3 years ago

How can i prove that nascent oxygen more reactive than normal oxygen ?

1 answer:

6 0

Nascent oxygen has much higher reactivity than the oxygen bubbled through the reaction mixture. It doesn't stay nascent for long (you are right about it being converted quick to just O2), which is why it has to be generated in situ

You might be interested in

How much money do you guys make off of this lame website?

MrRissso [65]

Answer:

so far from what i've seen its been 80 mil

Explanation:

the power of GoogIe

7 0

2 years ago

Read 2 more answers

Which gas law refers to the solubility of a gas changing with the pressure over the solution

Dmitry [639]

Answer:

the answer should be henry's law

6 0

3 years ago

A gas has a volume of 1.75L at -23C and 150.0kPa. At what temperature would the gas occupy 1.30L at 210.0kPa?

Katen [24]

Answer:

T2 = 260 K

Explanation:

Given data:

P1 = 150.0 k Pa

T1 = (-23+ 273.15) K = 250.15 K

V1 = 1.75 L

P2 = 210.0 kPa

V2 = 1.30 L

To find:

T2 = ?

Formula:

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

$T_2 = \frac{P_2 V_2 T_1}{P_1 V_1}$

Calculation:

T2 = (210.0 kPa) x (1.30 L) x (250.15 K) / (150.0 kPa) x (1.75 L)

T2 = 260 K

8 0

3 years ago

What is the symbol for an ion with a 3+ charge, 28 electrons, and a mass number of 71

zhenek [66]

Answer:

Ga3+ is a gallium cation

7 0

2 years ago

Iodine-131 decays with a half-life of 8.02

dlinn [17]

Radioactive material undergoes 1st order decay kinetics.

For 1st order decay, half life = 0.693/k

where k = rate constant

k = 0.693/half life = 0.693/8.02 = 0.0864 day-1

Now, for 1st order reaction,
k = $\frac{2.303}{t} X log \frac{initial.conc}{final.conc}$

Given: t = 6.01d, initial conc. = 5mg

∴0.0864 = $\frac{2.303}{6.01} X log \frac{5}{final.conc}$
∴ final conc. = 2.975 mg

3 0

3 years ago