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3 years ago

Is pyrophoric a chemical or physical property

Chemistry

1 answer:

Neporo4naja [7]3 years ago

5 0

Answer:

Pyrophoricity is a property of metals and oxides of lower oxidation states, including radioactive ones, in which they spontaneously ignite during or after stabilization.

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Did democritus belive atoms retain there identity as a chemical reaction

mihalych1998 [28]

No He believed tiny particles were invisible and couldn't be changed....So No The person that believed in this was Dalton .

4 0

3 years ago

Read 2 more answers

1. A 225-L barrel of white wine has an initial free SO2 concentration of 22 ppm and a pH of 3.70. How much SO2 (in grams) should

Alexandra [31]

Answer:

The appropriate answer is "9.225 g".

Explanation:

Given:

Required level,

= 63 ppm

Initial concentration,

= 22 ppm

Now,

The amount of free SO₂ will be:

= $Required \ level -Initial \ concentration$

= $63-22$

= $41 \ ppm$

The amount of free SO₂ to be added will be:

= $41\times 225$

= $9225 \ mg$

∵ 1000 mg = 1 g

So,

= $9225\times \frac{1}{1000}$

= $9.225$

Thus,

"9.225 g" should be added.

3 0

3 years ago

I have 345mL of a 1.5M NaCl solution. If I boil the water until the volume of the solution is 250mL, what will the molarity of t

Karo-lina-s [1.5K]

Answer is
C. 2.07 M

For explanation

M1V1 = M2V2

M2 = (M1V1)/V2

M2 = (1.5M x 345ml) / 250 ml

:. M2 = 2.07 M

8 0

3 years ago

How many grams of water do you need to weigh if a reaction requires 5 moles of h2o

rewona [7]

One mole of water weighs 18 grams. H₂O is composed of 2H= 2 and 1 0=16 adding gives you 18. number of moles= mass/ Relative Molecular Mass

Therefore, mass= Relative Molecular Mass×number of moles

= 18×5 moles

= 90 grams

5 0

3 years ago

The gas phase decomposition of sulfuryl chloride at 600 K SO2Cl2(g)SO2(g) + Cl2(g) is first order in SO2Cl2. During one experime

krok68 [10]

Answer: The rate constant for the reaction is $3.96\times 10^{-3}min^{-1}$

Explanation:

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

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample = 559 min

a = let initial amount of the reactant = $2.83\times 10^{-3}$

a - x = amount left after decay process = $3.06\times 10^{-4}$

$559min=\frac{2.303}{k}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}$

$k=\frac{2.303}{559}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}$

$k=3.96\times 10^{-3}min^{-1}$

The rate constant for the reaction is $3.96\times 10^{-3}min^{-1}$

6 0

3 years ago