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3 years ago

What does it mean when water is supercooled?

1 answer:

4 0

Supercooling<span>, a state where liquids do not solidify even below their normal freezing point. Means sometimes we have liquid water below 0 degree C.</span>

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What happens to the electron cloud at very high atomic numbers, when the innermost electrons would, using a non-relativistic mod

marin [14]

Answer:

tough question

Explanation:

8 0

2 years ago

What is the de Broglie wavelength, in cm, of a 11.0-g hummingbird flying at 1.20 x 10^2 mph?

KonstantinChe [14]

Answer:1.123 x 10^-31cm

Explanation:

mass of humming bird= 11.0g

speed= 1.20x10^2mph

but I mile = 1.6m

1km=1000

I mile = 1.6x10^3m

1.20x10^2mph= 1.6x10^3m /1mile x at 1.20 x 10^2

=1.932 x10^5m

recall that

1 hr= 60 min

1 min=60 secs, 1hr=3600s

Speed = distance/ time

=1.932 x10^5 / 3600= 5.366 x 10 ^1 m/s

m= a 11.0g= 11.0 x 10^-3kg

h=6.626*10^-34 (kg*m^2)/s

Wavelength = h/mu

= 6.626*10^-34/(11 x 10^-3 x 5.366x 10^1)

6.63x10^-34/ 590.26x 10 ^-3= 1.123 x10^-33m

but 1m = 100cm

1.123 x 10 ^-33 x 100 = 1.123 x 10^-31cm

de broglie wavelength of humming bird = 1.123 x 10 ^-31cm

5 0

3 years ago

Which of the following is non polar covalent bond?

vladimir2022 [97]

Answer:

N-Cl

Explanation:

Look at the chart below. Since N-Cl bond has a electronegativity difference of (3.0-3.0) zero, they are non-polar.

5 0

2 years ago

Compare the amount of heat required to vaporize a 200.-gram sample of H2O(ℓ) at its boiling point to the amount of heat required

stira [4]

The enthalpy of vaporization of H2O is higher than the enthalpy of fusion of H2O, therefore vaporizing the same mass of H2O would require more heat/energy than melting the same mass of H2O.

6 0

3 years ago

A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound

Karolina [17]

Answer:

The empirical formula of the compound is $C_{0.504}HO_{1.008}$ .

Explanation:

We need to determine the empirical formula in its simplest form, where hydrogen ( $H$ ) is scaled up to a mole, since it has the molar mass, and both carbon ( $C$ ) and oxygen ( $O$ ) are also scaled up in the same magnitude. The empirical formula is of the form:

$C_{x}HO_{y}$

Where $x$ , $y$ are the number of moles of the carbon and oxygen, respectively.

The scale factor ( $r$ ), no unit, is calculated by the following formula:

$r = \frac{M_{H}}{m_{H}}$ (1)

Where:

$m_{H}$ - Mass of hydrogen, in grams.

$M_{H}$ - Molar mass of hydrogen, in grams per mole.

If we know that $M_{H} = 1.008\,\frac{g}{mol}$ and $m_{H} = 0.2\,g$ , then the scale factor is:

$r = \frac{1.008}{0.2}$

$r = 5.04$

The molar masses of carbon ( $M_{C}$ ) and oxygen ( $M_{O}$ ) are $12.011\,\frac{g}{mol}$ and $15.999\,\frac{g}{mol}$ , then, the respective numbers of moles are: ( $r = 5.04$ , $m_{C} = 1.2\,g$ , $m_{O} = 3.2\,g$ )