Answer:
Manganese(II) phosphate | Mn3(PO4)2 - PubChem
Answer:
Magnese(ll) posphate M23 (p042) Molecular weight.
That is what the leters stand for!
IF THIS HELPED AND IF YOU DON'T MIND CAN YOU PLEASE MARK ME BRAINLIEST?Identify the person who made the correct statement.
Mike said petrified fossils are hard and heavy like rock.
Bobby said that petrified fossils have the same appearance as when they were alive.
Neither Mike nor Bobby is correct.
Mike is correct.
Bobby is correct.
Both Mike and Bobby are correct.
Answer: Both Mike and Bobby are correct.
Explanation:
Petrifcation can be defined as the process in which the organic material of the dead living being becomes fossil by the replacement of mineral deposition in the bony, hard material.
Thus although the body components gets decomposed wiped out due to this process. The body shape of the dead organism remains the same as that was in living.
Thus the statements made by Mike and Bobby both are correct. The fossils are hard and have the same appearance as when they were alive.
A balanced equation has
Answer:
A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge is the same for both the reactants and the products.In other words, the mass and the charge are balanced on both sides of the reaction.
Explanation:
Given the information you now know, what is the effect of hyperventilation on blood pH?pH? During hyperventilation, the rapid in the blood CO2CO2 concentration shifts the equilibrium to the which the concentration of H+,H+, thereby the blood pH.
Answer:
When hypercapnia processes occur, where the concentration of carbon dioxide gas increases in the blood, the protonization of the blood increases, this means that the H + ions increase in concentration, thus generating metabolic acidosis.
This metabolic acidosis is regulated by various systems, but the respiratory system collaborates by generating hyperventilation, to increase blood oxygen pressures, decrease CO2 emissions, and indirectly decrease acidity.
Explanation:
This method of regulating the body is crucial, since the proteins in our body will not be altered if they do not happen.
The enzymes, the red globules, and many more fundamental things for life ARE PROTEINS, that in front of acidic media these modify their structure by denaturing themselves and ceasing to fulfill their functions. This is the reason why it seeks to neutralize the blood pH when it comes to an increase in CO2.
Suppose you were preparing 1.0 L of a bleaching solution in a volumetric flask, and it calls for 0.21 mol of NaOCl. If all you had available was a jug of bleach that contained 0.78 M NaOCl, what volume of bleach would you need to add to the volumetric flask before you added enough water to reach the 1.0 L line
Answer: brainliesss plssssssss
0.256 L
Explanation:
We should use the following formula:
concentration (1) × volume (1) = concentration (2) × volume (2)
concentration (1) = 0.82 M NaOCl
volume (1) = ?
concentration (2) = 0.21 M NaOCl
volume (2) = 1 L
volume (1) = [concentration (2) × volume (2)] / concentration (1)
volume (1) = [0.21 / 1] / 0.82 = 0.256 L
Identify the state(s) of matter that each property describes.
Answer:solid,liquid,gas,plasma
Explanation:
This question seems incomplete. I believe the full question is as followed:
Identify the state(s) of matter that each property describes.
1.) takes the shape of its container:
O gas
O liquid
O solid
2.) fills all available space:
O gas
O liquid
O solid
3.) maintains its shape:
O gas
O liquid
O solid
4.) can be poured:
O gas
O liquid
O solid
5.) is compressible:
O gas
O liquid
O solid
6.) has a fixed volume:
O gas
O liquid
O solid
The answers to the 1st are gas and liquid.
The answer to the 2nd is gas.
The answer to the 3rd is solid.
The answer to the 4th is liquid.
The answer to the 5th is gas.
The answers to the 6th are liquid and solid.
8) What is the molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22011) in 35.5 mL of solution?
A) 3.52 M
B) 1.85 x 10-2M
C) 0.104 M
D) 0.0657 M
E) 1.85 M
Answer:
E) 1.85 M
Explanation:
M(C12H22O11) = 342.3 g/mol
22.5 g * 1mol/342.3 g = 0.0657 mol
35.5 mL = 0.0355 L
Molarity = mol solute/L solution = 0.0657 mol/0.0355L =1.85 mol/L = 1.85 M
The molarity of the aqueous solution is 1.85 M. The correct option is E) 1.85 M
From the question,
We are to determine the molarity (that is, concentration) of the given sucrose solution
First, we will determine the number of moles present in the given mass of sucrose
Mass of sucrose = 22.5 g
Using the formula
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
Molar mass of sucrose = 342.2965 g/mol
∴ Number of moles of sucrose present = [tex]\frac{22.5}{342.2965}[/tex]
Number of moles of sucrose present = 0.0657325 moles
Now, for the molarity (concentration) of the sucrose solution
From the formula
Number of moles = Concentration × Volume
Then,
[tex]Concentration = \frac{Number\ of\ moles}{Volume}[/tex]
From the question,
Volume = 35.5 mL = 0.0355 L
∴ [tex]Concentration = \frac{0.0657325}{0.0355}[/tex]
Concentration = 1.85 M
Hence, the molarity of the aqueous solution is 1.85 M. The correct option is E) 1.85 M
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Consider the following reaction. I– 2 H2O2 (l) 2 H2O (l) + O2 (g) A solution contains 15 mL 0.1 M KI, 15 mL of DI water and 5 mL of 3% H2O2. After the decomposition of H2O2 is complete, you titrate the solution with 0.1 M AgNO3. If the catalyst, I–, is not consumed in the reaction and is completely recovered, what volume of the 0.1 M AgNO3 is required to reach the end point?
Answer:
Explanation:
The given chemical reaction is:
[tex]2H_2O_{(l)} \to^{I^-}} 2H_2O_{(l)}+O_2_{(g)}[/tex]
From above equation [tex]I^-[/tex] serves as catalyst which is not consumed by the reaction and also it is completely recovered; as a result to that , the full volume of KI will definitely react with AgNO₃.
Given that :
the volume of potassium iodide [tex]V_{KI} = 15 \ ml[/tex]
the molarity of potassium [tex]M_{KI} = 0.1 \ M[/tex]
the volume of distilled water [tex]V_W = 15 \ mL[/tex]
The volume of 3% [tex]H_2O_2 \ \ V_{H_2O_2} = 5 \ mL[/tex]
Molarity of AgNO₃ [tex]M_{AgNO_3} = 0.1 \ M[/tex]
Let take an integral look with the reaction between KI and AgNO₃; we have
[tex]KI + AgNO_3 \to KNO_3 + AgI[/tex]
At the end point; the moles of KI will definitely be equal to the moles of AgNO₃
So;
[tex]M_{KI}V_{KI}= M_{AgNO_3}V_{AgNO_3} \\ \\ V_{AgNO_3} = \dfrac{M_{KI}V_{KI}}{M_{AgNO_3}} \\ \\ \\ V_{AgNO_3} = \dfrac{ 0.1*15}{0.1}[/tex]
[tex]V_{AgNO_3} = 15 \ ml[/tex]
Thus; the volume of 0.1 M AgNO₃ needed to reach the end point is 15 mL
Consider this reaction:
2Cl2O5 —> 2Cl2 + 5O2
At a certain temperature it obeys this rate law.
rate = (2.7.M^-1•s^-1) [Cl2O5]^2
Suppose a vessel contains Cl2O5 at a concentration of 0.600M. calculate how long it takes for the concentration of Cl2O5 to decrease by 94%. you may assume no other reaction is important. round your answer to two digits
Answer:
[tex]t=9.7s[/tex]
Explanation:
Hello,
In this case, we have a second order kinetics given the second power of the concentration of chlorine (V) oxide in the rate expression, thus, the integrated equation for the concentration decay is:
[tex]\frac{1}{[Cl_2O_5]}=kt+\frac{1}{[Cl_2O_5]_0}[/tex]
Thus, the final concentration for a 94% decrease is:
[tex][Cl_2O_5]=0.600M-0.600M*0.94=0.036M[/tex]
Therefore, we compute the time for such decrease:
[tex]kt=\frac{1}{[Cl_2O_5]}-\frac{1}{[Cl_2O_5]_0}=\frac{1}{0.036M}-\frac{1}{0.60M} =26.1M^{-1}[/tex]
[tex]t=\frac{26.1M^{-1}}{k}= \frac{26.1M^{-1}}{2.7M^{-1}*s^{-1}}\\\\t=9.7s[/tex]
Regards.
How many protons are in an ion with 36 electrons and a -1 charge
Answer:
Explanation:
There are 35 protons.
The number of electrons = 36 electrons gives a -1 charge.
Where did all the other minus charges go?
They must be balanced by 35 protons.
Determine the amount of heat (in kJ) associated with the production of 5.71 × 104 g of ammonia according to the following equation. N2(g) + 3H2(g) 2NH3ΔH°rxn = −92.6 kJ Assume that the reaction takes place under standard-state conditions at 25°C.
Answer:
[tex]Q=-3.11x10^5kJ[/tex]
Explanation:
Hello,
In this case, for the given reaction, we are given the standard enthalpy of reaction per mole of ammonia that is -92.6 kJ, it means, that forming one mole of ammonia will release 92.6 kJ of energy. In such a way, for the formation of 5.71x10⁴ g of ammonia, the following amount of heat will be released:
[tex]Q=5.71x10^4gNH_3*\frac{1molNH_3}{17gNH_3}*-92.6\frac{kJ}{molNH_3}\\ \\Q=-3.11x10^5kJ[/tex]
Best regards.
The amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ
We'll begin by calculating the number of mole in 5.71×10⁴ g of NH₃
Mass of NH₃ = 5.71×10⁴ g
Molar mass of NH₃ = 14 + (3×1) = 17 g/mol
Mole of NH₃ =?Mole = mass / molar mass
Mole of NH₃ = 5.71×10⁴ / 17
Mole of NH₃ = 3358.82 moles Finally, we shall determine the heat required to produce 3358.82 moles (i.e 5.71×10⁴ g) of NH₃. This can be obtained as follow:N₂(g) + 3H₂(g) —> 2NH₃(g) ΔH°rxn = −92.6 kJ
Since reaction took place at standard conditions, it means:
1 moles of NH₃ required −92.6 kJ
Therefore,
3358.82 moles of NH₃ will require = 3358.82 × –92.6 = –311026.732 KJ
Thus, the amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ
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Precision can be defined as the?
Answer:Precision can be defined as the. reproducibility of a measured value. Precision is how close the measured values are to each others. In contrast with accuracy, accuracy is the agreement between a measured value and an accepted value.
Explanation:
Indicate whether each of the following indicates that a physical or chemical change has taken place when a piece of magnesium metal is studied: (a) Can be cut into tiny pieces (b) Fizzling occurs when placed water (c) Light is emitted when burned (d) Turns to ash
Answer:
a) Can be cut into tiny pieces - Physical Change
b) Fizzling occurs when placed water -Chemical Change
c) Light is emitted when burned -Chemical Change
d) Turns to ash -Chemical Change
Explanation:
When researchers need to prepare many reactions for polymerase chain reaction (PCR) amplification, they often create a "master mix" solution. A master mix contains the reagents common to all the planned PCR amplifications, regardless of the target DNA. Making a master mix is a way to minimize the number of pipetting steps.Suppose a researcher needs to PCR amplify seven different genes of interest from different organisms. The researcher prepares a master mix and dispenses it to seven different PCR tubes, one for each gene of interest.Select the PCR components the researcher must add to each of the seven tubes of master mix to selectively amplify each gene of interest.dNTPsMg2+-Mg2+-based bufferprimersDNA polymeraseDNA template
Answer:
The master mix contains the following reagents: dNTPs, DNA Polymerase, PCR buffer and MgCl2.
Explanation:
The DNA templates are the gene fragments to amplify by PCR, thereby they have to be added separately in each tube. Moreover, the primer pairs are specific for each gene, thereby they have to be added separately in each tube.
Deoxynucleotide triphosphates (dNTPs) are the building blocks of the DNA molecules: dGTP, dATP, TTP, and dCTP.
The PCR buffer provides a suitable medium for the activity of the DNA polymerase, often it contains Tris-Hcl and KCl.
MgCl2 is a cofactor for the activity of the DNA Polymerase.
The DNA Polymerase is an enzyme that amplifies DNA by adding nucleotides to the 3' end.
When 25ml of sulphuric acid, was titrated with 0.0820 mol/L sodium hydroxide solution the end point was detected (with phenolphthalein) at 22.5ml . Calculate the concentration of sulphuric
acid in mol/L.
Answer:
the concentration of sulphuric acid is 14mol/l
consider an exceptionally weak acid, HA, with Ka= 1 x 10-20. you make 0.1M solution of the salt NA. what is the pH.
Answer:
[tex]pH=10.5[/tex]
Explanation:
Hello,
In this case, the dissociation of the given weak acid is:
[tex]HA\rightleftharpoons H^++A^-[/tex]
Therefore, the law of mass action for it turns out:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
That in terms of the change [tex]x[/tex] due to the reaction extent is:
[tex]1x10^{-23}=\frac{x*x}{0.1-x}[/tex]
Thus, by solving with the quadratic equation or solver, we obtain:
[tex]x=31.6x10^{-12}M[/tex]
Which clearly matches with the hydrogen concentration in the solution, therefore, the pH is:
[tex]pH=-log(-31.6x10^{-12})\\pH=10.5[/tex]
Regards.
List three ways the rate of solvation of sodium chloride in water may be
increased
Answer:
1) Increasing temperature
2) Stirring
3) Increasing surface area of salt by grinding it
Mass is:
measured in kilograms
measured using a scale
affected by gravity
all of the above
A pure sample of the R enantiomer of a compound has a specific rotation, [ α], of +20 °. A solution containing 0.2 g/mL of a mixture of enantiomers rotates plane polarized light by −2 ° in a 1 dm polarimeter. What is the enantiomeric excess (%ee) of the mixture?
Answer:
Explanation:
The specific rotation of the sample is -2 degrees/0.2 g/mL of mixture
This equals -10 degrees/g/mL of sample.
let the proportion of the R (+) enantiomer be x. The proportion of the S (-) enantiomer in the mixture will be given by (1-x).
specific rotation of the mixture = proportion of R enantiomer* its specific rotation + proportion of S enantiome * its specific rotation
i.e.
-10 = x *(+20) + (1-x)*(-20)
-10 = 20x-20 + 20x
-10+20 = 40x
+10 = 40 x
x=10/40 = 25%
Since the proportion of the other enantiomer is 1-x, it is 0.75 or 75%
So the mixture contains 25% R, 75% S, giving you an excess of 50%.
Answer:
10%
Explanation:
Enantiomeric excess is a way of describing how optically pure a mixture is by calculating the purity of the major enantiomer. It can range from 0%-100%. Enantiomeric excess ( ee ) can also be defined as the absolute difference between the mole fractions of two enantiomers.
Enantiomeric excess is also called optical purity. This is because chiral molecules cause the rotation of plane-polarized light and are said to be optically active. An enantiomerically pure sample has an enantiomeric excess of 100 percent
Enantiomeric excess = observed specific rotation/specific rotation of the pure enantiomer x 100
From the data given in the question;
observed specific rotation= -2°
specific rotation of the pure enantiomer = +20°
Therefore;
ee= 2/20 ×100
ee= 10%
iron oxide + oxygen equals to ?
Answer:
It's ferric oxide Fe2O3
Explanation:
I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me plz...
Question 8
1 pts
A closed flask contains a 0.25 moles of O2 which exerts a pressure of
0.50 atm. If 0.75 moles of CO, is added to the container what is the
total pressure in the flask?
Answer:
\large \boxed{\text{2.0 atm}}
Explanation:
We can use Dalton's Law of Partial Pressures:
Each gas in a mixture of gases exerts its pressure separately from the other gases.
0.25 mol of O₂ exerts 0.50 atm.
If you add 0.75 mol of CO, the total amount of gas is
0.25 mol + 0.75 mol = 1.00 mol
[tex]p_{\text{total}} = \text{1.00 mol} \times \dfrac{\text{0.50 atm}}{\text{0.25 mol}}= \textbf{2.0 atm}\\\\\text{The total pressure in the flask is $\large \boxed{\textbf{2.0 atm}}$}[/tex]
The pressure of the closed flask after the addition of 0.75 moles of CO has been 2 atm.
Partial pressure can be defined as the pressure exerted by each gas in a given solution.
The total moles of gas in the container by the addition of CO has been:
Total moles = moles of oxygen + moles of CO
Total moles = 0.25 + 0.75
Total moles = 1 mol.
By using Dalton's law of partial pressure:
Total pressure = total moles [tex]\rm \times\;\dfrac{pressure\;of\;oxygen}{moles\;of\;oxygen}[/tex]
Total pressure = 1 [tex]\rm \times\;\dfrac{0.50}{0.25}[/tex]
Total pressure = 2 atm.
The pressure of the closed flask after the addition of 0.75 moles of CO has been 2 atm.
For more information about partial pressure, refer to the link:
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The type of nuclear decay an unstable nucleus will undergo depends on its ratio of neutrons to protons. The radioisotope cobalt-65 has a ratio of neutrons to protons of 1.41, which is too high for a nucleus of this size. What nuclear changes could reduce this ratio
Answer:
Explanation:
In cobalt - 65 ,
no of protons is 27 ( p )
no of neutron = 65 - 27 ( n )
= 38
n / p ratio
= 38 / 27
= 1.41
If case of emission of alpha particle
no of proton p = 27 - 2 = 25
no of neutrons = 38 - 2 = 36
n / p ratio = 36 / 25
= 1.44
So it increases
In case of emission of beta particle
No of neutron n = 38 - 1 = 37
No of proton = 27 + 1 = 28
n / p ratio = 37 / 28
= 1.32
Hence ratio decreases.
Hence beta ray decay will result in decrease in n / p ratio.
. The pI is called ________________. The rule of calculating pI of an amino acid is that first, write the dissociation equation from fully protonated form to fully deprotonated form, label the charge of each form; second, identify the zwitterionic form (zero charge) and find the closest pKs (left and right side in the dissociation equation); third, average these two pKs. Write the dissociation equations for amino acids, glutamate, histidine, and calculate their pIs.
Answer:
The isoelectric point is that the pH at which the compound is in an electronically neutral form.
For diss equations, please find them in the enclosed file.
The pIs of 2 amino acids:
Glutamate: pI = 3,2Histidine: pI = 7,6Explanation:
Formula for the pI calculation: pI = (pKa1 + pKa2)/2
Given 3 pKa :
Acid glutamic with an acid sidechain:Use the lower 2 pKas (corresponding with 2 -COOH groups)
pKa1 = 2,19; pKa2 = 4,25; so pI = 3,2
Histidine with 2 amino groups:Use the higher 2 pKas ( -COOH group and -NH= group)
pKa1 = 6; pKa2 = 9,17; so pI = 7,6
Suppose the amount of a certain radioactive substance in a sample decays from to over a period of days. Calculate the half life of the substance. Round your answer to significant digit.
The given question is incomplete, the complete question is:
Suppose the amount of a certain radioactive substance in a sample decays from 1.30 mg to 100. ug over a period of 29.5 days. Calculate the half life of the substance Round your answer to 2 significant digits.
Answer:
The correct answer is 7.974 days.
Explanation:
Based on the given question, the concentration of a radioactive substance present in a sample get decays to 100 micro grams from 1.30 milligrams in 29.5 days. There is a need to find the half-life of the substance.
Radioactive decay is an illustration of first order reaction.
K = (2.303 / t) log [a/(a-x)]
Here a is 1.30 mg and (a-x) is 100 micrograms = 100 * 10^-3 mg or 0.1 mg, and t is 29.5 days. Now putting the values we get,
K = (2.303 /29.5)log (1.30/0.1)
= 2.303/29.5 log13
= 2.303/29.5 * 1.1139
K = 0.0869
The half-life or t1/2 is calculated by using the formula, 0.693 / K
= 0.693 / 0.0869
= 7.974 days.
Phosphofructokinase is a four‑subunit protein with four active sites. Phosphofructokinase catalyzes step 3 of glycolysis, converting fructose‑6‑phosphate to fructose‑1,6‑bisphosphate. Phosphoenolpyruvate (PEP) is the product of step 9 of glycolysis. The PEP concentration in the cell affects phosphofructokinase activity.Select the true statements about PEP regulation of phosphofructokinase.
1. PEP is a feedback inhibitor of phosphofructokinase.
2. The apparent affinity of phosphofructokinase for its substrate increases when PEP binds.
3. PEP is a positive effector of phosphofructokinase.
4. PEP inhibition of phosphofructokinase yields a sigmoidal velocity versus substrate curve.
5. PEP competes with fructose-6-phosphate for the active site of phosphofructokinase.
6. The binding of PEP to one phosphofructokinase subunit causes a conformation change that affects the ability of the substrate to bind to the other subunits.
Answer:
1. PEP is a feedback inhibitor of phosphofructokinase.
4. PEP inhibition of phosphofructokinase yields a sigmoidal velocity versus substrate curve.
6. The binding of PEP to one phosphofructokinase subunit causes a conformation change that affects the ability of the substrate to bind to the other subunits.
Explanation:
Phosphofructokinase-1, PFK-1, is an allosteric enzymes composed of four protein subunits.
Allosteric enzymes are enzymes that function through non-covalent binding of allosteric modulators which may be activators or inhibitors. They produce a characteristic velocity versus substrate sigmoidal curve. PFK-1 has a separate binding site for its substrate, fructose-6-phosphate and it's allosteric modulators: ATP, ADP or phosphoenolpyruvate, PEP.
The enzyme can exist in two conformations, the T-state (tense) or the R-state (resting). Binding of substrate causes a conformational change from T-state to R-state, whereas binding of allosteric inhibitors returns it to the T-state.
PEP, the product of step 9 in glycolysis, is an allosteric inhibitor of PFK-1. When it binds to the the allosteric site, it leads to conformational changes in PFK-1 from the R-state to the T-state which reduces the enzymes ability to bind the substrate. These changes are responsible for the sigmoidal velocity/substrate curve in allosteric enzymes.
Therefore, the true statements from the options above are 1, 4, 6.
Options 2,3 and 5 are wrong because PEP is a negative effector of PFK-1, thus its binding reduces the affinity of PFK-1 for its substrate. Also, PFK-1 being an allosteric enzyme has separate binding sites for its substrate and its modulators. Thus, there is no competition for active site binding by substrate and modulators.
Fill in the blank: If an atom is in column V (or 15), it will most likely ____________ to fulfill the octet rule.
Gain 3 electrons
Lose 5 electrons
Gain 5 electrons
Lose 5 protons
Gain 3 protons
Answer:
If an atom is in column V (or 15), it will most likely gain 3 electrons to fulfill the octet rule.
Explanation:
The octet rule defines the property that atoms have to complete their last energy level with eight electrons to achieve stability through an ionic, covalent or metallic bond.
The pair of electrons that are transferred or gained belong to the last shell of the atom. If an atom is in column V (or 15), it means that it has 5 electrons in its last shell. So an atom in this group is more likely to gain 3 electrons to achieve stability than to lose the 5 electrons it has.
If an atom is in column V (or 15), it will most likely gain 3 electrons to fulfill the octet rule.
A. Gain 3 electrons
What does Octet rule say?The octet rule defines the property that atoms have to complete their last energy level with eight electrons to achieve stability through an ionic, covalent or metallic bond.
The pair of electrons that are transferred or gained belong to the last shell of the atom. If an atom is in column V (or 15), it means that it has 5 electrons in its last shell. So an atom in this group is more likely to gain 3 electrons to achieve stability than to lose the 5 electrons it has.
Thus, correct option is A.
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Now construct a different electrochemical cell. You put a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver cathode in contact with a 0.0042 M solution of silver(I) nitrate. What is the value of the electric potential at the moment the reaction begins
Answer:
[tex]1.66~V[/tex]
Explanation:
We have to start with the half-reactions for both ions:
[tex]Zn^+^2~+2e^-~->Zn[/tex] V= -0.76
[tex]Ag^+~e^-~->~Ag[/tex] V= +0.80
If we want a spontaneous reaction (galvanic cell) we have to flip the first reaction, so:
[tex]Zn~->~Zn^+^2~+2e^-~[/tex] V= +0.76
[tex]Ag^+~+~e^-~->~Ag[/tex] V= +0.80
If we want to calculate ºE we have to add the two values, so:
ºE=0.76+0.80 = 1.56 V
Now, we have different concentrations. So, if we want to calculate E we have to use the nerts equation:
[tex]E=ºE~+~\frac{0.059}{n}LogQ[/tex]
On this case, Q is equal to:
[tex]Q=\frac{[Zn^+^2]}{[Ag^+]^2}[/tex]
Because the total reaction is:
[tex]Zn~+~2Ag^+~->~Zn^+^2~+~2Ag[/tex]
So, the value of "Q" is:
[tex]Q=\frac{[0.052 M]}{[0.0042]^2}=2947.84[/tex]
Now, we can plug all the values in the equation (n=2, because the amount of electrons transferred is 2). So:
[tex]E=1.56~V~+~\frac{0.059}{2}Log(2947.84)=1.66~V[/tex]
I hope it helps!
What is the molar mass of P2O5?
Answer:
142 grams
Explanation:
To find the molar mass of a molecule or compound, you simply need to add together the molar masses of all of the atoms that comprise it. Phosphorus has a molar mass of about 31, while oxygen has one of about 16, meaning that the molar mass of this molecule is:
2(31)+5(16)=62+80=142
Hope this helps!
Photochromic lenses contain Group of answer choices both AgCl and CuCl embedded in the glass. only AgCl embedded in the glass. neither AgCl nor CuCl embedded in the glass. only CuCl embedded in the glass.
Answer:
both AgCl and CuCl embedded in the glass
Explanation:
Photochromic lenses contain both AgCl and CuCl embedded in the glass.
They are light-sensitive lenses that adapt to environmental changes. They appear clear when in an apartment or a building and automatically darken when outside as a result of exposure to sunlight. The darkening is activated by the UV component of the sunlight.
Photochromic lenses are otherwise known as light-adaptive or intelligent lenses and they are formed by coating lenses with silver chloride compounds whose concentration ranges from 0.01 to 0.001 %. Copper (I) chloride is also included in addition to the silver halide.
In summary, photochromic lenses contain both AgCl and CuCl.
What would form a solution?
O A. Mixing two insoluble substances
O B. Mixing a solute and a solvent
O C. Mixing a solute and a precipitate
O D. Mixing two solutes together
Answer:
B. Mixing a solute and a solvent
Explanation:
Hello,
In this case, solutions are defined as liquid homogeneous mixtures formed when two substances having affinity are mixed. It is important to notice that the two substances are known as solute, which is added to other substance that is the solvent. Therefore, answer is B. Mixing a solute and a solvent.
Notice that when two insoluble substances are mixed no solution is formed. Furthermore, if two solutes together or a solute and a precipitate are mixed, no liquid homogeneous solution is formed, as commonly solutes are solid, nevertheless, when liquid, one should have to act as the solvent.
Best regards.
Answer:
B. Mixing a solute and a solvent
Explanation:
ap3x
Ni
Express your answer in condensed form in the order of orbital filling as a string without blank space between orbitals. For example, [He]2s22p2 should be entered as [He]2s^22p^2.
Answer:
[Ar]3d^84s^2
Explanation:
From the question given, we are asked to write the condensed form of electronic configuration of nickel, Ni.
To do this, we simply write the symbol of the noble gas element before Ni in a squared bracket followed by the remaining electrons to make up the atomic number of Ni.
This is illustrated below:
The atomic number of Ni is 28.
The noble gas before Ni is Argon, Ar.
Therefore, the condensed electronic configuration of Ni is written as:
Ni(28) => [Ar]3d^84s^2
Answer:
[Ar] 4s^23d^8
Explanation:
