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3 years ago

Why is water a liquid at room temperature instead of a gas?

Chemistry

2 answers:

maxonik [38]3 years ago

5 0

water has stronger intermolecular forces

svlad2 [7]3 years ago

3 0

Has strong hydrogen bonding which requires alot of energy to overcome hence high boiling point hence liq in room temp

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If an object has a mass of 180 g and a density of 6 g/cm^3, what is the volume of the object?

cupoosta [38]

Answer:

<h3>The answer is 30 cm³</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density} \\$

From the question

mass = 180 g

density = 6 g/cm³

We have

$volume = \frac{180}{6} \\$

We have the final answer as

Hope this helps you

5 0

3 years ago

Why is the distance between New York and London increasing each year?

Dmitriy789 [7]

Answer:

because the tectonic plates move 4 inches per year.

Explanation:

8 0

3 years ago

At 80 ∘C, Kc=1.87×10^−3 for the reaction PH3BCl3(s)⇌PH3(g)+BCl3(g)

daser333 [38]

Answer:

A) he equilibrium concentration of PH3 = 0.0432M

B) he equilibrium concentration of BCl3 = 0.0432M

C) what is the minimum mass of PH3BCl3(s) that must be added to the flask to achieve equilibrium = 1.69g

Explanation:

The detailed steps and appropriate calculation is as shown in the attached file.

6 0

3 years ago

Who was the first person to describe a chemical reaction as the recombination of atoms?

Cerrena [4.2K]

Niels Bohr is the first person to describe a chemical reaction as the combination of atoms.

4 0

3 years ago

calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?

inessss [21]

<u>Answer:</u>

$\Delta E=E_{final}-E_{initial}$

$\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}$

$\Delta E=-1312[0.111-1]KJ mol^{-1}$

$\Delta E=1166 KJ mol^{-1}$

$\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}$

$=193623 \times 10^{-23} \frac {J}{photon}$

$\Delta E=1.93623 \times 10^{-18} \frac {J}{photon}$

$\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}$

h is planck's constant

c is the speed of light

λ is the wavelength of light

$\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18} J/photon)}$

Wavelength

$\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m} =103 nm (Answer)$

<em>Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is </em><u><em>103 nm.</em></u>

7 0

3 years ago