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Law Incorporation [45]
3 years ago
13

Convert .5km to inches

Chemistry
2 answers:
olasank [31]3 years ago
8 0

Answer:

196,850.39370078

Explanation:

I simply put it in a conversion calculator:)

myrzilka [38]3 years ago
4 0

Answer:

196850.39370079 in

Explanation:

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Help pls before I end it all
qwelly [4]

Answer:

We might just have to end it together

Explanation:

I tried to answer it now I'm stuck in the same hole -_-

7 0
3 years ago
Read 2 more answers
The half-life of cesium-137 is 30 years. suppose we have a 200-mg sample.
Assoli18 [71]
a) To find  the mass after t years:

we will use this formula:

A = Ao / 2^n 

when A =the amount remaining

and Ao = the initial amount

and n = t / t(1/2)

by substitution:

∴ A = 200 mg/ 2^(t/30y)


b) Mass after 90 y :

by  using the previous formula and substitute t by 90 y

A = 200mg/ 2^(90y/30y)

∴ A = 25 mg

C) Time for 1 mg remaining:

when A= Ao/ 2^(t/t(1/2)

so, by substitution:

1 mg = 200 mg / 2^(t/30y)

∴2^(t/30y) = 200 mg  by solving for t

∴ t = 229 y 


7 0
3 years ago
NEED HELP QUICKLY!!! How many moles are in each of the following?
oksano4ka [1.4K]

Answer: a. 0.26mol

b. 0.000479mol

c. 1.12mol

Explanation: Please see attachment for explanation

6 0
3 years ago
Which statement is correct regarding the reaction below? 3A + 2B yields C + 2D The rate of formation of D is twice the rate of d
Vedmedyk [2.9K]

Answer:

The correct statements are:

The rate of disappearance of B is twice the rate of appearance of C.

Explanation:

Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.

3A + 2B → C + 2D

Rate of the reaction:

R=-\frac{1}{3}\times \frac{d[A]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}

-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{1}\times \frac{d[C]}{dt}

-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{2}\times \frac{d[D]}{dt}

The rate of disappearance of B is twice the rate of appearance of C.

\frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}

2\times \frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{1}\times \frac{d[B]}{dt}

8 0
3 years ago
Read 2 more answers
The hydrogen-line emission spectrum includes a line at a wavelength of 434 nm. What is the energy of this radiation? (h= 6.626 x
Andrews [41]
Wavelength = 434nm = 434 x 10⁻⁹m
planck's constant = <span>h= 6.626 x 10 ⁻³⁴ J
E =?
by using the formula;
E = hc /</span>λ
value for c is 3 x 10⁸ m/s
E = (6.626 x 10 ⁻³⁴ J)(3 x 10⁸ m/s) / 434 x 10⁻⁹m
E = 1.9878 x 10⁻²⁵ / 434 x 10⁻⁹m
E = 4.58 x 10⁻¹⁹ joules

6 0
3 years ago
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