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3 years ago

Which electromagnetic wave has the highest frequency?

Chemistry

2 answers:

Juli2301 [7.4K]3 years ago

4 0

Answer:

D) ultraviolet

Explanation:

Out of these, ultraviolet have the highest frequency.

The order of frequency from highest to lowest is gamma rays, x-rays, ultraviolet, visible light, infared, microwave, radio.

Natali5045456 [20]3 years ago

4 0

Answer: D) ultraviolet

Explanation:

edge 2021

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Does mint really cool your breath? doing a science fair and also is there three good questions i could use for the topic?

PilotLPTM [1.2K]

Answer:

Yes it does.

Explanation:

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3 years ago

Read 2 more answers

Which has the higher percentage of aluminum, Al2O, or Al(NO3)3

Rzqust [24]

Al₂O has a higher percentage of aluminium.

Explanation:

To solve this problem, we have to compare the molar mass of the aluminium in each compound to one another as percentage of the whole compound:

Molar mass of Al₂O = 2(27) + 16 = 70g/mol

Molar mass of Al(NO₃)₃ = 27 + 3[14 + 3(16)] = 213g/mol

Percentage by mass of Al in Al₂O = $\frac{2x27}{70}$ x 100 = 77%

Percentage by mass of Al in Al(NO₃)₃ = $\frac{27}{213}$ x 100 = 12.7%

Al₂O has a higher percentage of aluminium.

learn more:

Molar mass brainly.com/question/2861244

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3 years ago

If a reaction occurs, what will be the products of the unbalanced reaction below?

castortr0y [4]

Explanation:

I think the correct answer is B

Hope this helped u

8 0

3 years ago

The reaction is first order in cyclopropane and has a measured rate constant of k=3.36×10−5 s−1 at 720 k. if the initial cyclopr

sergey [27]

Answer:

0.0277 M.

Explanation:

The integral rate law of a first order reaction:

Kt = ln ([A₀]/[A]),

where, k is the rate constant of the reaction (k = 3.36 × 10⁻⁵ s⁻¹),

t is the time of the reaction (t = 235.0 min = 14100 s),

[A₀] is the initial concentration of cyclopropane ([A₀] = 0.0445 M)

∵ Kt = ln ([A₀]/[A]),

∴ (3.36 × 10⁻⁵ s⁻¹)(14100 s) = ln (0.0445 M)/[A]

Taking the exponential of both sides:

1.6 = (0.0445 M)/[A]

∴ [A] = (0.0445 M)/1.6 = 0.0277 M.

6 0

3 years ago

in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

dimulka [17.4K]

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{12}{4}=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{1}{4}=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{6}{4}=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{10}{4}=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_{H_2O}=X_{H_2O}\times p_T$

where,

$p_{H_2O}$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_{H_2O}$ = mole fraction of water vapor gas = 0.345

Now put all the given values in the above formula, we get:

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.345\times 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

3 0

3 years ago