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3 years ago

how much hydrogen gas is necessary to exert a pressure of 1.4 ATM at 430k if occupying a volume of 15.1 l

Chemistry

1 answer:

lora16 [44]3 years ago

7 0

Answer:

Explanation:

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Please help thaank youu

krek1111 [17]

First, let's start off by finding the mass of this whole hydrate.
(Note: the unit of measurement for mass will be amu)

Let's find the molecular mass of each element.
$Co=58.933$
$Cl=35.45$
$H=1.008$
$O=15.999$

Now, let's find the mass of each compound.

$CoCl_2=58.933+2(35.45)=129.833$
$H_2O=2(1.008)+15.999=18.015$

We have 6 molecules of H2O, so multiply 18.015 by 6 then add that with the weight of CoCl2.

$6(18.015)=108.09$
$129.833+108.09=237.923$

Now divide 108.09 (mass of all the H2O in the hydrate) by 237.923 (total mass of hydrate).

$\dfrac{108.09}{237.923}$

$\approx0.45431$

Turn that into a percentage and you get 45.431%.
Hope this helps! :)

6 0

3 years ago

In calcium fluoride, what is the closets noble gas that each ion becomes?

alekssr [168]

Calcium fluoride: CaF₂
Ca(2+) >>> Ar (argon)
F(-) >>> Ne (neon)

4 0

3 years ago

Jan drew a diagram to compare centipedes and millipedes.

andre [41]

Answer:

BBB millipedes have two sets of legs on each segment.

Explanation:

4 0

3 years ago

Read 2 more answers

Calculate the percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl on the assumption that t

stealth61 [152]

Answer:

1.089%

Explanation:

From;

ν =1/2πc(k/meff)^1/2

Where;

ν = wave number

meff = reduced mass or effective mass

k = force constant

c= speed of light

Let

ν =1/2πc (k/meff)^1/2 vibrational wave number for 23Na35 Cl

ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl

The between the two is obtained from;

ν' - ν /ν = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2

Therefore;

ν' - ν /ν = [meff/m'eff]^1/2 - 1

Substituting values, we have;

ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2 -1

ν' - ν /ν = -0.01089

percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;

ν' - ν /ν * 100

|(-0.01089)| × 100 = 1.089%

4 0

3 years ago

What is the relation between work done height mass and acceleration due to gravity

Zanzabum

Explanation:

Potential energy is the the relationship between work done height mass and acceleration due to gravity, because of this some objects also experience kinetic energy due to the factors mentioned above

6 0

1 year ago