Answer:
o diferença chave entre a radioatividade e a transmutação é que a radioatividade se refere à transmutação natural, enquanto a transmutação se refere à mudança de um elemento químico em outro por meios naturais ou artciais.Tanto a radioatividade quanto a transmutação são processos químicos que envolvem a mudança de núcleos atômicos para formar um novo elemento químico a partir de um elemento químico existente. A radioatividade é um tipo de processo de transmutação
Explanation:
mais informação em; ahttps://pt.strephonsays.com/radioactivity-and-transmutation-9500
5. Vilayanur Ramachandran belleves that in order to understand how our brains work, we need to look at brains that don't work as they should in order to see the differences
and effects that the disconnect of neurons can have on our memories, thoughts, and understanding.
True
O False
Answer:
true
Explanation:
What is the main function of burette?
Answer:
I think it A
Explanation:
Answer:
its d i think
Explanation:
pH de un jugo de naranja con [H+]=5.6×10-4M.
Explanation:
[tex]pH = - log[H {}^{ + } ] \\ = - log(5.6 \times {10}^{ - 4} ) \\ = 3.25[/tex]
The half-life of argon-39 is 269 years. It decays into krypton-39. After 1,076 years, what fraction of the original amount of argon-39 in a sample will still be argon
1/16
1/4
1/2
1/8
Answer:
1/16
Explanation:
From the question given above, the following data were obtained:
Half-life (t½) = 269 years
Time (t) = 1076 years
Fraction remaining =?
Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 269 years
Time (t) = 1076 years
Number of half-lives (n) =?
n = t / t½
n = 1076 / 269
n = 4
Thus, 4 half-lives has elapsed.
Finally, we shall determine the fraction of the original amount remaining. This can be obtained as follow:
Let N₀ be the original amount.
Let N be the amount remaining.
Number of half-lives (n) = 4
Fraction remaining (N/N₀ ) =?
N = 1/2ⁿ × N₀
N = 1/2⁴ × N₀
N = 1/16 × N₀
Divide both side by N₀
N/N₀ = 1/16
Thus, the fraction of the original amount remaining is 1/16
Please help:
An object rolled 15m/s for 3 seconds. How far did it roll?
3. Determine the enthalpy of formation for propane. 3C(s, gr) + 4H2(g) ---> C3H2(g) CzH3(g) AH = -2219.9 kJ C(s, gr) AH = -393.5 kJ H,(g) AH = -285.8 kJ a. -205.7 /mol b. -103.8 kJ/mol Å C. + 205.7 /mol d. + 103.8 /mol
Answer:
b I hope this is right good luck
Which of the following behaviors might indicate a patient is drug seeking?
A. A patient wants to avoid a specific medication because of
potential side effects.
B. A patient explains that she is from out of town and needs a
specific medicine because she left her prescription at home.
C. A patient fears her new prescription will conflict with another
medication she's currently taking.
D. A patient thinks she needs a smaller dose of her prescription
because it gives her headaches.
SUBMIT
Answer:
B is correct :)
Explanation:
Trust me I just took the test
water and Air are needed for iron rust answer true or false
Answer:True
Explanation:
PLZ GIVE ME BRAINLIEST, I NEED IT
the number of particles in a mole if substance
A.avografo number
b.molar volume
c.molar mass
d.molar ratio
if a lake has excess levels of phosphates and nirates in its water what will most likely result ?
Answer:
Eutrophication
Explanation:
Eutrophication results when there is excess levels of phosphates and nitrates in its water of lake.
When there is high level of nutrition level in the water body of lake, then the productivity of water plants increases thereby leading to algal boom and hence Eutrophication.
A gas in a sealed container has a pressure of 125 atm at a temperature of 303 K. If the pressure in the container is increased to 200 atm, what is the new temperature if the volume remains constant?
(Show work pls :)!)
Answer:
[tex]T_2=484.8K[/tex]
Explanation:
Hello there!
In this case, according to the the variable temperature and pressure and constant volume, it turns out possible for us to calculate the new temperature via the Gay-Lussac's law as shown below:
[tex]\frac{T_2}{P_2} =\frac{T_1}{P_1}[/tex]
Thus, by solving for the final temperature, T2, we obtain:
[tex]T_2 =\frac{T_1P_2}{P_1}[/tex]
So we plug in the given data to obtain:
[tex]T_2 =\frac{303K*200atm}{125atm}\\\\T_2=484.8K[/tex]
Best regards!
does anyone know thisss???
Which phase change is exothermic?
Answer:
B. Solidification
Hope this helped! :)
what is the difference between an acid and a base?
Answer:

A substance, mostly liquid that donates a proton or accepts an electron pair in reactions. An acid increases the concentration of H+ ions. A base is a substance that releases hydroxide (OH-) ions in aqueous solution, donates electrons and accepts protons.
[tex]{\underline{\underline{\bf{\red{\:\:Acid\:\:}}}}}[/tex]
Proton donor and electron pair acceptorIt can increase the concentration of H+ ion in aq. solution.pH value is <7Turn blue litmus paper into redAcid reacts with bases to form saltsAcid taste sour[tex]{\underline{\underline{\bf{\orange{\:\:Base\:\:}}}}}[/tex]
Proton acceptor and electron pair donorIt can increase the concentration of OH- ion in aq. solution.pH value is >7Turn red litmus paper into blueBases reacts with acids to form saltsBases taste soapyHow many grams of hydroxide pellets, NaOh are required to prepare 50.0ml of a 0.150 M solution ?
A.0.300g
B.3.00g
C.2.00g
D.200.g
E. 0.025 M
Which property do all metals have?
А They are hard.
B They conduct electricity.
С They form acidic oxides.
D They react with water.
A 31.25 mL aliquot of weak base that has a concentration of 0.683 M will be titrated with 0.434 M HCl. Calculate the pH of of the solution upon neutralization of half of the weak base. The Kb of the base is 1.5×10-6.
Answer:
pH = 8.18
Explanation:
The weak base, X, reacts with HCl as follows:
X + HCl → HX⁺ + Cl⁻
Where 1 mole of X with 1 mole of HCl produce 1 mole of HX⁺ (The conjugate acid of the weak base).
Now, using H-H equation for bases:
pOH = pKb + log [XH⁺] / [X]
Where pOH is the pOH of the buffer (pH = 14 -pOH)
pKb is -log Kb = 5.824
And [X] [HX⁺] are the molar concentrations of each specie
Now, at the neutralization of the half of HX⁺, the other half is as X, that means:
[X] = [HX⁺]
And:
pOH = pKb + log [HX⁺] / [X]
pOH = 5.824 + log 1
pOH = 5.824
pH = 14-pOH
pH = 8.18Gelatinization happens with what type of Carbohydrate?
What is the balanced net ionic reaction for the following -
CaCl2 + Na2S --> NaCl + CaS?
Answer:
CaCl2 + Na2S --> 2NaCl + CaS?
Explanation:
CaCl2 + Na2S --> 2NaCl + CaS?
A 108.9 g sample of water absorbs 114.6 calories of heat. The specific heat capacity of water is 1 cal/(g·°C). By how much did the temperature of this sample change, in degrees Celsius?
Answer:
The temperature of this sample changes by 1.052 degrees Celsius
Explanation:
As we know
[tex]Q = mc\Delta T[/tex]
Where m is the mass of the substance
c is the specific heat of the substance
and [tex]\Delta T[/tex] is the change in temperature
Substituting the given values in above equation, we get -
[tex]114.6 = 108.9 * 1 * \Delta T\\\Delta T = 1.052[/tex]degree Celsius
The temperature of this sample changes by 1.052 degrees Celsius
What happens when particles collide in sound waves?
Answer:
When sound is created, the air particles vibrate and collide with each other, causing the vibrations to pass between air particles. The vibrating particles pass the sound through to a person's ear and vibrate the ear drum. Light travels much faster than sound through air.
Explanation:
Imine formation from an aldehyde and an amine proceeds reversibly under slightly acidic conditions. The reaction is reversible due to acid-catalyzed hydrolysis of the imine. Imine formation generates an equivalent of water. Given these facts, explain why the imine can be isolated from the reaction mixture.
Answer:
Imine can be isolated from the reaction mixture as water is continuously removed from the reaction chamber
Explanation:
In this reaction, a non -aqueous solvent is not used (not mentioned in the question). Thus, we can say that there is continuous removal water under suitable reacting conditions and hence the imine formed is left behind.
- Calculate the Standard Enthalpy of the reaction below:
NH3(g) + HCI (g) → NH4Cl(s)
Using the following Enthalpy of Reactions:
2HCI(g) → H2(g) + Cl2(g)
AH = +184.6 KJ
2H2(g) + 1/2 N2(g) + 1/2 Cl2(g) → NH4Cl(s) deltaH = -314.4 kJ
N2(g) + 3 H2(g) → 2 NH3(g)
deltaH = +184.6 kJ
Answer:
Explanation:
We have the three equations:
[tex]NH_{3(g)} + HCl_{(g)} => NH_4Cl_{(s)} ..... \Delta H = ? (1)\\2HCl_{(g)} => H_{2(g)} + Cl_{2(g)} .... \Delta H = +184.6 kJ (2)\\2H_{2(g)} + 1/2N_{2(g)} + 1/2Cl_{2(g)} => NH_4Cl_{(s)} ..... \Delta H = -314.2 kJ (3)\\ N_{2(g)} + 3H_{2(g)} => 2NH_{3(g)} .... \Delta H = +184.6kJ (4)[/tex]
(can you double check that it is 184.6kJ for both equations 2 and 4 because it seems unlikely). We need to solve for equation 1 by addition and changing equations 2, 3 and 4. After possibly some trial and error, we can find that if we flip equations 4, multiply equation 3 by 2, add the equations together, and then finally divide by 2, we can get equation 1. We will get the answer of -314.2 kJ. However, I am again skeptical about the delta H values for equation 2 and 4 so double check that. This method might be super confusing and it is really hard to explain. So what I would suggest you to watch videos on Hess' law.
An isolated atom of a certain element emits light of wavelength 655 nm when the atom falls from its fifth excited state into its second excited state. The atom emits a photon of wavelength 435 nm when it drops from its sixth excited state into its second excited state. Find the wavelength of the light radiated when the atom makes a transition from its sixth to its fifth excited state.
Answer:
[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]
Explanation:
From the question we are told that
Light wavelength [tex]\lambda_l=655nm[/tex]
Light wavelength atom fall [tex]x_L=5th-2nd[/tex]
Photon wavelength [tex]\lambda_p=435nm[/tex]
Photon wavelength atom fall [tex]x_P=^th-2nd[/tex]
Generally the equation for the reciprocal of wavelength of emitted photon is is mathematically given by
[tex]\frac{1}{\lambda}=R(\frac{1}{ \lambda _f^2}-\frac{1}{\lambda _i^2} )[/tex]
Therefore for initial drop of 5th to 2nd
[tex]\frac{1}{\lambda_{5-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )[/tex]
Therefore for initial drop of 6th to 2nd
[tex]\frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]
Generally we subtract (5th to 2nd) from (6th to 2nd)
[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )-\frac{1}R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]
[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{5^2}-\frac{1}{6^2} )[/tex]
[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=\frac{1}{\lambda_{5-6}}[/tex]
[tex]\frac{1}{\lambda_{5-6}}=\frac{1}{4350nm}-\frac{1}{655nm}[/tex]
[tex]\frac{1}{\lambda_{5-6}}=1.33*10^{-3}[/tex]
Therefore for 6th to 5th stage is mathematically given by
[tex]\frac{1}{\lambda_{6-5}}=(1.33*10^{-3})^{-1}[/tex]
[tex]\frac{1}{\lambda_{6-5}}=751.879nm[/tex]
[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]
The reduction of iron(III) oxide () to pure iron during the first step of steelmaking, ()()() is driven by the high-temperature combustion of coke, a purified form of coal: ()()() Suppose at the temperature of a blast furnace the Gibbs free energies of formation of and are and , respectively. Calculate the maximum mass of pure iron that can be produced by the combustion of of coke. Round your answer to significant digits.
Answer:
hello your question is incomplete attached below is the complete question
answer : To 2 significant digits = 5500 kg
Explanation:
Given
C(s) + O2 (g) ----------> CO2 ( g )
mass of CO2 = 8.9 * 10^6 g
number of moles = 8.9 * 10^6 / 12 = 7.416 * 10 ^5 mol
same amount of moles is needed by O2 hence
attached below is the detailed solution
The initial volume of a gas is 21 L and the final volume is 38.7 L. If the initial temperature was 261 K, what is the final temperature (in Kelvin)?
Answer:
480.9K
Explanation:
V1=21L
V2=38.7L
T1=261K
T2=?
Using Charles law,
V1/T1=V2/T2
Inputting the given values,
21/261=38.7/T2
T2=480.9K
What is the empirical formula for C6H12?
Answer:
The empirical formula of C6 H12,
If I have 7.0 x 10^24 formula units of magnesium chloride, how many grams of chlorine would I have?
Answer:
824 g
Explanation:
First we convert 7.0 x 10²⁴ formula units of magnesium chloride (MgCl₂) into moles, using Avogadro's number:
7.0 x 10²⁴ formula units ÷ 6.023x10²³ formula units/mol = 11.62 mol MgCl₂There are two Cl moles per MgCl₂ moles, meaning that we have (2 * 11.62) 23.24 moles of Cl
Finally we convert 23.24 moles of chlorine into grams, using chlorine's molar mass:
23.24 mol * 35.45 g/mol = 824 gA 15.0 mL solution of Sr(OH)₂ is neutralized with 21.1 mL of 0.350 M HCl. What is the concentration of the original Sr(OH)₂ solution?
Answer:
0.246 M
Explanation:
The reaction that takes place is:
Sr(OH)₂ + 2HCl → SrCl₂ + 2H₂OFirst we calculate how many HCl moles reacted, using the given volume and concentration:
0.350 M * 21.1 mL = 7.385 mmol HClThen we convert HCl moles into Sr(OH)₂ moles, using the stoichiometric coefficients of the balanced reaction:
7.385 mmol HCl * [tex]\frac{1mmolSr(OH)_2}{2mmolHCl}[/tex] = 3.692 mmol Sr(OH)₂Finally we calculate the concentration of the original Sr(OH)₂ solution:
3.692 mmol Sr(OH)₂ / 15.0 mL = 0.246 MIdentify the term that matches each electrochemistry definition.
a. The electrode where reduction occurs ___________
b. An electrochemical cell powered by a spontaneous redox reaction ___________
c. The electrode where oxidation occurs__________
d. An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction ____________
e. A chemical equation showing either oxidation or reduction ___________
Answer: a. Cathode
b. Galvanic cell
c. Anode
d. Electrolytic cell
e. half reaction
Explanation:
Galvanic cell or Electrochemical cell is defined as a device which is used for the conversion of the chemical energy produced in a spontaneous redox reaction into the electrical energy.
Electrolytic cell is a device where electrical energy is used to drive a non spontaneous chemical reaction.
In the electrochemical cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode. Thus the electrons are produced at anode and travel towards cathode.
The balanced two-half reactions will be:
Oxidation half reaction : [tex]M\rightarrow M^{n+}+ne^-[/tex]
Reduction half reaction : [tex]N^{n+}+ne^-\rightarrow N[/tex]
Thus the overall reaction will be: [tex]M+N^{n+}\rghtarrow M^{n+}+N[/tex]