Which particle has a mass of 9.11 x 10^-28g and charge of -1?
A. electron
B. proton
C. neutron
QUESTION:- Which particle has a mass of 9.11 x 10^-28g and charge of -1?
OPTIONS:-
A. electron
B. proton
C. neutron
ANSWER:-
CHARGE ON PROTRON IS +1 AND IT HAS MASS OF [tex]1.6 \times 10 {}^{ - 27} [/tex] SO IT CANNOT BE URE ANSWER
THERE IS NO CHARGE ON NEUTRON AND HAS MASS ALMOST EQUAL TO THE PROTON SO IT ALSO CANNOT BE URE ANSWER
MASS OF THE ELECTRON:- [tex]9.11 \times 10^{ - 28} [/tex]
CHARGE ON ELECTRON:- [tex] -1[/tex]
SO URE ANSWER IS ELECTRON
How can this product be achieved using the starting material shown?
Answer:
this product can be achieved using the starting material shown is by use of NaOH as catalyst.
Answer:
By using NaOH as catalyst.
Explanation:
This product can be achieved using the starting material shown is by the use of the NaOH as catalyst.
What is the specific rotation of 13g of a molecule dissolved in 10 mL of solvent that gives an observed rotation of 23 degrees in a sample tube of 10 cm.
Answer:
[tex]\alpha=17.7[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=13g[/tex]
Volume [tex]V=10mL[/tex]
Angle [tex]\theta=23[/tex]
Sample Tube=10cm
Generally the equation for concentration is mathematically given by
[tex]C=m/v[/tex]
[tex]C=\frac{13}{10}\\C=1.3g/mL[/tex]
Therefore the Specific Rotation
[tex]\alpha=frac{\theta }{m*l}[/tex]
[tex]\alpha=frac{23 }{1.3*1.0}[/tex]
[tex]\alpha=17.7[/tex]
How many grams of NaF should be added to 500 mL of a 0.100 M solution of HF to make a buffer with a pH of 3.2
Answer:
2.25g of NaF are needed to prepare the buffer of pH = 3.2
Explanation:
The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:
pH = pKa + log [A-] / [HA]
Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF).
The moles of HF are:
500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF
Replacing:
3.2 = 3.17 + log [A-] / [0.0500moles]
0.03 = log [A-] / [0.0500moles]
1.017152 = [A-] / [0.0500moles]
[A-] = 0.0500mol * 1.017152
[A-] = 0.0536 moles NaF
The mass could be obtained using the molar mass of NaF (41.99g/mol):
0.0536 moles NaF * (41.99g/mol) =
2.25g of NaF are needed to prepare the buffer of pH = 3.2A uniform plastic block floats in water with 50.0 % of its volume above the surface of the water. The block is placed in a second liquid and floats with 23.0 % of its volume above the surface of the liquid.
What is the density of the second liquid?
Express your answer with the appropriate units.
Answer:
density of second liquid = 650 kg/m³
Explanation:
Given that:
The volume of the plastic block submerged inside the water = 0.5 V
The force on the plastic block = [tex]\rho_1V_1g[/tex]
[tex]= 0.5p_1 V_g[/tex]
when the block is floating, the weight supporting the force (buoyancy force) is:
W [tex]= 0.5p_1 V_g[/tex]
[tex]\rho Vg = 0.5p_1 V_g[/tex]
[tex]\rho = 0.5 \rho _1[/tex]
where;
water density [tex]\rho _1[/tex] = 1000
[tex]\rho = 0.5 (1000)[/tex]
[tex]\rho = 500 kg/m^3[/tex]
In the second liquid, the volume of plastic block in the water = (100-23)%
= 77% = 0.7 V
The force on the plastic block is:
[tex]= 0.77p_2 V_g[/tex]
when the block is floating, the weight supporting the force (buoyancy force) is:
[tex]W = 0.77p_2 V_g[/tex]
[tex]\rho Vg = 0.77 \rho_2 V_g \\ \\ \rho = 0.77 \rho_2 \\ \\ 500 = 0.77 \rho_2 \\ \\ \rho_2 = 500/0.77[/tex]
[tex]\mathbf{ \rho_2 \simeq 650 \ kg/m^3}[/tex]
Determine the equilibrium constant, Keq, at 25°C for the reaction
2Br- (aq) + I2(s) <--> Br2(l) + 2I- (aq)
Eocell = (0.0257/n) lnKeq, Calculate Eocell from Use this equation to calculate K value.
Eo (I2/I-) = +0.53, Eo (Br2/Br-) = +1.07,
Explanation:
The given chemical reaction is:
[tex]2Br^- (aq) + I_2(s) <-> Br_2(l) + 2I^- (aq)[/tex]
[tex]E^ocell=oxidation potential of anode + reduction potential of cathode\\[/tex]
The relation between Eo cell and Keq is shown below:
[tex]deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell[/tex]
The value of Eo cell is:
Br- undergoes oxidation and I2 undergoes reduction.
Reduction takes place at cathode.
Oxidation takes place at anode.
Hence,
[tex]E^ocell= (-1.07+0.53)V\\=-0.54V[/tex]
F=96485 C/mol
n=2 mol
R=8.314 J.K-1.mol-1
T=298K
Substitute all these values in the above formula:
[tex]ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3[/tex]
Answer:
Keq=6.13x10^33
Consider the reaction “2 SO2 (g) + O2 (g) = 2 SO3 was 0.175 M. After 50 s the concentration of SO2 Date: (g)”. Initial concentration of SO2 (g) (g) became 0.0500 M. Calculate rate of the reaction
Answer:
The answer is "[tex]1.25 \times 10^{-3} \ \frac{m}{s}[/tex]"
Explanation:
Calculating the rate of the equation:
[tex]=-\frac{1}{2} \frac{\Delta [SO_2]}{\Delta t} =-\frac{\Delta [O_2]}{\Delta t}= +\frac{1}{2} \frac{\Delta [SO_3]}{\Delta t}\\\\=\frac{\Delta [SO_2]}{\Delta t}=\frac{0.0500-0.175\ M}{505}= -2.5 \times 10^{-3} \ \frac{m}{s}\\\\[/tex]
Rate:
[tex]=\frac{-2.5 \times 10^{-3}}{2}=1.25 \times 10^{-3} \ \frac{m}{s}[/tex]
A student prepares a aqueous solution of acetic acid . Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource.
Answer:
10.71%
Explanation:
The dissociation of acetic acid can be well expressed as follow:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:
Then:
The I.C.E Table is expressed as follows:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Initial 0.0014 0 0
Change - x +x +x
Equilibrium (0.0014 - x) x x
Recall that:
Ka for acetic acid CH₃COOH = 1.8×10⁻⁵
∴
[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]
[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]
[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]
By rearrangement:
[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]
Multiplying through by (-) and solving the quadratic equation:
[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]
[tex](-0.00015 + x) (0.000168 + x) =0[/tex]
x = 0.00015 or x = -0.000168
We will only consider the positive value;
so x=[CH₃COO⁻] = [H⁺] = 0.00015
CH₃COOH = (0.0014 - 0.00015) = 0.00125
However, the percentage fraction of the dissociated acetic acid is:
[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]
= 10.71%
a chemist combines 4.9 g of nitrogen gas with 9.4 grams of nitrogen gas to form 11.4 g of ammonia 2.9 g of nitrogen is remaining
what is speed lathe machine.
Answer:
A speed lathe is a type of lathe that is designed to operate much faster than its common counterpart.
What is true of all matter?
A. It pushes or pulls on objects.
B. You can see it.
C. It gives off heat energy.
D. It has mass.
What is the energy change when 78.0 g of Hg melt at −38.8°C
Answer:
The correct answer is - 2.557 KJ
Explanation:
In this case, Hg is melting, the process is endothermic, so the energy change will have a positive sign.
we can calculate this energy by the following formula:
Q = met
where, m = mass,
e = specific heat
t = temperature
then,
Q = 78*0.14* (273-38.8)
here 0.14 = C(Hg)
= 2.557 Kj
How much heat capacity, in joules and in calories, must be added to a 75.0-g iron block with a specific heat of 0.499J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C?
Answer:
56511.75 J
13506.3 Calories
Explanation:
Applying,
Q = cm(t₂-t₁).................. Equation 1
Where Q = amount of heat, m = mass of the iron, c = specific heat capacity of the iron, t₁ = initial temperature, t₂ = final temperature.
From the question,
Given: m = 75 g, c = 0.499 J/g.°C, t₂ = 1535°C, t₁ = 25°C
Substitute these values into equation 1
Q = 75(0.499)(1535-25)
Q = 75(0.499)(1510)
Q = 56511.75 J
Q in Calories is
Q = (56511.75×0.239)
Q = 13506.3 Calories
Consider the following reaction:
CO(g)+2H2(g)⇌CH3OH(g)
A reaction mixture in a 5.15-L flask at a certain temperature initially contains 26.6 g CO and 2.36 g H2. At equilibrium, the flask contains 8.63 g CH3OH.
Part A
Calculate the equilibrium constant (Kc) for the reaction at this temperature.
Answer:
26.6
Explanation:
Step 1: Calculate the molar concentrations
We will use the following expression.
M = mass solute / molar mass solute × liters of solution
[CO]i = 26.6 g / (28.01 g/mol) × 5.15 L = 0.184 M
[H₂]i = 2.36 g / (2.02 g/mol) × 5.15 L = 0.227 M
[CH₃OH]e = 8.63 g / (32.04 g/mol) × 5.15 L = 0.0523 M
Step 2: Make an ICE chart
CO(g) + 2 H₂(g) ⇄ CH₃OH(g)
I 0.184 0.227 0
C -x -2x +x
E 0.184-x 0.227-2x x
Since [CH₃OH]e = x, x = 0.0523
Step 3: Calculate all the concentrations at equilibrium
[CO]e = 0.184-x = 0.132 M
[H₂]e = 0.227-2x = 0.122 M
[CH₃OH]e = 0.0523 M
Step 4: Calculate the equilibrium constant (Kc)
Kc = [CH₃OH] / [CO] [H₂]²
Kc = 0.0523 / 0.132 × 0.122² = 26.6
How many moles of Al are necessary to form 45.0 g of AlBr, from this
reaction:
2 Al(s) + 3 Br_(1) ► 2 AlBr_(s)?
Answer:
0.169 mole of Al
Explanation:
We'll begin by by calculating the number of mole in 45 gof AlBr₃. This can be obtained as follow:
Mass of AlBr₃ = 45 g
Molar mass of AlBr₃ = 27 + (3×80)
= 27 + 240
= 267 g/mol
Mole of AlBr₃ =?
Mole = mass /molar mass
Mole of AlBr₃ = 45 / 267
Mole of AlBr₃ = 0.169 mole
Finally, we shall determine the number of mole of Al needed to produce 45 g (i.e 0.169 mole) of AlBr₃. This can be obtained as illustrated below:
2Al + 3Br₂ —> 2AlBr₃
From the balanced equation above,
2 moles of Al reacted to produce 2 moles of AlBr₃.
Therefore, 0.169 mole of Al will also react to produce 0.169 mole of AlBr₃.
Thus, 0.169 mole of Al is needed for the reaction.
what is perodic table list all the elements up to 30 with their valencies
Answer:
a table of the chemical elements arranged in order of atomic number, usually in rows, so that elements with similar atomic structure (and hence similar chemical properties) appear in vertical columns.
What direction would equilibrium moves towards based on the following if we increased the volume of the container.
[tex]2A_{(g)} + 5B_{(g)} + 12C_{(g)}[/tex] ↔ [tex]14AC_{(g)} + 5B_{(s)}[/tex]
Answer choices:
a) reactants
b) no change
c) products
d) decrease in volume
Please help!
To answer this question, we will first find out the number of gaseous moles on each side of the equilibrium
on the left:
we have 2 moles of A, 5 moles of B and 12 moles of C
which gives us a grand total of 19 gaseous moles
on the right:
here, we have 14 moles of AC gas, we will not count the number of moles of B because it's a solid
giving us 14 gaseous moles on the right
Where does the reaction shift?
more gaseous moles means more space taken, because gas likes to fill all the space it can
if we have more volume, more gas can move around without colliding (reacting) with each other
Hence more volume favors the side with more gaseous moles
here, the left has more gaseous moles. So we can say that the reaction will shift towards the left, or the reactants side
Answer:
Explanation:
given reversible chemical reaction:
2A(g) + 5B(g) + 12C(g) ↔ 14AC(g) + 5B(s)
chemicals in solid form do not take up a lot of volume so change in container volume has no effect
look at chemicals in gas form only:
the total no. of moles of reactants in gas form = 2 + 5 + 12 = 19
the total no. of moles of products in gas form = 14
so an increase in volume of the container will favor the reaction direction with higher volume n high volume means higher no. of moles
the ans is the equilibrium will move towards a) reactants
Choose a Metal to Make Cookware
Aluminum
Copper
Iron
Lead
0.90
0.35
0.44
0.12.
Specific heat (J/g°C)
Cost ($ per lb)
1.00
5.00
0.10
1.00
Safety risk
slight
slight
none
significant
Density (g/cm3)
2.70
8.92
7.87
11.30
Considering only specific heat, V lead would
be the most ideal for use in cookware.
COMPLETE
However, compared to the other options, lead is
too
Answer:
lead
Explanation:
All metals are good conductors of heat and electricity. This property allows the heat to flow through them. The metals which are used to make cookware are Aluminum, Copper and Iron. The correct options are A, B and C.
What are metals?The metals are defined as the substances which are formed naturally below the surface of the earth. Most of the metals are lustrous or shiny. They are very strong and durable, so they are used to make many things like satellites, automobiles, cooking utensils, etc.
Malleability is the property of metals which allows them to be beaten into flat sheets. Aluminium sheets are used in the manufacture of aircrafts and utensils because of its light weight and strength.
So here aluminium, copper and iron are used to make utensils, since they are good conductors of heat and electricity.
Thus the correct options are A, B and C.
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Read the chemical equation.
N2 + 3H2 - 2NH3
Using the volume ratio, determine how many liters of NH3 is produced if 4.2 liters of H2 reacts with an excess of N2
if all measurements are taken at the same temperature and pressure?
A 2.8 liters
B 3.2 liters
C 5.4 liters
D 6.3 liters
Answer:
A 2.8 liters
Explanation:
Step 1: Write the balanced equation
N₂ + 3 H₂ ⇄ 2 NH₃
Step 2: Establish the appropriate volume ratio
At the same temperature and pressure, the volume ratio of H₂ to NH₃ is 3:2.
Step 3: Calculate the volume of ammonia produced from 4.2 L of hydrogen
4.2 L H₂ × 2 L NH₃/3 L H₂ = 2.8 L
Write a net ionic equation for the overall reaction that occurs when aqueous solutions of carbonic acid and sodium hydroxide are combined. Assume excess base.
Answer:
[tex]H_2CO_3(aq)+2OH^-(aq)\rightarrow (CO_3)^{2-}(aq)+2H_2O(l)[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to set up this net ionic equation, by firstly setting up the complete molecular equation as follows:
[tex]H_2CO_3(aq)+2NaOH(aq)\rightarrow Na_2CO_3(aq)+2H_2O(l)[/tex]
Thus, since carbonic acid is weak it merely ionizes whereas sodium hydroxides ionizes for the 100 % as it is strong; thus, we can write the complete ionic equation:
[tex]H_2CO_3(aq)+2Na^+(aq)+2OH^-(aq)\rightarrow 2Na^+(aq)+(CO_3)^{2-}(aq)+2H_2O(l)[/tex]
Whereas sodium ions act as the spectator ones to be cancelled out for us to obtain:
[tex]H_2CO_3(aq)+2OH^-(aq)\rightarrow (CO_3)^{2-}(aq)+2H_2O(l)[/tex]
Regards!
Generally the vapor pressure of a liquid is related to: I. the amount of liquid II. atmospheric pressure III. temperature IV. intermolecular forces
Answer:
Atmospheric pressure, because a liquid is said to be boiling when the vapour pressure equals the atmospheric pressure.
Question In nickel-cadmium batteries: Select the correct answer below: the anodes are nickel-plated and the cathodes are cadmium-plated the anodes are cadmium-plated and the cathodes are nickel-plated both the anodes and cathodes are plated with a nickel-cadmium alloy none of the above
Answer:
the anodes are cadmium-plated and the cathodes are nickel-plated
Explanation:
Nickel cadmium battery works on the principle as by the other cell. There is anode and a cathode which is separated by a separator (spiral shaped inside the case). The anode is negative and is cadmium plated while the cathode is positive and is nickel plated. An electrolyte is also used.
So the correct answer is : "The anodes are cadmium-plated and the cathodes are nickel-plated."
A sample of oxygen occupies 1.00 L. If the temperature remains constant, and the pressure on the oxygen is decreased to one third the original pressure, what is the new volume
Answer:
3.00 L
Explanation:
P₁V₁ = P₂V₂
V₁ = 1.00 L
P₁ = (x) atm
P₂ = [tex]\frac{1}{3}[/tex] · (P₁) = [tex]\frac{x}{3}[/tex]
V₂ = unknown
(x atm)(1.00 L) = ( [tex]\frac{x}{3}[/tex] atm)(V₂)
divide both sides by ( [tex]\frac{x}{3}[/tex] atm)
( 1.00x )( [tex]\frac{3}{x}[/tex] ) = V₂
x cancels out
(1.00)(3) = V₂
V₂ = 3.00 L
Step 7: Measuring the Volume of Air Near 60°C
Answer:
57------6.9
330------.87
Explanation:
Answer:
The required volume of air is 3.64 L
Explanation:
Ideal gas equation:-
The relation between pressure, volume, and temperature of the gas is known as the ideal gas equation and it is given as,
PV=RT
Where R=gas constant
Now,
Let the volume of air is, V
According to the question we have
Temperature, T= 60°C= (60+273) K= 333 K
Atmospheric pressure, P= 760 mm
And gas constant, R= 8.314 J/mole K
Substitute values in ideal gas equation and we get
760V= 8.314(333)
V= 2768.562/760
V= 3.642 L
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For each of the following compounds, indicate the pH at which 50% of the compound will be in a form that possesses a charge and at which pH more than 99% of the compound will be in a form that possesses a charge.
ClCH2COOH (pKa = 2.86)
CH3CH2NH+3 (pKa = 10.7)
Express your answer using two decimal places
a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.
b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.
c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.
d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.
Answer:
a. 2..86 b. 4.86 c. 10.7 d. 8.7
Explanation:
a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.
Using the Henderson-Hasselbalch equation,
pH = pKa + log[A⁻]/[HA]
where [A⁻] = concentration of conjugate base (or charged form) and [HA] = concentration of acid.
At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1
So, pH = pKa + log[A⁻]/[HA]
pH = pKa + log1
pH = pKa = 2.86
b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.
Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.99x while the acidic concentration is remaining 1 % (1 - 0.99)x = 0.01x
Using the Henderson-Hasselbalch equation,
pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base (or charged form) = 0.99x and [HA] = concentration of acid = 0.01x.
pH = pKa + log0.99x/0.01x
pH = pKa + log0.99/0.01
pH = 2.86 + log99
pH = 2.86 + 1.996
pH = 4.856
pH ≅ 4.86
c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.
Using the Henderson-Hasselbalch equation,
pH = pKa + log[A⁻]/[HA]
where [A⁻] = concentration of conjugate base and [HA] = concentration of acid.
At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1
So, pH = pKa + log[A⁻]/[HA]
pH = pKa + log1
pH = pKa = 10.7
d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.
Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.01x while the acidic concentration is remaining 99 % (1 - 0.01)x = 0.99x (which possesses the charge).
Using the Henderson-Hasselbalch equation,
pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base = 0.01x and [HA] = concentration of acid = 0.99x.
pH = pKa + log0.01x/0.99x
pH = pKa + log1/99
pH = 10.7 - log99
pH = 10.7 - 1.996
pH = 8.704
pH ≅ 8.7
Write a balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution. Be sure to add physical state symbols where appropriate.
Answer:
O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e
Explanation:
An oxidation reaction reaction refers to a reaction in which electrons are lost. In this case, we are about to see the full balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution.
The full equation is;
O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e
So, two electrons were lost in the process.
Identify the phase of the copper product after each reaction in the copper cycle.
The addition of HNO3 HNOX3 to Cu ______________
The addition of H2SO4 HX2SOX4 to CuO ____________ The addition of Z n Zn to C u S O 4 CuSOX4 Choose... The addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 Choose... The heating of C u ( O H )
Answer:
addition of HNO3 HNOX3 to Cu - Aueous
addition of H2SO4 HX2SOX4 to CuO - Aqueous
addition of Z n Zn to C u S O 4 CuSOX4 - Solid
addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 - Solid
heating of C u ( O H ) - Solid
Explanation:
Copper when introduced with acids form an aqueous solution and fumes are released in air during the chemical reaction. When NaOH is added to copper then solid copper product is released. Copper dissolves on HNO but does not dissolves in HCL.
What is an emission spectrum?
A. The total amount of energy emitted by an element
B. The products created when an element is burned
C. The energy absorbed when an electron gains energy
D. The colors of light given off when an element loses energy
Answer:
D
Explanation:
The electromagnetic radiation is emitted due to a particle moves from a higher to a lower energy state
An emission spectrum is the colors of light given off when an element loses energy. Therefore, option D is correct.
What is emission spectrum ?The electromagnetic radiation spectrum produced when an electron changes from a high energy state to a lower energy state is known as the emission spectrum of a chemical element or chemical compound.
An emission spectrum is the range of radiations that are released in different places when electrons jump back and forth between higher and lower energy levels to achieve stability.
Since what you are seeing is the direct radiation produced by the source, this form of spectrum is also known as an emission spectrum. You can see all the colors in the Sun's spectrum because light from the Sun is produced at practically all energies in the visible spectrum.
Thus, option D is correct.
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What is "X" in the following reaction?
How can a Bose-Einstein condensate be formed? A. B super-heating a gas. B. By super-cooling certain types of solid. C. By super-cooling certain types of plasma. D. By super-heating a plasma
Answer:
C. By super-cooling certain types of plasma.
Explanation:
Bose-Einstein condensate is a state of matter whereby atoms or particles become cooled to a very low energy state leading to their condensation to give a single quantum state.
Note that plasma refers to atoms that have had some or even all of its electrons stripped away leaving only positively charged ions. Simply put, plasma is ionized matter.
When certain types of plasma are super cooled, Bose-Einstein condensate are formed.
