Answer:
P4O10
Explanation:
(P2O5)n=283.89
62+80=283.89
142n=283.89
n=2
P2O10
The empirical and molecular formula are related by the formula, molecular formula=2×empirical formula ,meaning double the atoms present in empirical formula,hence molecular formula is P₄O₁₀.
What is empirical formula?Empirical formula of a compound is defined as the simplest whole number ratio of atoms which are present in a compound.It does not make any mention of the arrangement of atoms or the number of atoms. The empirical formula gives information about the ratio of number of atoms which are present in a compound.
Molecular formula is determined from the empirical formula by dividing the molar mass of a compound by the empirical formula mass. The resultant which should be a whole number or very close to the whole number , then the subscripts are multiplied by the whole number to get the molecular formula.
For the compound P₂O₅ the subscripts of phosphorous and oxygen are multiplied by 2 and hence the molecular formula P₄O₁₀.
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What is a similarity between a fish embryo and a human embryo in the late stages of development?
They both have tails
They both have gills.
They both have spines.
They both have arms.
Answer:In the LATER stages of development they would have a spine
Explanation:
is chemical energy stored in coal, gas, and oil?
- Explain why the term greenhouse effect is used to describe the theory of global
warming.
Does the greenhouse effect affect life on Earth? If yes, explain how?
What are the possible effects of a buildup of greenhouse gases in our atmosphere?
Question 17 of 20
The reactants of a chemical equation have 1 Satom and 40 atoms. Which
set of atoms must also be found in the equation's products so that the
equation models the law of conservation of mass?
A. 4 S and 40
B. 1 S and 40
C. 1 S and 10
O D. 4 S and 10
Answer:
B) 1S and 40
hope it helps
Answer:
b is ur answer
Explanation:
1S and 4O
Solution remained colorless.
During the experimentation, the test tube was gently heated in a Bunsen burner flame for 60 seconds. What was the reason for this specific
procedure?
A)
Heating was done to initiate the combustion of the metal in water.
B)
Heating was done to confirm that no chemical reaction would take
place in acid
0
Heating was done to precipitate the chemical change in each test
tube containing water
D)
Heating helped released the hydrogen contained in water
molecules so students would have a positive H+ test.
g An oxidized silicon (111) wafer has an initial field oxide thickness of d0. Wet oxidation at 950 °C is then used to grow a thin film gate of 500 nm in 50 minutes. What is the original field oxide thickness d0 (in nm)?
Answer:
Explanation:
From the information given:
oxidation of oxidized solution takes place at 950° C
For wet oxidation:
The linear and parabolic coefficient can be computed as:
[tex]\dfrac{B}{B/A} = D_o \ exp \Big [\dfrac{-\varepsilon a}{k_BT} \Big][/tex]
Using [tex]D_o[/tex] and [tex]E_a[/tex] values obtained from the graph:
Thus;
[tex]\dfrac{B}{A} = 1.63 \times 10^8 exp \Big [ \dfrac{-2.05}{8.617 \times 10^{_-5}\times 1173}\Big] \\ \\ = 0.2535 \ \ \mu m/hr[/tex]
[tex]B= 386 \ exp \Big [-\dfrac{0.78}{8.617 \times 10^{-3} \times 1173} \Big] \\ \\ = 0.1719 \ \mu m^2/hr[/tex]
So, the initial time required to grow oxidation is expressed as:
[tex]t_{ox} = \dfrac{x}{B/A}+ \dfrac{x^2}{B} - t_o (initial)[/tex]
[tex]where; \\ \\ t_{ox} = 2 \ hrs;\\ \\ x = 0.5 \\ \\ B/A = 0.2535 \\ \\ B = 0.1719[/tex]
∴
[tex]2= \dfrac{0.5}{0.2535}+ \dfrac{0.5^2}{0.1719} - t_o (initial)[/tex]
[tex]2 = 3.4267 - t_o (initial) \\ \\ t_o(initial) = 3.4267 - 2 \\ \\ t_o(initial) = 1.4267 \ hr[/tex]
NOW;
[tex]1.4267 = \dfrac{d_o}{0.2535} + \dfrac{d_o^2}{0.1719} \\ \\ 1.4267 = 3.9448 \ d_o + 5.8173 \ d_o^2 \\ \\ d_o^2 + 0.6781 \ d_o = 0.2453[/tex]
[tex]d_o = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]d_o = \dfrac{-(0.6781) \pm \sqrt{(0.6781)^2-4(1)(-0.245)}}{2(10)}[/tex]
[tex]d_o = \dfrac{-(0.6781) \pm \sqrt{0.45981961+0.98}}{20}[/tex]
[tex]d_o = \dfrac{-(0.6781) \pm \sqrt{1.43981961}}{20}[/tex]
[tex]d_o = \dfrac{-(0.6781) + \sqrt{1.43981961}}{20} \ OR \ \dfrac{-(0.6781) - \sqrt{1.43981961}}{20}[/tex]
[tex]d_o =0.02609 \ OR \ -0.0939[/tex]
Thus; since we will consider the positive sign, the initial thickness [tex]d_o[/tex] is ;
≅ 0.261 μm
How many Calories are in 5,926 joules
A scientist places 10 ml of water in a test tube and heat it with flaming burner for 2 minutes. The liquid boils and escapes as a steam. This experiment is a good example of a
A chemical change
B physical change
C chemical property
D physical property
HELP FAST 100 PTS Calculate the amount of heat needed to lower the temperature of 50.0g of ice from -40 °C to -100 °C.
Answer:
[tex]\Large \boxed{\sf -6000 \ J}[/tex]
Explanation:
Use formula
[tex]\displaystyle \sf Heat \ (J)=mass \ (kg) \times specific \ heat \ capacity \ (Jkg^{-1}\°C^{-1}) \times change \ in \ temperature \ (\°C)[/tex]
Specific heat capacity of ice is 2,000 J/(kg °C)
Substitute the values in formula and evaluate
[tex]\displaystyle \sf Heat \ (J)=0.05 \ kg \times 2000 \ Jkg^{-1}\°C^{-1} \times (-100\°C-(-40 \°C))[/tex]
[tex]Q=0.05 \times 2000 \times (-100-(-40)) =-6000[/tex]
When 577 J of energy is added to 32.3 g of aluminum at 17.4ºC, the temperature increases to 46.6ºC. What is the specific heat of aluminum?
Answer:
0.612 J/g°C
Explanation:
Using the formula as follows:
Q = m × c × ∆T
Where;
Q = amount of heat (Joules)
m = mass of substance (g)
c = specific heat capacity (J/g°C)
∆T = change in temperature (°C) i.e. final - initial temperature
According to the information provided in this question;
Q = 577 J
m = 32.3 g
c = ?
Final temp = 46.6ºC, initial temp. = 17.4°C
∆T = (46.6 - 17.4) = 29.2°C
Using Q = m × c × ∆T
c = Q ÷ m.∆T
c = 577 ÷ (32.3 × 29.2)
c = 577 ÷ 943.16
c = 0.6117
c = 0.612 J/g°C
Two trials are run, using excess water. In the first trial, 7.8 g of Na2O2(s) (molar mass 78 g/mol) is mixed with 3.2 g of S(s). In the second trial, 7.8 g of Na2O2(s) is mixed with 6.4 g of S(s). The Na2O2(s) and S(s) react as completely as possible. Both trials yield the same amount of SO2(aq). Which of the following identifies the limiting reactant and the heat released, q, for the two trials at 298 K?
Limiting Reactant q
A. S 30. kJ
B. S 61 kJ
C. Na2O2 30. kJ
D. Na2S2 61 kJ
Answer:
The answer is "Option C".
Explanation:
Given equation:
[tex]2Na_20_2 (s)+S(s)+2H_2O \longrightarrow 4NaOH(aq)+SO_2(aq)[/tex]
[tex]\to \Delta H^{\circ}_{rxn} (298\ K) = -610 \frac{kJ}{mol}[/tex]
[tex]\to Na_2O_2 \ Mass = 7.8 \ g\\\\ \to Na_2O_2 \ Molar \ mass = 78 \frac{g}{mol}[/tex]
[tex]Na_2O_2[/tex] Has been the reactant which is limited since the two experiments are equal to[tex]Na_2O_2[/tex] for relationship between stress amounts.
[tex]Na_2O_2, n =\frac{Mass of Na_2O_2}{Molar mass of Na_2 O_2}=\frac{7.8 \ g}{78 \frac{g}{mol}} =0.1 \ mol \\\\q=\Delta H^{\circ}_{rxn} \times n = \frac{ -610 \ kJ}{ 2 \ mol \ Na_2 O_2} \times 0.1 \ mol \ Na_2O_2= 30.5 \ KJ\\\\[/tex]
Limiting reactant =[tex]Na_2O_2[/tex]
[tex]q=30.5 \ kJ \approx 30 \ kJ[/tex]
why iodide ions cannot be determined by Mohr method
Answer:
Exactly the same approach can be used for determination of bromides. Other halides and pseudo halides, like I- and SCN-, behave very similarly in the solution, but their precipitate tends to adsorb chromate anions making end point detection difficult.
Hope it helps!
Mohr method is the titration method, used for the determination of the chloride ion concentration in solution. It can not be used for iodine ions, as the alkaline titration is not suitable for iodine.
What is titration?Titration is the analytical process used for the determination of the concentration of a compound, by reacting it with titrant.
The mohr method is the titration method, that determines the concentration of the cations in the solution by titrating them with the salt. The salt formed precipitate in the reaction and the end point is determined.
The limitation to the use of mohr method, is its inability for the analysis of the chlorides of cations, as the reaction is carried out at high pH in alkaline conditions.
The iodide slats are suitably precipitated in the alkaline solution, and thus limits the use of the mohr method.
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1°C if the pressure is 149.3 kPa? How many moles of hydrogen sulfide H2S are contained in a 327.3 mL bulb at 48.
Answer:
Number of moles of H₂S gas = 0.0183 moles
Note: The question is incomplete. The complete question is given below:
How many moles of hydrogen sulfide H2S are contained in a 327.3 mL bulb at 48.1°C if the pressure is 149.3 kPa?
Explanation:
The following values are given in the question :
Volume of H₂S gas = 327.3 M = 0.3273 L
Temperature of gas = 48.1°C = (273.15 + 48.1) K = 321.25 K
Pressure of gas = 149.3 kPa
1 kPa = 0.00987 atm; 149.3 kPa = 149.3 × 0.00987 atm = 1.474 atm
Molar gas constant, R = 0.0821 liter·atm/mol·K.
number of moles, n = ?
Using PV = nRT
n = PV/RT
n = (0.3273 × 1.474)/(0.0821 × 321.25)
n = 0.0183 moles
Therefore, number of moles of H₂S gas = 0.0183 moles
A student uses a sample of KOH stock solution and dilutes it to a total of 120 mL. If the diluted solution is 0.60 M KOH and its original concentration was 2.25 M, what was the volume of the original sample? *
1.4 mL
89 mL
32 mL
5.5 mL
(DON'T POST LINKS PLEASE)
Answer:
5.5
Explanation:
i think so?????????
The diluted solution of volume 120 ml has the molarity of 0.60 M. Then, the volume of the original solution with a molarity of2.25 M is 32 ml.
What is molarity ?The molarity of a solution is the ratio of the number of moles of its solute particles to the volume of solution in liters.
To solve the given problem, we can use the formula for dilution:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
We are given that the diluted solution has a concentration of 0.60 M and a total volume of 120 mL. We are also given that the original concentration was 2.25 M, and we want to find the original volume.
Using the formula for dilution, we can write:
2.25 M x V1 = 0.60 M x 120 mL
Simplifying, we get:
V1 = (0.60 M x 120 mL) / 2.25 M
V1 = 32 mL
Therefore, the original volume of the KOH stock solution was 32 mL.
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is the sun the only star in our solar system
Answer:
Hey mate......
Explanation:
This is ur answer.......
The largest star, and indeed the only star in our solar system, is the sun. The sun is a bit under a million miles across. About 110 Earths put side by side would equal the size of the sun. The sun has 99.8 percent of the mass of our solar system.
Hope it helps!
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What is a hot spot? {Must be in your own words} Plz hurry
Answer:
Hot spot is like someone who has data or min on there phone turn on their phone wifi so you can use it pretty much
Answer:
a hot spot is a form of wifi that u can use anywhere at anytime. it connects to near satalites or wifi towers. it allows you to use devices r games without the need for wifi.
Explanation:
Definition of Acid, base and
salt
Answer:
nenrhj4rhty4bdwkwwa
Explanation:
An FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated temperature and in an atmosphere wherein the surface carbon concentration is maintained at 1.0 wt%. If after 51 h the concentration of carbon is 0.35 wt% at a position 3.9 mm below the surface, determine the temperature at which the treatment was carried out. You will need to use data in the two tables below to solve this problem.
Answer:
Explanation:
[tex]\text{From the information given:}[/tex]
[tex]C_o = 0.20 \ wt\% \\ \\ C_s = 1 \ wt\% \\ \\ t = 51 \ h \\ \\ x = 3.9 \ mm \\ \\ C_x = 0.35 \ wt\%[/tex]
[tex]\text{Using Fick's 2{nd} \ law \ of \ diffusion;} \\ \\ \dfrac{C_x- C_o}{C_s-C_o}= 1 - erf ( \dfrac{x}{2\sqrt{Dt}})[/tex]
[tex]\dfrac{0.35-0.20}{1-0.20}= 1 - erf ( \dfrac{x}{2\sqrt{Dt}})[/tex]
[tex]0.1875 = 1 - erf ( \dfrac{x}{2\sqrt{DT}}) \\ \\ erf ( \dfrac{x}{2\sqrt{DT}}) = 1 - 0.1875 \\ \\ erf ( \dfrac{x}{2\sqrt{DT}}) = 0.8125[/tex]
[tex]\text{To find the value of Z by Obtaining Data from Tabulation of Error Function}[/tex] [tex]\text{Table Values:}[/tex]
Z erf(z)
0.90 → 0.7970
0.95 → 0.8209
? → 0.8225
∴
[tex]\dfrac{z-0.90}{0.95-0.90}= \dfrac{0.8125-0.7970}{0.8209-0.7970}[/tex]
[tex]\dfrac{z-0.90}{0.05}= \dfrac{0.0155}{0.0239}[/tex]
[tex]z = 0.9324[/tex]
[tex]\text{To determine the diffusion coefficient;}[/tex]
[tex]erf (0.9324) = 0.8125 = erf (\dfrac{x}{2\sqrt{Dt}}) \\ \\[/tex]
[tex]\dfrac{x}{2 \sqrt{Dt}}= 0.9324 \\ \\ \dfrac{3.9 \times 10^{-3}}{2 \times \sqrt{D\times 51 \times 3600}} = 0.92324 \\ \\ \sqrt{D} = 4.88 \times 10^{-6} \\ \\ D = \sqrt{4.88 \times 10^{-6}} \\ \\ D = 2.38 \times 10^{-11} \ m^2 /s[/tex]
Use the following balanced reaction to solve:
P4 (s) + 6H2 (g) → 4PH3 (g)
How many grams of phosphorus trihydride will be formed by reacting 60 L of Hydrogen gas with an excess of P4?
Answer: 60.7 g of [tex]PH_3[/tex] will be formed.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}[/tex]
[tex]\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles[/tex]
The balanced chemical reaction is
[tex]P_4(s)+6H_2(g)\rightarrow 4PH_3(g)[/tex]
[tex]H_2[/tex] is the limiting reagent as it limits the formation of product and [tex]P_4[/tex] is the excess reagent.
According to stoichiometry :
6 moles of [tex]H_2[/tex] produce = 4 moles of [tex]PH_3[/tex]
Thus 2.68 moles of [tex]H_2[/tex] will produce=[tex]\frac{4}{6}\times 2.68=1.79moles[/tex] of [tex]PH_3[/tex]
Mass of [tex]PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g[/tex]
Thus 60.7 g of [tex]PH_3[/tex] will be formed by reactiong 60 L of hydrogen gas with an excess of [tex]P_4[/tex]
if a compound formula mass was experimentally determined to be 58, the chemical formula could be?
Answer:
NaCl
Explanation:
If you follow traditional rounding, Na atomic mass is 23 and Cl atomic mass 35, 23 +35 = 58.
I was taught that Cl is one of the elements you round to 35.5, but this works if you’re not in an advanced chem class.
Identify the techniques used in the work-up and characterization of benzoic acid. The analytical method used to confirm the structure and functional groups of the product NMR spectroscopy The technique used to separate the pure product from any excess reagent, impurities, and byproducts Recrystallization The quick, numeric analysis used to characterize the product and assess the purity Melting point.
Answer:
Explanation:
[tex]\text{From the list of the options given; we are to identify the suitable techniques} \\ \\ \text{for the characterization of benzoic acid.}[/tex]
[tex]\text{The analytical method used to confirm the structure and functional groups}\\ \\ \text{present in the product is} \ \ \mathbf{IR \ spectroscopy.}[/tex]
[tex]\text{The technique used to separate pure products from any excess reagents,} \\ \\ \text{impurities, and byproducts is}\ \ \mathbf{Recrystallization.}[/tex]
[tex]\text{The quick, numeric analysis done to characterize the product and assess the purity is}[/tex][tex]\mathbf{melting \ point.}[/tex]
Explain this method (Froth floatation method)..........
Answer:
froth flotation is a technique commonly used in the mining industry. In this technique, particles of interest are physically separated from a liquid phase as a result of differences in the ability of air bubbles to selectively adhere to the surface of the particles, based upon their hydrophobicity.
Explanation:
Froth floatation method is commonly used to concentrate sulphide ore such as galena (PbS), zinc blende (ZnS) etc. (ii) In this method, the metaalic ore particles which are perferentially wetted by oil can be separated from gangue. (iii) In this method, the crushed ore is suspended in water and mixed with frothing agent such as pine oil, eucalyptus oil etc. (iv) A small quantity of sodium ethyl xanthate which act as a collector is also added. (v) A froth is generated by blowing air through this mixture. (vi) The collector molecules attach to the ore particles and make them water repellent. (vii) As a result, ore parrticles, wetted by the oil, rise to the surface along with the froth. (viii) The froth is skimmed off and dried to recover the concentration ore. (ix) The gangue particles that are preferentially wetted by water settle at the bottom.
What is the other product for this reaction ? H3PO4 + Ca(OH)2 —> H20 + _________
*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆
Answer: h3po4 + ca(oh)2 = h2o + ca3(po4)2
Explanation:
I hope this helped!
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Help this is for marks, who ever answers get brainliest
Answer:
Coal
Explanation:
thx for points :D
Answer:
Coal
Explanation:
your welcome<3
How many joules are in 1.81 calories need to know ASAP
Please help me as soon as possible!!! How many grams of H3PO4 are in a 4 L sample of a 17 M solution?
Answer:
about 6664 grams of hydro phosphoric acid are in a sample of 4 L of 17 M solution
Explanation:
grams of H3PO4 = 4 L | 17 mol | 98 grams H3PO4
| 1 L | 1 mol
A separatory funnel contains the two immiscible liquids water and toluene. Use the given densities to determine which layer is on top and which is on the bottom in the binary mixture.
Solvent Density (g/mL)
toluene 0.87
water 0.998
Drag and drop each label into the box to indicate the position of the liquid in the mixture.
Top layer
Bottom layer
Answer:
Top layer TolueneBottom layer WaterExplanation:
When two non-miscible liquids are put together, the one with the higher density will be on the bottom, while the one with the lower density will be on top.
Meaning that in this problem's case toluene would be on the top layer and water in the bottom layer.
I need help fast pls someone
Answer:
I would say A. I'm no expert, but it can't be C obviously, and I think wind would hit all of it, wearing off the top as well like the great pyramids. B would be my next choice, but A i think would be best.
2.0M Propionic Acid HC3H5O2 Dissolves In Distilled Water. If It Has A Ka Of 1.3*10-5, What Is The Final PH?
The base ethylamine C2H5NH2 has a Kb of 5.6*10-4. What will the pH be in .53 molar solution?
Answer: first you have to calculate the amount ionized. We will say it is x mol / L
then % ionization = (amount ionized / initial concentration) * 100%
x can be calculated using an ice chart
HC3H5O2 -----> H+ + C3H5O2-
Initial HC3H5O2 = 0.250
change = -x
equilibrium = 0.250 - x
initial H+ = 0
change = +x
equilibrium = x
C3H5O2- initial = 0
change = +x
equilibrium = x
Ka = [H=][C3H5O2-] / HC3H5O2]
1.3 * 10 ^ -5 = [x][x] / (0.250 - x)
So 1.3 * 10 ^ -5 * (0.250 - x) = x ^ 2
3.25 * 10^ -6 - (1.3 * 10^-5)x = x^2 now this is a quadratic equation and you have to rearrange it and solve for x
x^2 + 1.3 * 10^-5)x - 3.25 * 10^ -6 = 0
use the equation x = {-b (+ or -)[b^2 - 4.a.c] ^ 1/2} / 2a
you should get x = 1.80 * 10 ^ -3 or x = -1.80* 10^-3
but x can not be negative..
so x = 1.80 * 10 ^ -3
so percent ionization = (1.80 * 10 ^ -3 / 0.250) * 100%
=0.72 %
the other way which is more easier is
assuming that x is very small and therefore 0.250 - x is approximately equals to 0.250
then 1.3 * 10^-5 = x^2 / 0.250
so x^2 = 1.3 * 10^-5 * 0.250
x = 1.80 * 10 ^-3
then percent ionization is = (1.80 * 10 ^ -3 / 0.250) * 100%
=0.72 %
if the percent ionization is > 5 % you can not do that approximation. in such a case you have to solve the quadratic equation. that is why I showed both methods.
now you can do the parts b and c
b answer : percent ionization = 1.27 %
c answer : 2.54%
good luck
A student placed 2 drops of an unknown sample in a test tube and added 2 mL of ethanol to the test tube while mixing gently. They added 2 drops of potassium permanganate reagent to the test tube and mixed the contents of the test tube gently. The initial color was a deep purple but then changed to a yellow color which precipitates as a brown solid. What compound is most likely the unknown
Answer:
2-propanol
Explanation:
From the given information
2 drops of an unknown sample were said to be placed in a test tube followed by the addition of 2 ml of ethanol then gentle mixing. They then initiate a further addition of 2 drops of potassium permanganate reagent (KMNO₄) to the test tube and mixed the contents of the test tube thoroughly. After adding 2 drops of potassium permanganate reagent, the reagent oxidizes the secondary alcohols(2-propanol) to ketone(i.e acetone) and no further reaction will take place since there are no reactive C-H bonds left. The diagram attached below shows how the reaction proceeds.
