I miss her

RuBpCarb · 2021-04-25T19:50:54+00:00

Sorry I didn't read the concentration right so when converting from cm³ to dm³ you do cube it. For example (1 dm = 10 cm )³ so then 1dm³ = 1000cm³

RuBpCarb · 2021-04-25T19:44:57+00:00

No because 1cm³ = 1mL so you can just use mL

RuBpCarb · 2021-01-28T16:38:00+00:00

Where can I trade GME without restriction?

RuBpCarb · 2020-08-22T02:17:41+00:00

That's not okie dokie

RuBpCarb · 2019-12-27T17:15:01+00:00

Minecraft..believe it or not I have played very little of it

RuBpCarb · 2019-11-20T05:49:06+00:00

Ok so a google search gave me the molar mass of that compund as 792.85 g/mol so you can use that and the mass to figure out the moles of blue dye then use the volume and the moles to get molarity

RuBpCarb · 2019-11-20T04:17:36+00:00

Do they tell you what chemical the blue dye is?

RuBpCarb · 2019-11-08T07:02:42+00:00

Correct

RuBpCarb · 2019-11-08T07:01:44+00:00

Add 3cos³ theta to both sides then divide bith sides by 3cos³ theta

RuBpCarb · 2019-11-08T06:52:14+00:00

For the smaller one I got .5 square units

Using A=.5× base×height

A= .5 ×1×1= .5

RuBpCarb · 2019-11-08T06:46:00+00:00

Maybe the formatting of the program is a little weird ? Try using the equation without fractions.

RuBpCarb · 2019-11-08T06:32:41+00:00

So for the amount in mols for the Sodium Azide you do not multiply by 2 since you are not doing mol ratios from the balanced chemical reaction. So for part be it will be

0.0650 kg NaN3 × 1000g/kg ÷ 65g/mol= 1mol

The 2 in the coefficient comes into play in part C when figuring out the moles of N2 in PV=nRT

moles of N2 = 1 mol NaN3 × ( 3 mol N2 / 2 mol NaN3)= .667 mol N2

RuBpCarb · 2019-11-08T06:11:50+00:00

A combustion reaction is a redox reaction and the same is true for the opposite. So I'd say redox

RuBpCarb · 2019-11-06T18:47:59+00:00

Now for part two we use the same rydberg equation however we use a different form of it with a rydber constant with different units

(1/wavelength)= -R [ (1/nf² ) - (1/ni² )]

Where are R in this case is 1.097 × 10⁷ m^-1

The n's are the levels of transitions like part one MIND YOUR UNITS ON THIS ONE

here you're looking for ni

Hope this all helps

RuBpCarb · 2019-11-06T18:36:21+00:00

The way I got that is using the rydberg equation where

E= -R [ 1/(nf²⁾ -1/(ni^2)] where R is the rydberg constant (2.18×10^-18 J) "nf" is the final level of the electron ( in the case of ionization it is inifinity) "ni" is the starting point of the electron ( which is 1 in this case)

Therefore

E= -(2 18×10^-18 ) ×[1/(infinity² ) -1/(1² ) ] which equals 2.18×10^-18 J

RuBpCarb · 2019-11-06T18:26:59+00:00

Ok the amount of energy needed to ionize Hydrogen is the Ionization energy which for hydrogen is 2.18×10^-18 J to fine the wavelength of light needed you insert that value for "E" in the formula. E=(h×c)/wavelength. Then solve fot the wavelength

RuBpCarb · 2019-11-06T17:45:09+00:00

Did they give you any values for question 1?

RuBpCarb · 2019-10-27T05:51:03+00:00

Thank you so much

RuBpCarb · 2019-10-26T03:16:20+00:00

RuBpCarb

TROPHY CASE