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3 years ago

At room temp which of the following is the MOST compressible?

Chemistry

2 answers:

Rom4ik [11]3 years ago

6 0

Answer:

A-one liter of oxygen, a gas

Explanation:

Gas is the most compressible state of matter.

rosijanka [135]3 years ago

5 0

I think the answer is C, Hope I helped you

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Be sure to answer all parts.

tangare [24]

Answer:

(a) cesium bromide (CsBr): 9.15 grams

(b) calcium sulfate (CaSO4): 5.85 grams

(d) lithium dichromate (Li2Cr2O7): 9.88 grams

(e) potassium oxalate (K2C2O4): 7.15 grams

Explanation:

(a) cesium bromide (CsBr):

Molar mass of CsBR = 212.81 g/mol

Number of moles = molarity * volume

Number of moles = 0.100 M *0.43 L

Number of moles = 0.043 moles

Mass of CsBr required = moles * Molar mass

Mass of CsBr required = 0.043 moles * 212.81 g/mol

Mass of CsBr required = 9.15 grams

(b) calcium sulfate (CaSO4):

Molar mass of CaSO4 = 136.14 g/mol

Mass of CaSO4 required = moles * Molar mass

Mass of CaSO4 required = 0.043 moles * 136.14 g/mol

Mass of CaSO4 required = 5.85 grams

(c) sodium phosphate (Na3PO4):

Molar mass of Na3PO4 = 163.94 g/mol

Mass of Na3PO4 required = moles * Molar mass

Mass of Na3PO4 required = 0.043 moles * 163.94 g/mol

Mass of Na3PO4 required = 7.05 grams

(d) lithium dichromate (Li2Cr2O7):

Molar mass of Li2Cr2O7 = 229.87 g/mol

Mass of Li2Cr2O7 required = moles * Molar mass

Mass of Li2Cr2O7 required = 0.043 moles * 229.87 g/mol

Mass of Li2Cr2O7 required = 9.88 grams

(e) potassium oxalate (K2C2O4):

Molar mass of K2C2O4 = 166.22 g/mol

Mass of K2C2O4 required = moles * Molar mass

Mass of K2C2O4 required = 0.043 moles * 166.22 g/mol

Mass of K2C2O4 required = 7.15 grams

7 0

3 years ago

PLEASE HELLLLOOPLPPPPP

vovangra [49]

Answer:

the answer is d.Class there u go i hope its riht

4 0

3 years ago

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The energy required to ionize a mole of potassium ions is 419 kJ/mol . What is the longest wavelength of light capable of this i

liraira [26]

First divide the ionization energy by Avogadro's number to get the energy per atom of potassium;
419 kj/mol / 6.023 x 10²³
= 4.19 x 10⁵ / 6.023 x 10²³ = 6.96 x 10⁻¹⁹
E = hc/λ
where lambda (λ) is the wavelength, h is Planck's constant, c is the speed of light
E = 6.96 x 10⁻¹⁹ j/atom
h =6.63x10⁻³⁴ Js
c = 3 x 10⁸ m/s
λ = ?
λ = hc/E = (6.63x10⁻³⁴ x 3 x 10⁸ ) / 6.96 x 10⁻¹⁹ = 285.8nm = 286nm
The longest wavelength of light capable of this ionization is 286nm.

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3 years ago

Create a time that marks important events during the Precambrian Supereon.

Maurinko [17]

I really hope this help!!

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3 years ago

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Is SiO2 London Dispersion, Dipole-Dipole, or Hydrogen Bonding?

s2008m [1.1K]

Answer:

It is Dispersion