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3 years ago

Can someone explain me this with answer? I will make brain list

Chemistry

1 answer:

Ostrovityanka [42]3 years ago

4 0

Answer:

Sawdust floats, sand dissolves, and 3rd dissolves

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Consider this reaction: →2NH3g+N2g3H2g At a certain temperature it obeys this rate law: rate =2.11·M−1s−1NH32 Suppose a vessel c

enot [183]

Answer: It takes 5.89 s for the concentration of $NH_3$ to decrease by 88.0%.

Explanation:

$2NH_3(g)\rightarrow N_2(g)+3H_2(g)$

As the units of rate constant is $M^{-1}s^{-1}$ , the kinetics must be second order.

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

$a_0$ = initiaal concentration = 0.590 M

a= concentration left after time t = $0.590-\frac{88}{100}\times 0.590=0.0708M$

$\frac{1}{0.0708}=2.11\times t+\frac{1}{0.590}$

$t=5.89s$

Thus it takes 5.89 seconds for the concentration of $NH_3$ to decrease by 88.0%.

6 0

3 years ago

Which of the following is true about lanthanides?

Pavlova-9 [17]

The correct option is A.
The lanthanides are a group of transition metals which are found in the periodic table. They are fifteen in number and they are all radioactive. They are usually refereed to as the 'rare earth' metals because they are not common. Lanthanides have fairly high melting points.

8 0

3 years ago

What have a definite shape and volume.

Over [174]

Solids they both have shape and volume

8 0

3 years ago

True or false an atom is the smallest particle of a whole element

alina1380 [7]

True, as an atom has the same chemical properties as the first bulk element. Dalton’s Atomic Theory explains all :))

5 0

4 years ago

a sample of carbon dioxide gas occupies a volume of 20 L at standard temperature and pressure (STP). what will be the volume of

fredd [130]

Volume of Argon = 59.972 L

<h3>Further explanation</h3>

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.08206 L.atm / mol K

T = temperature, Kelvin

At STP(0 C, 1 atm), 1 mol gas= 22.4 L

so for 20 L

$\tt \dfrac{20}{22.4}=0.893$

For Argon

P = 1 atm

T = 273 K x 3 =819 K (tripled)

n = 0.893 (the same number of moles)

R = 0.082 L/atm.mol K

$\tt V=\dfrac{nRT}{P}=\dfrac{0.893\times 0.082\times 819}{1}=59.972~L$

7 0

3 years ago