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lisabon 2012 [21]
2 years ago
14

Help please !! im not very good at science ):

Chemistry
1 answer:
frez [133]2 years ago
4 0
Vas happenin!!

1-
A: 4

B: 48

C: 10


2-

A: 20

B: 18

C: 27

D: 16

Hope this helps!


-Zayn Malik
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Michelle is trying to find the average atomic mass of a sample of an unknown
GREYUIT [131]

The average atomic mass of her sample is 114.54 amu

Let the 1st isotope be A

Let the 2nd isotope be B

From the question given above, the following data were obtained:

  • Abundance of isotope A (A%) = 59.34%
  • Mass of isotope A = 113.6459 amu
  • Mass of isotope B = 115.8488 amu
  • Abundance of isotope B (B%) = 100 – 59.34 = 40.66%
  • Average atomic mass =?

The average atomic mass of the sample can be obtained as follow:

Average \: atomic \: mass \:  =  \frac{mass \: of \: A \times A\%}{100}  + \frac{mass \: of \: B \times B\%}{100}  \\  \\ Average \: atomic \: mass \:  =  \frac{113.6459\times 59.34}{100} + \frac{115.8488\times 40.66}{100} \\  \\ Average \: atomic \: mass \:  = 114.54 \: amu  \\  \\

Thus, the average atomic mass of the sample is 114.54 amu

Learn more about isotope: brainly.com/question/25868336

3 0
2 years ago
What is the same for all of the drilling sites we examined?
yarga [219]

Answer:

the same is what is this question like what did u exame

4 0
3 years ago
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What is the (OH-) in a solution with a pOH of 6.48
inn [45]
Through manipulation of equations, we are able to obtain the equation:

-pOH= log [ OH^{-}]

Then we can transform the equation into:

[ OH^{-}]= 10^{-pOH}

Then we are able to plug in the pOH and directly get [OH-]:

[ OH^{-}] = 10^{-6.48}

[ OH^{-}]=3.31* 10^{-7} M
3 0
3 years ago
A force of 350 N causes a body to move with an acceleration of 10 m/s2. What's the mass of the body?
Scrat [10]
You have to use the equation F=ma and solve for m to get m=F/a.
m=mass in kg
F=force (in this case 350N)
a=acceleration (in this case 10m/s²)
when you plug everything in you should find that m=35kg

I hope this helps.
3 0
3 years ago
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The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-
grin007 [14]

Answer:

Total pressure 5.875 atm

Explanation:

The equation for above decomposition is

2N_2O \rightarrow 2N_2 + O_2

rate constant k =  1.94\times 10^{-4} min^{-1}

Half life t_{1/2} = \frac{0.693}{k} = 3572 min

Initial pressure N_2 O = 4.70 atm

Pressure after 3572 min = P

According to first order kinematics

k = \frac{1}{t} ln\frac{4.70}{P}

1.94\times 10^{-4} = \frac{1}{3572} \frac{4.70}{P}

solving for P we get

P = 2.35 atm

2N_2O \rightarrow 2N_2 + O_2

initial           4.70                         0             0

change        -2x                          +2x           +x

final             4.70 -2x                     2x           x

pressure ofO_2 after first half life  = 2.35 = 4.70 - 2x

                                                          x = 1.175

pressure of N_2 after first half life  =  2x = 2(1.175) = 2.35 ATM

Total pressure  = 2.35 + 2.35 + 1.175

                          = 5.875 atm

5 0
3 years ago
Read 2 more answers
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