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2 years ago

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Gino made a table to describe parts of the electromagnetic spectrum. A 3-column table with 4 rows. The first column labeled wave

has entries ultraviolet, radio waves, infrared, x-rays. The second column labeled frequency has entries high, very low, low, high. The third column labeled Wavelength has entries long, very long, long, short. What mistake did Gino make? X-rays should have a low frequency and a long wavelength. Infrared light should have a high frequency, not a low frequency. Radio waves should have a very high frequency and a very short wavelength. Ultraviolet light should have a short wavelength, not a long wavelength.

2 answers:

IgorC [24]2 years ago

7 0

Answer:

D. Ultraviolet light should have a short wavelength, not a long wavelength.

Explanation:

just took the quiz on Ed

xeze [42]2 years ago

5 0

Answer:

d

Explanation:

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A student uses visible spectrophotometry to determine the concentration of CoCl2(aq) in a sample solution. First the student pre

k0ka [10]

Answer:

$4\times 10^{-9} J$ is the approximate energy of one photon of this light.

Explanation:

Energy of the photon can be calculated by

$E=h\nu=\frac{h\times c}{\lambda}$ (Planck's equation)

where,

E = energy of photon

h = Planck's constant = $6.63\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of light =

$\nu$ = frequency of the light

we have , $\lambda =510 nm=510\times 10^{-9}m$

Now put all the given values in the above formula, we get the energy of the photons.

$E=\frac{(6.63\times 10^{-34}Js)\times (3\times 10^8m/s)}{510\times 10^{-9}m}$

$E=3.9\times 10^{-19}J\approx 4\times 10^{-9} J$

$4\times 10^{-9} J$ is the approximate energy of one photon of this light.

5 0

2 years ago

You have 49.8 g of O2 gas in a container with twice the volume as one with CO2 gas. The pressure and temperature of both contain

IrinaVladis [17]

Answer:

34.2 g is the mass of carbon dioxide gas one have in the container.

Explanation:

Moles of $O_2$ :-

Mass = 49.8 g

Molar mass of oxygen gas = 32 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{49.8\ g}{32\ g/mol}$

$Moles_{O_2}= 1.55625\ mol$

Since pressure and volume are constant, we can use the Avogadro's law as:-

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

Given ,

V₂ is twice the volume of V₁

V₂ = 2V₁

n₁ = ?

n₂ = 1.55625 mol

Using above equation as:

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

$\frac {V_1}{n_1}=\frac {2\times V_1}{1.55625}$

n₁ = 0.778125 moles

Moles of carbon dioxide = 0.778125 moles

Molar mass of $CO_2$ = 44.0 g/mol

Mass of $CO_2$ = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g

<u>34.2 g is the mass of carbon dioxide gas one have in the container.</u>

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1 year ago

A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Lit

My name is Ann [436]

Answer:

$M=0.213M$

Explanation:

Hello,

In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:

$n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-$

$n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-$

$n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-$

We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:

$n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-$

Finally, we compute the molarity:

$M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M$

Regards.

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2 years ago

At low temperatures and high pressures ethene gas C2H4(g), does not behave like an ideal gas. Use chemical principles to explain

Luba_88 [7]

Answer:

The particles of an ideal gas have no volume and no attractions for each other. In a real gas, however, the molecules do have a measurable volume. The molecules of real cases have intermolecular attractions for each other.

An ideal gas behaves like a real gas under the conditions of low temperature and high pressure.

This is because at low temperature and high pressure molecules of gas will have negligible kinetic energy and strong force of attraction.

Thus ethene gas does not behave like an ideal gas at low temperatures and high pressures.

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1 year ago

Use electron configurations to account for the stability of the lanthanide ions Ce⁴⁺ and Eu²⁺.

Citrus2011 [14]

Answer:

Explanation:

The Ce metal has electronic configuration as follows

[Xe] 4f¹5d¹6s²

After losing 4 electrons , it gains noble gas configuration ,. So Ce ⁺⁴ is stable.

Eu has electronic configuration as follows

[ Xe ] 4 f ⁷6s²

[ Xe ] 4 f ⁷

Its outermost orbit contains 2 electrons so Eu²⁺ is stable. Its +3 oxidation state is also stable.

Ce⁺²

7 0

2 years ago