position of an electron moving at 2.00 × 106 m/s with

Kruka [31]

Just use the Heisenberg Uncertainty principle:

ΔpΔx = h/2*pi 

Δp = the uncertainty in momentum 
Δx = the uncertainty in position 
h = 6.626e-34 J s (plank's constant) 

Hint: 

to calculate Δp use the fact that the uncertainty in the momentum is 1% (0.01) so that 

Δp = mv*(0.01) 

m = mass of electron 
v = velocity of electron 

Solve for Δx 

Δx = h/(2*pi*Δp) 

And that is the uncertainty in position.

6 0

4 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

Answer: The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago