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3 years ago

What is the frequency of light that is associated with green in the visible spectrum with a

Chemistry

1 answer:

Delicious77 [7]3 years ago

8 0

Answer:

5.5 × 10^14 Hz

Explanation:

The wavelength can be represented using the formula;

λ = v/f

Where;

λ = wavelength (m)

v = speed of light (3 × 10^8m/s)

f = frequency (Hz)

According to this question, a light associated with the visible spectrum has a wavelength of 550 nm.

550nm = 550 × 10^-9m

f = v/λ

f = 3 × 10^8 ÷ 550 × 10^-9

f = 0.0055 × 10^ (8+9)

f = 0.0055 × 10^17

f = 5.5 × 10^14 Hz

The frequency is 5.5 × 10^14 Hz

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When water vapor condenses, how much heat energy will be released into the atmosphere?

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How many electrons in a silicon atom have n = 3 as one of their quantum numbers? How many electrons in a potassium atom have l =

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4 years ago

If the vapor pressure of ethanol at 34.7degree C is 100

saw5 [17]

Answer:

we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:

ln (P2/P1) = $\frac{ΔvapH}{R}([tex]\frac{1}{T1}$ - $\frac{1}{T2}$ )

where

P1 and P2 are the vapour pressures at temperatures T1 and T2

Δ vapH = the enthalpy of vaporization of the ETHANOL

R = the Universal Gas Constant

In this problem,

P 1 = 100 mmHg

; T 1 = 34.7 °C = 307.07 K

P 2 = 760mmHg

T 2 =T⁻²=?

Δ vap H = 38.6 kJ/mol

R = 0.008314 kJ⋅K -1 mol -1

ln ( 760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1 /0.008314 )

0.0004368=(0.00325 - T⁻²)

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6 0

3 years ago

How many molecules of Mg3N2 (magnesium nitride) are formed when excess Mg (magnesium)

dybincka [34]

Explanation:

3Mg(s) + N2(g) = Mg3N2(s)

First check that the equation is balanced. In this case, it is.

Assuming that magnesium is the limiting reactant:

First find the molecular weight using the Periodic Table.

We find that the atomic mass of magnesium is approximately

24.3g, so the molecular weight is just 24.3g\mol

2. Next we need the mole to mole ratio. As there are 3

magnesiums for 1 magnesium nitride (shown by the coefficients), the

mole to mole ratio is 1 mol Mg3N2\3 mol Mg.

3. We need the amount of the substance, in grams. Since you have not

stated it in the question, I'll just do 10g AS AN EXAMPLE. Note that

depending on the amount, the LIMITING REAGENT MAY DIFFER.

4. Finally, we need the molecular weight of Mg3N2, which we can easily

calculate to be around 100.9\mol.

5. Putting this all together, we have 10gMg⋅ (mol Mg\24.3gMg)

(1mol Mg3N2\ 3mol Mg) (100.9g Mg3N2\mol Mg3N2)

the units will cancel to leave gMg3N2 (grams of magnesium nitride):

10gMg ⋅ (mol Mg\24.3gMg) (1mol Mg3N2\3mol Mg)

(100.9g Mg3N2\mol Mg3N2)

Doing the calculation yields approximately 13.84g.

Assuming that nitrogen is the limiting reactant:

Similarly, following the above steps but with 10g of nitrogen yields 36.04g

In conclusion, as we produce less amount of Mg3N2 when we assumed that Mg was the limiting reagent, magnesium is the limiting reagent and nitrogen is the excess.

Note: This is in THIS CASE, where we have 10g of both. The answer may vary depending on the amount of each substance.

7 0

3 years ago

Read 2 more answers