Answer:
volcanic eruption.
mark my answer as brainlest......
How many moles of silver are equivalent to 2.408 x 10^24 atoms
The mole is used to measure small particles like atoms and molecules. The 4 moles of silver is equivalent to [tex]\bold{2.408 x 10^2^4 }[/tex] atoms.
Given here,
The number of atoms
[tex]\bold{2.408 x 10^2^4 }[/tex]
Number of moles = ?
1 mol of substance = [tex]\bold{ 6.02 x10^2^3}[/tex]
Hence,
moles of silver,
[tex]\bold {= \dfrac {2.408 x 10^2^4 } { 6.02 x10^2^3} }}\\\\\bold {= 4 mol}[/tex]
Therefore, the 4 moles of silver is equivalent to [tex]\bold{2.408 x 10^2^4 }[/tex] atoms.
To know more about mole concept, refer to the link:
https://brainly.com/question/20483253
En la electrólisis del Cloruro de plomo se han depositado 2.6 gramos del metal en 20 minutos el peso atómico del plomo es 206 calculo la intensidad de la corriente eléctrica empleada AYUDENNNNN PLISSSS es para hoy!!!!!
Answer:
2.01 A
Explanation:
Para esta pregunta debemos empezar por la semi-reaccion del plomo:
[tex]Pb^+^2~_(_a_q_)~+~2e^-~->~Pb_(_s_)[/tex]
Se intercambian dos electrones en la semi-reacción del plomo. Si tenemos en cuenta la ecuación:
[tex]n=\frac{I*t}{z*F}[/tex]
Donde:
n= Moles depositados
I= Intensidad de corriente (en Amperios)
z= Numero de electrones intercambiados
F= Constante de faraday = 96484 C/mol
t=tiempo (en segundos)
Que conocemos de esta ecuación?
Los moles (se pueden calcular a partir de la masa atómica del plomo, 207.2 g/mol)
[tex]2.6~g~Pb\frac{1~mol~Pb}{207.2~g~Pb}=0.0125~mol~Pb[/tex]
El tiempo (hay que convertirlo a segundos):
[tex]20~min\frac{60~s}{1~min}=1200~s[/tex]
Z (Numero de electrones)
De acuerdo a la semi-reacción son intercambiados 2 electrones.
Por lo tanto podemos resolver para "I":
[tex]I=\frac{n*z*F}{t}=\frac{0.0125~mol*2*96484\frac{C}{mol}}{1200~s}[/tex]
[tex]I~=~2.01~\frac{C}{s}=2.01~A[/tex]
Espero que sea de ayuda!
Consider an electrochemical cell based on the spontaneous reaction 2AgCl(s) + Zn(s) → 2Ag(s) + 2Cl– + Zn2+. If the zinc ion concentration is kept constant at 1 M, and the chlorine ion concentration is decreased from 1 M to 0.001 M, the cell voltage should:
Answer:
there is an increase by 0.18 V in the cell voltage.
Explanation:
The given equation of the reaction can be well written as
[tex]2AgCl_{(s)} + Zn _{(s)} \to 2Ag_{(s)} + 2 Cl^- _{(aq)}+ Zn^{2+}_{(aq)}[/tex]
By application of Nernst Equation ; we have the expression
[tex]E_{cell} = E^0- \dfrac{0,059}{n}log (\dfrac{[product]}{[reactant]})[/tex]
here in the above equation;
n = number of electrons transferred in the equation of the reaction
n = 2
Also;
[tex]E^0 = E_{cathode} - E_{anode}[/tex]
[tex]E^0 = E_{Ag^+/Ag} - E_{Zn^+/Zn}[/tex]
[tex]E^0 = +(0.80 \ V) - (-0..76 \ V)[/tex]
[tex]E^0 = (0.80 \ V +0..76 \ V)[/tex]
[tex]E^0 = 1.56 \ V[/tex]
If the zinc ion concentration is kept constant at 1 M; we have:
[tex]E_{cell} = E^0- \dfrac{0.059}{n}log (\dfrac{[product]}{[reactant]})[/tex]
[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log ({[Zn^{2+} ]}{[Cl^{2-}]})[/tex]
[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log (1)[/tex]
Since log(1) = 0
Therefore;
[tex]E_{cell} = 1.56\ V[/tex]
When the chlorine ion concentration is decreased from 1 M to 0.001 M; we have;
[tex]E_{cell} = E^0- \dfrac{0.059}{n}log (\dfrac{[product]}{[reactant]})[/tex]
[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log ({[Zn^{2+} ]}{[Cl^{2-}]})[/tex]
[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log ({[1*0.001^2}]})[/tex]
[tex]E_{cell} = 1.56 - 0.0295 \ * \ log ({[1*10^{-6}}]})[/tex]
[tex]E_{cell} = + 1.737 \ V[/tex]
The change in voltage = [tex]E_{cell} - E^0[/tex]
=( 1.737 - 1.56 )V
= 0.177 V
≅ 0.18 V
Thus; from the following observation; there is an increase by 0.18 V in the cell voltage.
The voltage of the cell increased by 0.18 V.
The equation of the reaction is; 2AgCl(s) + Zn(s) → 2Ag(s) + 2Cl– + Zn2+
We know that;
E°cell = 1.36 - (-0.76) = 2.12 V
If the cells are both at 1M concentration the Ecell = E°cell = 2.12 V
When the concentration of Cl- decreased from 1 M to 0.001 M
Ecell = E°cell - 0.0592/n log Q
Substituting values;
Ecell = 2.12 V - 0.0592/2 log (1 × (0.001)^2)
Ecell = 2.298 V
Increase in voltage = 2.298 V - 2.12 V = 0.18 V
Learn more: https://brainly.com/question/165414
Which consists of only one type of atom?
Answer:
A chemical element
Explanation:
A chemical element consists of only one type of atom.
How does the government control scientific research
Answer:
The government allocates a budget for research every year. The spending of that money is determined by government priorities. Some of the money is spent directly, in government-funded research centers.
Other money is distributed to other research institutions.
Money spent by other institutions for research has no government oversight.
Explanation:
how many atoms are contained in 2.70g of aluminum provided that 32g of sulphur equals 6.02 × 10^(23)atoms
Answer:
[tex]1.63 \times {10}^{24} [/tex]
one atom of an element = 6.02 \times {10}^{23} atom
The mass of one atom of sulphur = 32g
The mass of one atom of aluminium = 27g
so one atom of aluminium = 6.02 \times {10}^{23}
27g of AL = 6.02 \times {10}^{23} atom
2.70g of AL = X atoms
Then you cross multiply ........
and get the answer
.
A 75 lb (34 kg) boy falls out of a tree from a height of 10 ft (3 m). i. What is the kinetic energy of the boy when he hits the ground? Round your answer to the nearest joule. ii. What is the speed of the boy when he hits the ground? Round your answer to two significant figures. iii. Using the conversion factors of 1 m = 1.094 yd and 1 mi = 1760 yd, calculate the speed of the boy in miles per hour when he hits the ground.
Answer:
Kinetic energy of boy just before hitting the ground is [tex]\approx[/tex]1000 J.
Speed of boy just before hitting the ground is 7.67 m/s
or 17.16 mi/hr.
Explanation:
Given that:
Mass of boy = 75lb = 34 kg
Height, h = 10ft = 3m
To find:
Kinetic energy of boy when he hits the ground.
As per law of conservation of energy The potential energy gets converted to kinetic energy.
[tex]\therefore[/tex] Kinetic energy at the time boy hits the ground = Initial potential energy of the boy when he was at the Height 'h'
The formula for potential energy is given as:
[tex]PE = mgh[/tex]
Where m is the mass
g is the acceleration due to gravity, g = 9.8 [tex]m/s^2[/tex]
h is the height of object
Putting all the values:
PE = [tex]34 \times 9.8 \times 3 \approx 1000\ J[/tex]
Hence, Kinetic energy is [tex]\approx[/tex]1000 J.
Formula for Kinetic energy is:
[tex]KE = \dfrac{1}{2}mv^2[/tex]
where m is the mass and
v is the speed
Putting the values and finding v:
[tex]1000 = \dfrac{1}{2}\times 34 \times v^2\\\Rightarrow v^2 = 58.82\\\Rightarrow v = 7.67\ m/s[/tex]
Given that:
1 m = 1.094 yd and 1 mi = 1760 yd
[tex]\Rightarrow 1609\ m = 1\ mi[/tex]
Converting 7.67 m/s to miles/hour:
[tex]\dfrac{7.67 \times 3600}{1609}=17.16\ mi/h[/tex]
what is the molar mass of magnesium tartrate
Answer:
172.385 g/mol
Explanation:
Magnesium Tartrate is C4H4MgO6
C - 12.01 g/mol
H - 1.01 g/mol
Mg - 24.305 g/mol
O - 16.00 g/mol
12.01(4) + 1.01(4) + 24.305 + 16(6) = 172.385 g/mol
Answer:
172.38
Explanation:
[tex]C_4H_4MgO_6\\C=12.01\\H=1.01\\Mg=24.30\\O=16.00\\\\4(12.01)+4(1.01)+24.30+6(16.00)\\48.04+4.04+24.30+96\\=172.38[/tex]
C = 12.01
H=1.01
Mg=24.30
O =16.00
4(12.01)+4(1.01)+24.30+6(16.00)
48.04 +4.04+24.30+96
=172.38