Answer:
Approximately [tex]279\; \rm mg[/tex] of fluorine would be produced.
Explanation:
Look up the relative atomic mass of these calcium and fluorine on a modern periodic table:
[tex]\rm Ca[/tex]: [tex]40.078[/tex].[tex]\rm F[/tex]: [tex]18.998[/tex].In other words, the mass of each mole of calcium atoms is (approximately) [tex]40.078\; \rm g[/tex]. Similarly, the mass of each mole of fluorine atoms is (approximately) [tex]18.998\; \rm g[/tex].
Calculate the number of moles of calcium atom that were produced:
[tex]\begin{aligned} &n({\rm Ca}) \\ &= \frac{m({\rm Ca})}{M({\rm Ca})} \\ &= \frac{294\; \rm mg}{40.078\; \rm g \cdot mol^{-1}} \\ &= \frac{0.294\; \rm mg}{40.08 \; \rm g \cdot mol^{-1}}\\ & \approx 7.3357 \times 10^{-3}\; \rm mol\end{aligned}[/tex].
Every formula unit of calcium fluoride, [tex]\rm CaF_2[/tex], contains twice as many fluorine atoms as calcium atoms.
Hence, if the sample contained approximately [tex]7.33570 \times 10^{-3}\; \rm mol[/tex] calcium atoms, it would contain twice as many fluorine atoms:
[tex]\begin{aligned}n({\rm F}) &= 2\, n({\rm Ca}) \\ &\approx 2 \times 7.33570\times 10^{-3}\; \rm mol \\ &\approx 1.4671 \times 10^{-2}\; \rm mol\end{aligned}[/tex].
Calculate the mass of that many fluorine atoms:
[tex]\begin{aligned}& m({\rm F}) \\ &= n({\rm F}) \cdot M({\rm F}) \\ &\approx 1.4671 \times 10^{-2}\; \rm mol \times 18.998\; \rm g \cdot mol^{-1} \\ &\approx 278.72 \times 10^{-3}\; \rm g \\&\approx 279\; \rm mg\end{aligned}[/tex].
279 mg of Fluorine was formed in the decomposition reaction
The equation of the reaction is;
CaF2 -----> Ca + 2F
Mass of calcium formed = 294 mg
Number of moles of Ca in 294 mg = 294 × [tex]10^-3[/tex] g/40 g/mol = 0.00735 moles
From the equation, the products are in ration 1 : 2
Since 1 mole of Ca is to 2 moles of F
0.00735 moles of Ca is to 0.00735 moles × 2 moles/1 mole
= 0.0147 moles of F
Mass of F = 0.0147 moles × 19 g/mol
= 0.279g or 279 mg of F
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If 0.000066 moles of 0.01 M solution of carbonic acid dissociates, then what is Ka of carbonic acid
Answer: 4.4 x 10^-7
Explanation:
0.000066 x 0.0000066 / 0.01
= 4.4 x 10^-7
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.
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:
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.
im not sure if it is a molecule or an atom
Answer:
the first one
Explanation:
I looked it up:P
Only the 2nd answer is right, uncheck the 1st one
Explanation:
If we had a chemical formula like [tex]3(O_{2} )[/tex] the 3 means there are 3 molecules of [tex]O_{2}[/tex]. The 2 means there are two atoms of oxygen in each molecule.
Classify the following as either an element, compound, homogeneous mixture or
heterogeneous mixture:
sodium chloride
Sodium chloride is the chemical name for salt and is composed of sodium and chlorine, so it is a compound.
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!
<!> Brainliest is appreciated! <!>
- Zack Slocum
*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆
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
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]
Exit ticket: Name the 2 factors that Terrestrial Biomes are separated by ___________ & ________
Answer:
temperature and rainfall
Explanation:
Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 300 cm3, the pressure is 180 kPa, and the pressure is increasing at a rate of 30 kPa/min. At what rate is the volume decreasing at this instant?
Answer:
The gas is decreasing at a rate of 50 cubic centimeters per minute.
Explanation:
The Boyle's Law is represented by the following expression:
[tex]P\cdot V = k[/tex] (1)
Where:
[tex]P[/tex] - Pressure, in kilopascals.
[tex]V[/tex] - Volume, in cubic centimeters.
[tex]k[/tex] - Proportionality constant, in kilopascal-cubic centimeters.
By definitions of rate of change and implicit differentiation, we derive the following differential equation:
[tex]\dot P \cdot V + P\cdot \dot V = 0[/tex] (2)
Where:
[tex]\dot P[/tex] - Rate of change of the pressure, in kilopascals per minute.
[tex]\dot V[/tex] - Rate of change of the volume, in cubic centimeters per minute.
Then, we clear the rate of change of the volume within (2):
[tex]P\cdot \dot V = -\dot P\cdot V[/tex]
[tex]\dot V = -\left(\frac{\dot P}{P} \right) \cdot V[/tex]
If we know that [tex]P = 180\,kPa[/tex], [tex]\dot P = 30\,\frac{kPa}{min}[/tex] and [tex]V = 300\,cm^{3}[/tex], then the rate of change of the volume is:
[tex]\dot V = -\left(\frac{\dot P}{P} \right) \cdot V[/tex]
[tex]\dot V = -50\,\frac{cm^{3}}{min}[/tex]
The gas is decreasing at a rate of 50 cubic centimeters per minute.
Solubility is the mass in a given volume of a substance.
True
False
Answer: True
Explanation:
Consider the reactivity of phenol, bromobenzene, toluene, and nitrobenzene toward electrophilic aromatic substitution.
The most reactive compound is:_____
because the _______ character of the ______ increases the rate of the reaction.
The least reactive compound is ______
because the _______ character of the ________ decreases the rate of the reaction.
Answer:
The most reactive compound is: Phenol because the electron donating character of the alcohol group increases the rate of the reaction.
The least reactive compound is nitrobenzene
because the electron withdrawing character of the nitro group decreases the rate of the reaction.
Explanation:
PHENOL-: Any of a group of organic compounds with a hydroxyl (OH) group bound to a carbon atom in an aromatic ring is known as phenol. The word phenol is also the basic name for its simplest member, monohydroxybenzene (C6H5OH), also known as benzenol or carbolic acid, in addition to being the common name for the entire family.ELECTRON DONATING CHARACTER-: Via the carbon atom it is bound to, an electron donating group (EDG) has the net effect of increasing electron density in a molecule. EDGs alter a molecule's reactivity by increasing electron density on neighboring carbon atoms: EDGs make nucleophiles stronger.ALCOHOL GROUP-:Each of a class of organic compounds that have one or more hydroxyl (OH) groups bound to an alkyl group's carbon atom (hydrocarbon chain)is called Alcohol. Alcohols are organic compounds of water [tex](H_2O)[/tex] in which one of the hydrogen atoms has been substituted by an alkyl group, which in organic structures is usually expressed by R.NITROBENZENE-:The organic compound nitrobenzene has the chemical formula [tex](C_6H_5NO_2)[/tex]. It's a pale yellow oil that's insoluble in water and smells like almonds. Greenish-yellow crystals form when it freezes. It is made on a wide scale as a precursor to aniline from benzene. It is sometimes used as a solvent in the laboratory, especially for electrophilic reagents.ELECTRON WITHDRAWING GROUP-: An electron withdrawing group (EWG) is a type of group that reduces electron density in a molecule by bonding to a carbon atom. EWGs alter a molecule's reactivity by reducing electron density on neighboring carbon atoms.NITRO GROUP-: The nitro group is one of the most widely used explosophores (functional groups that combine to form a compound explosive). In addition, the nitro group is a heavy electron-withdrawing group. CH bonds alpha (adjacent) to the nitro group may be acidic due to this property.- 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?
1L = 1000 mL
Molarity (M) = moles of solute / L of solution
Calculate the moles of NaCl required to prepare 250 mL of 1.5 M aqueous solution.
Answer:· 44 g NaCl Explanation: The problem provides you with the molarity and volume of the target
Explanation:
Help this is for marks, who ever answers get brainliest
Answer:
Coal
Explanation:
thx for points :D
Answer:
Coal
Explanation:
your welcome<3
5
Select all the correct answers.
Which sentences describe chemical changes?
The temperature of water rises when heated.
When mixed, water and limestone give off heat.
A mixture of red paint and green paint becomes black.
On adding yellow turmeric to a transparent soap solution, the mixture turns bright red.
When ground into powder, grains of wheat become flour.
Answer: The answers are:
When mixed, water and limestone give off heat.
On adding yellow turmeric to a transparent soap solution, the mixture turns bright red.
Definition of Acid, base and
salt
Answer:
nenrhj4rhty4bdwkwwa
Explanation:
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]
Please help I have no idea where to start
Answer:
Download Ptable, it will show all of the elements, and you can click on them, and it will show all the information you need about them!
Explanation:
I hope this helps!!!!!
help people i really need this
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
can u identify a chemical equation for the saponification of olive oil as per your experiment?
I
You have three containers each of which is connected to a fourth empty container. Three of the containers contain pure gases. In the first container oxygen is at 2 atm, nitrogen at 4 atm is in the second container, and argon at 1 atm is in the third. When you open the valve and mix the gases in the forth container, what is the pressure, PT, in the forth container
Answer: The total pressure is 7 atm
Explanation:
According to Dalton's law, the total pressure is the sum of individual pressures.
[tex]p_{total}=p_A+p_B+p_C...[/tex]
Given : [tex]p_{total}[/tex] =total pressure of gases = ?
[tex]p_{O_2}[/tex] = partial pressure of oxygen = 2 atm
[tex]p_{N_2}[/tex] = partial pressure of nitrogen = 4 atm
[tex]p_{Ar}[/tex] = partial pressure of argon = 1 atm
putting in the values we get:
[tex]p_{total}=2atm+4atm+1atm[/tex]
[tex]p_{total}=7atm[/tex]
Thus the total pressure is 7 atm
Predict the product of this reaction
Al(s) + N2(g)—->
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]
In order to determine the answer to a chemistry problem, a student first converted the given percentages to mole by assuming the sample size was 100 grams. Then the student divide by the smallest mole amount of the elements present. Finally the student used the ratios to write a formula. Which of the following best describes the problem the student likely solved?
a. Determined percent composition
b. None of the above
c. Determining molecular formula
d. Determining empirical formula
Answer: d. Determining empirical formula
Explanation:
Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.
Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.
Step 1: If percentage are given then we are taking total mass is 100 grams. So, the mass of each element is equal to the percentage given.
Step 2 : convert given masses into moles
Step 3 : For the mole ratio, divide each value of moles by the smallest number of moles calculated
Step 4: The mole ratio gives the empirical formula of the compund.
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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Consider the fatty acid. A carboxylate salt is attached to a 19 carbon chain, where double bonds are present between carbons 5 and 6, between carbons 8 and 9, between carbons 11 and 12, and between 14 and 15. The carboxylate carbon is carbon 1. All double bonds are cis. Which of the designations are accurate for the fatty acid
Answer:
ω-3 fatty acid and 19:4(Δ5,8,11,14)
Explanation:
From the image attached below:
We can see the structure of the delta nomenclature showing the numbers of carbons, numbers of double bonds, and the locations of the double bonds in the structure of the carboxylate salt.
Also, we can see the structure of the fatty acid where; the location of the first double bond from the methyl end is shown.
Hence, the designation which are accurate for these structures are:
ω-3 fatty acid and 19:4(Δ5,8,11,14)
During a dry compression test, the first puff of the test shows 50 PSI and the maximum reading after 4 strokes is 150 PSI. This indicates A. that the cylinders are good. B. weak piston rings. C. a bad spark plug. D. a hole in the piston.
One of the basic engine-diagnostic procedures that can be run is an engine compression test. All of the cylinders must have the same compression for the engine to run smoothly. There are just three possible ways for air to leak out of an engine and cause compression loss.
Exhaust or intake valve Engine rings (or piston, if there is a hole) An engine's head gasketThe procedures for conducting a compression testThe engine should be warmed to its typical operating temperature before testing in order to achieve the best results. The following procedures should be followed for an accurate compression test:
STEP 1: Dispose of each spark plug. As a result, the engine can be turned up to the same speed. Label all spark plug wires accurately.STEP 2 is to block the throttle open. This enables the engine to pull in the most air possible. Additionally, this step guarantees reliable compression test results.STEP 3 After inserting a compression gauge into a spark plug hole, start the engine. Four compression strokes of engine cranking should be performed in total. A puffing sound is produced after each compression stroke.EXAMPLE-If the ultimate, maximum value is 150 PSI, for instance, the reading following the first puff should be greater than 75 PSI. Low first-puff readings could be a sign of weak piston rings.STEP 4 Keep track of the highest readings, then compare the outcomes.Learn more about compression test here https://brainly.com/question/13274092
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