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kompoz [17]
2 years ago
7

An engineer is investigating the ways in which electroplating is currently

Chemistry
1 answer:
stiks02 [169]2 years ago
5 0

Answer:

Background research

Explanation:

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Which is a discontinued variation: height, eye colour or weight
Natasha2012 [34]

Answer:

Eye colour

Explanation:

6 0
2 years ago
Dinitrogen pentoxide decomposes in the gas phase to form nitrogen dioxide and oxygen gas. The reaction is first order in dinitro
vitfil [10]

Answer : The partial pressure of O_2 is, 222.93 torr

Explanation :  

Half-life = 2.81 hr = 168.6 min

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{168.6min}

k=4.11\times 10^{-3}min^{-1}

Now we have to calculate the partial pressure of O_2

The balanced chemical reaction is:

                           2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)

Initial pressure   760                0             0

At eqm.             (760-2x)            4x            x

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{P_o}{P_t}

where,

k = rate constant

t = time passed by the sample  = 215 min

a = initial pressure of N_2O_5 = 760 torr

a - x = pressure of N_2O_5 at equilibrium = (760-2x) torr

Now put all the given values in above equation, we get:

215=\frac{2.303}{4.11\times 10^{-3}}\log\frac{760}{760-2x}

x=222.93torr

The partial pressure of O_2 = x = 222.93 torr

7 0
3 years ago
Select the correct answer.
sladkih [1.3K]
C not sure 100 percent but my best guess
7 0
3 years ago
EASYYYYY!!!!!!!!
nlexa [21]

crop rotation, green manure, and bone meal

Explanation:

I just looked it up. hope it helps

6 0
2 years ago
I need this answer quick please show work
Ainat [17]

Answer:

The answer to your question is 25.2 g of acetic acid.

Explanation:

Data

[Acetic acid] = 0.839 M

Volume = 0.5 L

Molecular weight = 60.05 g/mol

Process

1.- Calculate the number of moles of acetic acid

    Molarity = moles / volume

-Solve for moles

    moles = Molarity x volume

-Substitution

    moles = (0.839)(0.5)

-Result

    moles = 0.4195

2.- Calculate the mass of acetic acid using proportions and cross multiplications

                   60.05 g ----------------------- 1 mol

                        x        ----------------------- 0.4195 moles

                        x = (0.4195 x 60.05) / 1

                        x = 25.19 g

3.- Conclusion

25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M

               

4 0
3 years ago
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