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1 month ago

Which of the following statements is true about the relationships between photon energy, wavelength, and frequency?

Chemistry

1 answer:

KiRa [2.9K]1 month ago

8 0

Answer: The Answer is A.

Explanation:

The energy of a photon is directly related to its electromagnetic frequency, meaning it is inversely related to the wavelength. A higher frequency results in greater energy for the photon. Conversely, a longer wavelength corresponds to lower energy levels.

Hope this Helps!

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If a reaction occurs, what will be the products of the unbalanced reaction below?Cu(s) + Ni(NO3)2(aq)

Alekssandra [3086]

Cu(NO3)2 --> MM187.5558 NiNO3 *COEF2* --> 120.6983

3 0

25 days ago

Acrylonitrile () is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. If 1

alisha [2963]

2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)

To begin, this isn't really a chemistry forum, but anyway.

This represents a limiting reagent scenario.

Set it up as a Dimensional Analysis issue.

Begin with your desired outcome.

Your goal is to find the mass of acrylonitrile (C3H3N)

so you should initiate with that (I'll abbreviate Acrylonitrile as ACL for convenience)

(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)

If you calculate that, you will find that 15 grams of C3H6 yields 18.9 grams of acrylonitrile produced.

Utilize the same approach for the remaining two reactants.

So, I figured it out, and for

oxygen, I calculated 11.04 grams

and for ammonia, I calculated 15.29 grams

This means that the maximum possible production is 11.04 grams, since to create any additional amount, more O2 would be necessary, but with only 10 grams available, that's the upper limit for this reaction.

The other two reactants are in excess.

Please rate as brainliest!

3 0

1 month ago

Sea water's density can be calculated as a function of the compressibility, B, where p = po exp[(p - Patm)/B]. Calculate the pre

lions [2927]

Answer:

A pressure of 137.14 MPa exists 10,000 m beneath the ocean surface.

At this same depth, the density measures 2039 kg/m3.

Explanation:

P0 and ρ0 symbolize the pressure and density at sea level (indicative of atmospheric conditions). With an increase in ocean depth, both pressure and density likewise rise.

The relationship between pressure and density can be expressed as:

$\frac{dP}{dy}=\rho*g=\rho_0*g*e^{(P-P_0)/\beta\\\\$

By rearranging

$\frac{dP}{e^{(P-P_0)/\beta}}= \rho_0*g*dy\\\\\int\limits^{P}_{P_0} {e^{-(P-P_0)/\beta}}dP =\int\limits^y_0 {\rho_0*g*dy}\\\\(-\beta*e^{-(P-P_0)/\beta})-(\beta*e^0)=\rho_0*g*(y-0)\\\\-\beta*(e^{-(P-P_0)/\beta}-1)=\rho_0*g*y\\\\e^{-(P-P_0)/\beta}=1-\frac{\rho_0*g*y}{\beta}\\\\-\frac{P-P_0}{\beta} =ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\$

This equation allows for computation of P at 10,000 m beneath the ocean's surface:

$P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-200MPa*ln(1-\frac{1027kg/m^3*9.81m/s^2*10,000m}{200MPa})\\\\P-P_0=-200MPa*ln(1-\frac{1027*9.81*10,000Pa}{200*10^6Pa})\\\\P-P_0=-200MPa*ln(1-0.5037)\\\\P-P_0=-200MPa*(-0.6857)=137.14MPa$

The density found at a depth of 10,000 m in the ocean is

$\rho=\rho_0*e^{(P-P_0)/\beta}\\\rho=1027kg/m^3*e^{(137.14/200)}=1027*e^{0.686}kg/m^3\\\rho=1027*1.985 kg/m^3\\\rho=2039\,kg/m^3$

4 0

1 month ago

(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o

Tems11 [2777]

Answer :

The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.

Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100$

Here, Xa is the electronegativity of atom A and Xb is that of atom B.

Given: The compound is TiAl₃.

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 (as shown in the provided image)

Substitute these values into the formula:

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100$

$Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100$

$Percent Ionic character = 1 - e^(^-^0^.^1^) * 100$

The value of e⁻¹ equals 0.90.

Therefore, percent ionic character = (1 - 0.90) × 100

Percent Ionic Character = 10%

Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.

8 0

2 months ago

Find the age t of a sample, if the total mass of carbon in the sample is mc, the activity of the sample is a, the current ratio

Alekssandra [3086]

N₀ signifies the quantity of C-14 atoms per kg of carbon in the original sample at time = 0 seconds, when the carbon composition matched that in today’s atmosphere. As time progresses to ts, the number of C-14 atoms per kg declines to N, due to radioactive decay. λ indicates the decay constant.
Hence, we have N = N₀e - λt, which is the equation for radioactive decay. Rearranging gives us N₀/N = e λt, or In(N₀/N) = - λt, which becomes equation 1.
The sample contains mc kg of carbon, leading to an activity measured as A/mc decay per kg. The variable r represents the initial mass of C-14 in the sample at t=0 relative to the total mass of carbon which is calculated as [(total number of C-14 atoms at t = 0) × ma] / total mass of carbon. Thus, N₀ equates to r/ma, which becomes equation 2.
The activity of the radioactive element is directly related to the atom count at the moment. The activity equation A = dN/dt = λ(N) indicates that: A = λ₁(N × mc). Rearranging provides N = A / (λmc), represented in equation 3.
By integrating equations 2 and 3, we can solve for t yielding
t = (1/λ) In(rλmc/m₀A).

6 0

1 month ago