Answer: Units. The SI derived unit of frequency is the hertz (Hz), named after the German physicist Heinrich Hertz. One hertz means that an event repeats once per second.
The frequency is the number of waves that pass a point in space during any time interval, usually one second. We measure it in units of cycles (waves) per second, or hertz. The frequency of visible light is referred to as color, and ranges from 430 trillion hertz, seen as red, to 750 trillion hertz, seen as violet.
Frequency of an electromagnetic wave is the number of wave cycles per unit time. The SI unit of frequency is Hertz or Hz. Frequency is represented by the symbol v called neu.
What is electromagnetic wave?Waves are propagation or transfer of energy through air or vacuum. An electromagnetic wave is associated with a combined electric field and magnetic field.
Waves are associated with characteristic frequency and wavelength. Wavelength of a wave is the distance between its two consecutive crests or troughs.
Frequency of a wave is the number of wave cycles per unit time. Its is expressed in the unit of Hertz or Hz. Frequency and wavelength are in inverse relation. The shortest waves have greater energy and greater frequency.
Frequency is inverse of time of propagation. So that it can also be used in s⁻¹ which is equal to one hertz.
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Write the balanced reaction and solubility product expression (KSP) for dissolving silver chromate: Ag2CrO4(s). Include all charges, stoichiometric coefficients, and phase subscripts.
Answer:
2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓
Ksp = [2s]² . [s] → 4s³
Explanation:
Ag₂CrO₄ → 2Ag⁺ + CrO₄⁻²
Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.
2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓ Ksp
That's the expression for the precipitation equilibrium.
To determine the solubility product expression, we work with the Ksp
Ag₂CrO₄ (s) ⇄ 2Ag⁺ (aq) + CrO₄⁻² (aq) Ksp
2 s s
Look the stoichiometry is 1:2, between the salt and the silver.
Ksp = [2s]² . [s] → 4s³
Can a closed system gain particles
1.
How many kJ are required to completely vaporize a 250 gram sample of water at 100°C?
general formula
Answer:
565 kJ
Explanation:
[tex]H_v[/tex] = Heat of vaporization of water at 100°C = 2260 J/g
m = Mass of water = 250 g
Heat required is given by
[tex]Q=mH_v[/tex]
[tex]\Rightarrow Q=250\times 2260[/tex]
[tex]\Rightarrow Q=565000\ \text{J}[/tex]
[tex]\Rightarrow Q=565\times 10^3\ \text{J}[/tex]
[tex]\Rightarrow Q=565\ \text{kJ}[/tex]
The heat required to vaporize the water is 565 kJ.
10. Determine the frequency of light that has a wavelength of 400nm. c=X*v. C=3.0x108
Answer:
7.5 x 10^14 Hz
Explanation:
2. A green laser pointer emits light at 554 nm. Calculate the frequency of this light and the energy of a photon of this light.
Answer:
Explanation:
Wavelength (λ) = speed/frequency
(λ) = c/f
Where;
c = speed (3× 10^8m/s)
f = frequency (Hz)
(
Calculate the molar solubility of mercury (I) bromide, Hg2Br2, in 1.0 M KBr. The Ksp for Hg2Br2 is 5.6 X 10−23. (Hint: How would the Br− concentration from the sparingly soluble compound itself compare to the Br− concentration that comes from the KBr?
Answer:
The correct answer is 5.6 × 10⁻²³ M.
Explanation:
As a highly soluble salt, KBr dissolves easily in water, while Hg₂Br₂ is very less soluble in comparison to KBr.
Let the solubility of Hg₂Br₂ is S mol per liter.
Therefore,
KBr (s) (1.0 M) ⇒ K⁺ (aq) (1M) + Br⁻ (aq) (1M)
Hg₂Br₂ (s) (1-S) ⇔ Hg₂⁺ (aq) (S) + 2Br⁻ (aq) (2S)
Net [Br-] = (2S + 1) M
Ksp = S (2S + 1)²
Ksp = S (4S² + 1 + 4S)
Ksp = 4S³ + S + 4S²
As the solubility is extremely less, therefore, we can ignore S² and S³. Now,
Ksp = S = 5.6 × 10⁻²³ M
Hence, the solubility of Hg₂Br₂ is 5.6 × 10⁻²³ M.
Globular proteins play a critical role in virtually all chemical processes.
True or false
Answer:
True
Explanation:
A gas with a volume of 525 mL at a temperature of -25°C is heated to 175°C.
What is the new volume, in milliliters, of the gas if pressure and number of moles
are held constant?
Answer:
Volume V2 = 948.13 ml
Explanation:
Given:
Volume V1 = 525 ml
Temperature T1 = -25°C + 273.15
Volume V2 = ?
Temperature T1 = 175°C + 273.15
Computation:
V1 / T1 = V2 / T2
525 / [-25°C + 273.15] = V2 / [175°C + 273.15]
Volume V2 = 948.13 ml
The new volume, in milliliters, of the gas is 948 L
From the question,
We are to determine the new volume of the gas.
From Charles' law which states that the volume of a fixed mass of gas is directly proportional to the temperature (in Kelvin) provided that the pressure remains constant.
That is,
V ∝ T
Then,
V = kT
Therefore, we can write that
[tex]\frac{V_{1} }{T_{1}} = \frac{V_{2}}{T_{2}}[/tex]
Where [tex]V_{1}[/tex] is the initial volume
[tex]T_{1}[/tex] is the initial temperature
[tex]V_{2}[/tex] is the final volume
and [tex]T_{2}[/tex] is the final temperature
From the given information,
[tex]V_{1} = 525 \ mL[/tex]
[tex]T_{1} = -25 ^{\circ}C = -25 + 273.15 \ K = 248.15 \ K[/tex]
[tex]T_{2} = 175 ^{\circ} C = 175 +273.15 \ K =448.15 \ K[/tex]
Putting the values into the formula, we get
[tex]\frac{525}{248.15} = \frac{V_{2} }{448.15}[/tex]
∴ [tex]V_{2} = \frac{525 \times 448.15}{248.15}[/tex]
[tex]V_{2} = \frac{235278.75}{248.15}[/tex]
[tex]V_{2} = 948.13 \ L[/tex]
V₂ ≅ 948 L
Hence, the new volume, in milliliters, of the gas is 948 L
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According to the ideal gas law, a 10.68 mol sample of methane gas in a 0.8295 L container at 501.9 K should exert a pressure of 530.3 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure
Answer:
The answer is "152.28%".
Explanation:
Formula for ideal gas:
PV=nRT
equation:
[tex]\to (P+\frac{an^2}{v^2})(v-nb)=nRT \\\\a=2.253 \frac{L^2 atm}{mol^2}\\\\b=4.278 \times 10^{-2} \frac{L}{mol}\\\\n=10.68 \ mol\\\\v= 0.8295 \ L\\\\\to (P+\frac{an^2}{v^2})= \frac{nRT}{(v-nb)}\\\\\to (P)= \frac{nRT}{(v-nb)} - \frac{an^2}{v^2} \\\\[/tex]
[tex]= \frac{10.68 \times 0.0082 \times 501.9}{0.8295 - (10.68 \times 4.278 \times 10^{-2})} - \frac{2.253 \times 10.68^2}{0.8295^2}\\\\=\frac{43.9543944}{0.4568904} - \frac{256.982587}{0.68807025}\\\\ = 96.2033661- 373.483067\\\\=-277.279701 \\\\ =-277.279701 \ atm[/tex]
The pressure of gas:
[tex]= (530.3 + 277.279701) \ atm\\\\= 807.579701 \ atm \\\\[/tex]
Calculating the pressure percentage:
[tex]=\frac{807.579701}{530.3} \times 100\\\\= 152.28 \%[/tex]
hhhheeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeelllllllllllllllllllllllllllllllppppppppppppppppppp
Answer:
What do you need help with?
Explanation:
If I have an unknown quantity of gas at STP with a volume of 41 liters, how many moles of gas do
have?
Answer:1 mol
Explanation:
the conversion from mol to volume of gas at STP is 22.4L, meaning that the amount of mols in a gas of volume 22.4L at STP is one mol.
(7th grader sorry if its wrong)
which of the following requires the most energy to break a bond? a. breaking cl-br bond. b. breaking a n-p bond. c. breaking a o-o bond. d. none of the above.
A. Breaking a Cl-Br bond
What is the acceleration of gravity on Saturn if your mass is 75 kg and your weight is 800n
Answer:
The answer is 10.67 m/s²Explanation:
The acceleration of gravity of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
f is the force
m is the mass
From the question we have
[tex]a = \frac{800}{75} = \frac{32}{3} \\ = 10.666666...[/tex]
We have the final answer as
10.67 m/s²Hope this helps you
If the earth were to spin half as fast as it does now, how many hours would be in one day?
Answer:
48 hours
Explanation:
Because if the world would spin regularly, then it would take 24 hours. But if it spins half as fast, then it would be 2x more.
Would someone mind helping me? I really need this answer but I'm so confused. I would appreciate any help :) and if you get the answer right, ill give you brainliest.
Answer:
A
Explanation:
liquids thake the shape of what holds them so shape is changing and volume is the same
Determine the differences In electronegativity ionic radius atomic radius and first ionization energy for oxygen and beryllium
Answer:
Electronegativity = 1.87.
Ionic radius = 109 pm.
Atomic radius = -39 pm
First ionization energy = 410 kJ/mol
Explanation:
Hello!
In this case, since electronegativity, ionic radius, atomic radius and first ionization energy are periodic properties that have specific trends, we can summarize it by realizing that oxygen and beryllium belong the same period 2 and differ in group, 6A and 2A respectively.
In such a way, the required comparison is written below:
Electronegativity = 3.44 (oxygen) - 1.57 (beryllium) = 1.87.
Ionic radius = 140 pm (oxygen)- 31 pm (beryllium) = 109.
Atomic radius = 73 pm (oxygen) - 112 pm (beryllium) = -39 pm
First ionization energy = 1310 kJ/mol (oxygen) - 900 kJ/mol (beryllium) = 410 kJ/mol
It means that electronegativity, ionic radius and first ionization energy increases from left to right whereas the atomic radius from right to left.
Best regards!
Oxygen and Beryllium are the elements that pertain to the same period and different groups 6A and 2A.
Ionic radius, atomic radius, electronegativity and first ionization energy are periodic qualities that have selective trends.
Differences in the trends are:
Ionic radius = 109 pmAtomic radius = -39 pmElectronegativity = 1.87First ionization energy = 410 kJ/molThis can be explained as:
Ionic radius:= 140 pm (oxygen)- 31 pm (beryllium)
= 109.
Atomic radius:= 73 pm (oxygen) - 112 pm (beryllium)
= -39 pm
Electronegativity:= 3.44 (oxygen) - 1.57 (beryllium)
= 1.87
First ionization energy:= 1310 kJ/mol (oxygen) - 900 kJ/mol (beryllium)
= 410 kJ/mol
Therefore, these data tell that ionic radius, electronegativity and first ionization energy increases from left to right in a period whereas, the atomic radius increases from right to left in a period.
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PLZZ ANSWER QUICKLY: True or False: MOST organelles found in plant cells are DIFFERENT than organelles found in animal cells
Answer:
Organelles allow for various functions to occur in the cell at the same time. Despite their fundamental similarities, there are some striking differences between animal and plant cells (see Figure 1). Animal cells have centrosomes (or a pair of centrioles), and lysosomes, whereas plant cells do not.
Explanation:
How much kilometers is in 6 m
Answer: 0.006 km
Explanation:
divide the length value by 1000
so 6/1000 = 0.006
CHEMWORK The preparation of NO2(g) from N2(g) and O2(g) is an endothermic reaction: N2(g) + O2(g) + NO2(g) (unbalanced) The enthalpy change of reaction for the balanced equation (with lowest whole-number coefficients) is AH = 67.7 kJ. If 304 ml N2 (9) at 100ºC and 3.32 atm and 410 ml 0,(9) at 100°C and 3.32 atm are mixed, what amount of heat is necessary to synthesize the maximum yield of NO2(g) ?
Heat required : 1.523 kJ
Further explanationReaction
N₂+2O₂⇒2NO₂ AH = 67.7 kJ
mol of N₂:(use Pv=nRT)
[tex]\tt n=\dfrac{3.32\times 0.304}{0.082\times 373}=0.033[/tex]
mol of O₂:
[tex]\tt n=\dfrac{3.32\times 0.41}{0.082\times 373}=0.045[/tex]
Limiting reactant :
N₂ : O₂
[tex]\tt \dfrac{0.033}{1}\div \dfrac{0.045}{2}=0.033:0.0225[/tex]
Limiting reactant : O₂(smaller ratio)
mol NO₂ = mol O₂ = 0.045
2 mol ⇒ 67.7 kJ, so for 0.045 mol, heat required :
[tex]\tt \dfrac{0.045}{2}\times 67.7=1.523~kJ[/tex]
The heat required to produce the maximum yield of NO2 is 1.52 kJ.
The balanced reaction equation is;
N2(g) + 2 O2(g) -> 2 NO2(g)
The number of moles of each gas is obtained from the ideal gas equation as follows;
For N2;
P = 3.32 atm
V = 304 ml or 0.304 L
T = 100ºC + 273 = 373 K
n = ?
R = 0.082 atmLK-1mol-1
From;
PV = nRT
n = PV/RT
n = 3.32 atm × 0.304 L/0.082 atmLK-1mol-1 × 373 K
n = 0.033 moles
For O2;
PV = nRT
n = PV/RT
V = 410 ml or 0.410 L
P = 3.32 atm
R = 0.082 atmLK-1mol-1
T = 100°C + 273 = 373 K
n = 3.32 atm × 0.410 L/0.082 atmLK-1mol-1 × 373 K
n = 0.045 moles
1 mole of N2 reacts with 2 moles of O2
0.033 moles of N2 reacts with 0.033 moles × 2 moles/1 mole
= 0.066 moles
2 moles of O2 yields 2 moles of NO2
0.045 moles O2 yields 0.045 moles × 2 moles/2 moles = 0.045 moles
If 2 mole of NO2 requires 67.7 kJ
0.045 moles of NO2 requires 0.045 moles × 67.7 kJ/2 mole
= 1.52 kJ
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At STP, 32 grams of O2 would occupy the same volume as:
At STP 32 g of O₂ would occupy by the same volume as 4 g of He
Further explanationComplete question
At STP 32 g of O₂ would occupy by the same volume as:
4.0 g of He8.0 g of CH₄64 g of H₂32 g of SO₂Standard Conditions
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
So the gas will have the same volume if the number of moles is the same
mol of 32 grams of O₂ :
[tex]\tt \dfrac{32}{32}=1[/tex]
He[tex]\tt mol=\dfrac{4}{4}=1[/tex]
CH₄[tex]\tt mol=\dfrac{8}{16}=0.5[/tex]
H₂[tex]\tt mol=\dfrac{64}{2}=32[/tex]
SO₂[tex]\tt mol=\dfrac{32}{64}=0.5[/tex]
So mol of 4 g He = mol of 32 g O₂
STP is the standard temperature and pressure of the gas to compare with the experimental data. At STP 32 gm of oxygen is equivalent to 4 gm of helium.
What is STP?STP is the standard temperature and pressure of 0 degrees Celsius and 1 atm. At this temperature and pressure, the volume of the gas is 22.4 L/mol.
So, if the moles of the 32 gm oxygen are the same as the other element then the volume will also be the same. Here, the moles of oxygen are 1 mol.
Moles of 4 gm helium is 1, 8 gm methane is 0.5 mol, 64 gm dihydrogen is 32 mol, and 32 gm sulfur dioxide is 0.5 mol.
Therefore, the mole of 4 gm helium is equivalent to the mole of 32 gm oxygen.
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Jet streams generally flow toward the
a. north
b. south
c. east
d. west
please answer thank you
Answer:
The major jet streams on Earth are westerly winds that flow west to east.
If 8.793 of heat are added to a 2.00g sample of metal and the temperature increases by 4.9°C. What is the specific heat?
The specific heat of metal = 0.897 J/g° C
Further explanationHeat absorbed can be calculated using the formula:
Q = mc∆TQ = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
Q = 8.793 J
m = 2 g
∆t = 4.9 °C
[tex]\tt c=\dfrac{Q}{m.\Delta t}\\\\c=\dfrac{8.793}{2\times 4.9}\\\\c=0.897~J/g^oC[/tex]
Calculate the mass of Na2O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction.
4NA + O2 ⟶ 2Na2O
Answer:
The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.
Explanation:
You know the balanced reaction:
4 NA + O₂ ⟶ 2 Na₂O
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) react and are produced:
Na: 4 molesO₂: 1 moleNa₂O: 2 molesBeing:
Na: 23 g/oleO: 16 g/molethe molar mass of the compounds participating in the reaction is:
Na: 23 g/moleO₂: 2*16 g/mole= 32 g/moleNa₂O: 2*23 g/mole +16 g/mole= 62 g/moleThen by stoichiometry of the reaction they react and are produced:
Na: 4 moles* 23 g/mole= 92 gO₂: 1 mole*32 g/mole= 32 gNa₂O: 2 moles* 62 g/mole= 124 gThen you can apply the following rule of three: if 92 grams of Na produce 124 grams of Na₂O, 4 grams of Na, how much mass of Na₂O does it produce?
[tex]massofNa_{2}O =\frac{4 grams of Na*124 gramsofNa_{2}O }{92 grams of Na}[/tex]
mass of Na₂O=5.39 g
The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.
Chlorine dioxide is used as a disinfectant and bleaching agent. In water, it reacts to form chloric acid (HClO3), according to the following balanced equation: 6 ClO2 + 3 H2O → 5 HClO3 + HCl If excess ClO2 is mixed with 18.0 mL of H2O (d = 0.998g/1ml ) how many grams of HClO3 are formed?
would be very helpful if the answer was explained with steps
Answer:
b
Explanation:
b
Calculate the total Calories in one cup of watermelon containing 0.9 g protein, 0.2 g fat, and 11 g carbohydrat
Answer:
49.4 cal
Explanation:
Step 1: Given data
Composition of 1 cup of watermelon: 0.9 g protein, 0.2 g fat, and 11 g carbohydrate.
Step 2: Calculate the calories provided by 0.9 g of protein
1 g of protein yields 4 cal.
0.9 g × (4 cal/1 g) = 3.6 cal
Step 3: Calculate the calories provided by 0.2 g of fat
1 g of fat yields 9 cal.
0.2 g × (9 cal/1 g) = 1.8 cal
Step 4: Calculate the calories provided by 11 g of carbohydrate
1 g of carbohydrate yields 4 cal.
11 g × (4 cal/1 g) = 44 cal
Step 5: Calculate the total number of calories
3.6 cal + 1.8 cal + 44 cal = 49.4 cal
Complete and balance the equation for the single displacement reaction Cl2 + NaCl + Br2
Answer:
NaBr + Cl2 = NaCl + Br2 - Chemical Equation Balancer.
Explanation:
Scientists use models to help them represent things in nature that they cannot observe directly. They can also use models to make predictions. For example, scientists use Bohr models of atoms to predict chemical reactions between atoms, You can use models in the same way, but you must first understand what each model component represents. Which element is represented by this Bohr model?
A. chlorine (Cl)
B.oxygen (O)
C. selenium (Se)
D. sulfur (S)
Answer:
I think it’s chlorine (Cl)
Explanation:
I’m literally doing the same question right now
can a water cycle ever end
Answer:
No It cannot.
Explanation:
A water cycle does not have an end as it moves all around the earth
The process:
A water cycle can start by water evaporating through the sun's heat from a water bed, such as: Ocean, sea, pond etc.
The heat from the sun will turn the water into water vapour and will go up into the atmosphere.
The water vapor gets together with other vapors and forms clouds.
These clouds will move around the earth, once they are so full of water they drop the water to Earth in some form of precipitation. It could be rain, snow, sleet, or hail.
These water will form ponds or go back in the ocean and the process wil repeat again.
How many moles are in 1.2 x 1023 Rps?
Explain why the atomic radius decreases across a period and increases down a group ?
Answer:
In general, atomic radius decreases across a period and increases down a group. ... Down a group, the number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital. This results in a larger atomic radius.