Answer: v = 200 ml · 0.45 M/ 1.88 M
Explanation: when solution is diluted, amount of substance
is same and n= c · V v1 = v2 · c2 / c1
The appropriate answer is "47.87 mL".
Given values are:
Concentration of undiluted HCl, [tex]C_1=1.88 \ M[/tex]Concentration of diluted HCl, [tex]C_2=0.45 \ M[/tex]Volume of diluted HCl, [tex]V_2=200 \ mL[/tex]By using the dilution equation, we get
→ [tex]C_1V_1=C_2V_2[/tex]
or,
→ [tex]V_1=\frac{C_2V_2}{C_1}[/tex]
By substituting the given values, we get
→ [tex]=\frac{0.45\times 200}{1.88}[/tex]
→ [tex]=\frac{90}{1.88}[/tex]
→ [tex]= 47.87 \ mL[/tex]
Thus the above answer is correct.
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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.
Exit ticket: Name the 2 factors that Terrestrial Biomes are separated by ___________ & ________
Answer:
temperature and rainfall
Explanation:
Solubility is the mass in a given volume of a substance.
True
False
Answer: True
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.
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
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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Predict the product of this reaction
Al(s) + N2(g)—->
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:
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]
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
How many Calories are in 5,926 joules
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
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]
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
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
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.
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
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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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]
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.
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.
is chemical energy stored in coal, gas, and oil?
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]
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 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
Definition of Acid, base and
salt
Answer:
nenrhj4rhty4bdwkwwa
Explanation:
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
- 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?
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.