Answer:
[tex]L_{\text{fusion}[/tex] is latent heat of fusion.
Explanation:
Latent heat is basically termed as hidden heat. It is defined as the amount of heat required to change 1 g of a substance at the temperature of its melting point from the solid to the liquid state at constant temperature.The formula that relates heat and latent heat of fusion is given by :[tex]Q=mL_{\text{fusion}}[/tex]
Here,
m is mass of substance
[tex]L_{\text{fusion}[/tex] is latent heat of fusion.
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
.
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!
Which consists of only one type of atom?
Answer:
A chemical element
Explanation:
A chemical element consists of only one type of atom.
What biological sample do home pregnancy tests use to detect pregnancy?
Answer:
The biological sample that pregnancy tests use to detect pregnancy is urine
Answer:
yes
Explanation:
the urine in the the tester will find hormones to determine if pregnant or not
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.
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Primary succession is most likely caused by?
Answer:
volcanic eruption.
mark my answer as brainlest......
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:
A student weighs out a 6.64 g sample of , transfers it to a 500. mL volumetric flask, adds enough water to dissolve it and then adds water to the 500. mL tick mark. What is the molarity of cobalt(II) fluoride in the resulting solution
Answer:
the molarity of cobalt(II) fluoride in the resulting solution is = 0.137 M
Explanation:
Given that :
a student dissolves 6.64 g of CoF₂ into 500 mL of water
volume of the solution(water) = 500 mL = 0.50 L
The standard molar mass of CoF₂ is 96.93 g/mol
number of moles of CoF₂ = mass of CoF₂/molar mass of CoF₂
number of moles of CoF₂ = 6.64 g/96.93 g/mol
number of moles of CoF₂ = 0.0685 mol
The molarity of any given substance is known to be as the number of moles of solute dissolved in one litre of solution.
Thus ;
Molarity of cobalt(II) fluoride CoF₂ in the resulting solution is = number of moles / Volume in (L)
Molarity of cobalt(II) fluoride CoF₂ = 0.0685 mol/ 0.50 L
= 0.137 M
Thus ; the molarity of cobalt(II) fluoride in the resulting solution is = 0.137 M
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]
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
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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
Phloem contains specialized cells that move sugars and other nutrients both
up and down in plants. Which object is the best model of one of these cells?
A. Water balloon
Answer:Drinking straw
Explanation: my quiz said so
The object that represents the best model of one of the phloem cells is the drinking straw.
What is Phloem?Phloem may be defined as a type of vascular tissue in plants that significantly conducts the passage of sugars and other metabolic products from the leaves to the roots. This mediates the migration of sugar which is synthesized in the leaves during photosynthesis.
Apart from this, the phloem also regulates the movement of nutrients from the roots to the upper part of the body. This function of phloem cells shares homology with the object known as a drinking straw that mediates the transport of any fluid in both directions.
Therefore, a drinking straw is an object that represents the best model of one of the phloem cells.
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Your question seems incomplete. The most probable complete question is as follows:
Can openerWater balloonDrinking StrawKitchen strainerCalculate the payback time if she insulates the loft with 300 mm insulation when: area of loft space = 100m^2 cost of roll of 100 mm thick loft insulation that covers 8.3m^2 = £20 cost of putting the insulation into the loft = £120 savings per year = £80 per 100 mm Payback Time = installation cost divided by annual savings
Answer: Payback time = 0.0075
Explanation: Since payback time is calculated as:
payback time = [tex]\frac{installation cost}{annual savings}[/tex]
First determine the installation cost:
100 mm thick insulation covers 8.3 m². Then 300 mm covers 24.9 m².
To cover 8.3m² costs £20. Then, the cost to cover 24.9 m² is:
cost = [tex]\frac{20*24.9}{8.3}[/tex]
cost = £60
The cost of putting the insulation is £120, so the total cost is:
total cost = £60 + £120
total cost = £180
Savings per year per 100 mm thick is £80. For 300 mm, the value of annual savings is:
savings = [tex]\frac{300*80}{100}[/tex]
savings = 240
payback time = [tex]\frac{installation cost}{annual savings}[/tex]
payback time = [tex]\frac{180}{240}[/tex]
payback time = 0.75
Identify the correct statement regarding the strength of chemical bonds.
Weak bonds require more surface area to form than strong bonds.
A weak bond forms from a smaller change in energy compared to a strong bond.
Strong bonds occur with high temperature and weak bonds with low temperature.
Weak bonds require more heat to form than strong bonds.
Answer:
B.
Explanation:
A weak bond forms from a smaller change in energy compared to a strong bond.