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creativ13 [48]
4 years ago
9

A deficiency of ADH in the body causes which disease

Chemistry
1 answer:
Annette [7]4 years ago
8 0
<span>καλή ιστορία αλλά όχι</span>
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Which of the following is found in both a cell and a virus?
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I dont know maybe b
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3 years ago
How many grams of C are there in 0.234 moles of caffeine?
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3 years ago
The rate constant for the first-order decomposition of N2O5 (g) to NO2 (g) and O2 (g) is 7.48 * 10-3 s-1 at a given temperature.
alina1380 [7]

Answer:

Explanation:

In this problem, we have a first-order decomposition reaction with a given rate constant. The rate law for a first-order reaction like this is

r

a

t

e

=

k

[

A

]

, where k is the rate constant and [A] is the concentration of the reactant (renamed as A, for brevity). To find the dynamics of the reaction with time, we can integrate the rate law to get an expression for [A](t):

rate = −d[A]dt = k[A]

[A]f =[A]i e−kt

We want the total pressure of the reaction chamber to be 0.145 atm, with a starting reactant pressure of 0.110 atm. To solve for the time this reaction takes, we need the reaction equation:

2N2O5(g) → 4NO2(g) + O2(g)

Using the stoichiometry of the reaction equation, we can determine the final pressure of the reactant. This requires us to rewrite the total pressure equation in terms of the change in pressure of the reactant.

Pf=0.145atm

Pi=0.110atm = pN2O5

iPf =pN2O5

f +pNO2 + pO2pNO2 = 4pO2

This comes from the stoichiometry.

pNO2 = 2(pN2O5i − pN2O5f )

This comes from the stoichiometry.

pNO2 = −2ΔpN2O5Pf = (pN2O5i + ΔpN2O5) − 2ΔpN2O 5 − 12ΔpN2O5

0.145atm =(0.110atm + ΔpN2O5) − 2.5ΔpN2O5 = 0.110atm − 1.5ΔpN2O5

ΔpN2O5 = −0.0233atm

pN2O5f = 0.110atm − 0.0233atm =

0.0867atm

This is our final pressure! Now we can use the integrated rate law.

8 0
3 years ago
Greg has 36 DVDs in his collection.
sp2606 [1]

Answer:

14

Explanation:

4 0
4 years ago
vanadium has an atomic mass of 50.9415 amu. it has two common isotopes.one isotopes has a mass of 50.9440 amu and a relative abu
Kaylis [27]

Explanation:

Average atomic mass of the  vanadium = 50.9415 amu

Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975

Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu

Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025

Atomic mass of Isotope (II) of vanadium ,m' = ?

Average atomic mass of vanadium =

m × abundance of isotope(I) + m' × abundance of isotope (II)

50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025

m'= 49.944 amu

The atomic mass of isotope (II) of vanadium is 49.944 amu.

5 0
3 years ago
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