Answer:
0.0238M SbCl3, 1.07M H+, 1.14M Cl-
Explanation:
The total volume of the solution is:
4mL + 5.00mL + 12.0mL = 21mL
As the volume of the SbCl3 is 5.00mL, the dilution factor is:
21mL / 5.00mL = 4.2 times
The concentration of SbCl3 is:
0.10M SbCl3 / 4.2 times = 0.0238M SbCl3
The concentration of H+ = [HCl]:
4.5M / 4.2 times = 1.07M H+
The initial concentration of Cl- is:
3 times SbCl3 + HCl = 0.10M*3 + 4.5M =
3 times SbCl3 because 1 mole of SbCl3 contains 3 moles of Cl-
4.8M Cl- / 4.2 times = 1.14M Cl-
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Answer:
what do you need help with
Please help me ASAP I’ll mark Brainly
Answer:
cell
chloroplast and cell wall
nucleus
life processes
cell membrane
shape and size
vacuole
Hope it helps
Ethylene glycol flows at 0.01 kg/s through a 3-mm diameter, thin-walled tube. The tube is coiled and submerged in well-stirred water bath maintained at 25°C. If the fluid enters the tube at 85°C, what heat rate and tube length are required for the fluid to leave at 35°C?
Answer:
heat rate= 1281W
length = 15.8m
Explanation:
we have this data to answer this question with
Tmi = 85 degrees
Tmo = 35 degrees
Ts = 25 dgrees
flow rate = 25 degrees
using engine oil property from table a-5
Tm = Tmo - TMi/2 = 333k
u =0.522x10⁻²
k = 0.26
pr = 51.3
cp = 2562 J/kg.k
mcp(Tmo-Tmi) =
0.01 x 2562(35-85)
= 1281 W
we find the change in Tim
= [(35-25)-(85-25)]/ln[(35-25)/(85-25)]
= -50/ln0.167
= -50/-1.78976
= 27.9°c
we finf the required reynold number
4x0.01/πx0.003x0.522x10⁻²
= 0.04/0.00004921
= 812.8
= 813
we find approximate correlation
NuD = hd/k
NuD = 3.66
3.66 = 0.003D/0.26
cross multiply
0.003D = 3.66x0.26
D = 3.66x0.26/0.003
= 317.2
As = 1281/317x27.9
= 0.145
As = πDL
L = As/πD
= 0.145/π0.003
= 0.145/0.009429
L = 15.378
Calculate the vapor pressure (in torr) at 298 K in a solution prepared by dissolving 46.8 g of the non-volatile non-electrolye glucose in 117 g of methanol. The vapor pressure of methanol at 298 K is 122.7 torr. Enter your answer to 2 decimal places.
Answer: The total partial pressure of the solution is 131.37 torr.
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
For glucose:Given mass of glucose = 46.8 g
Molar mass of glucose = 180 g/mol
Plugging values in equation 1:
[tex]\text{Moles of glucose}=\frac{46.8g}{180g/mol}=0.26 mol[/tex]
For methanol:Given mass of methanol = 117 g
Molar mass of methanol = 32 g/mol
Plugging values in equation 1:
[tex]\text{Moles of methanol}=\frac{117g}{32g/mol}=3.66 mol[/tex]
Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex] .....(2)
where n is the number of moles
Putting values in equation 2:
[tex]\chi_{methanol}=\frac{3.66}{0.26+3.66}=0.934[/tex]
Raoult's law is the law used to calculate the partial pressure of the individual gases present in the mixture. The equation for Raoult's law follows:
[tex]p_A=\chi_A\times p_T[/tex] .....(3)
where [tex]p_A[/tex] is the partial pressure of component A in the mixture and [tex]p_T[/tex] is the total partial pressure of the mixture
We are given:
[tex]p_{methanol}=122.7torr\\\chi_{methanol}=0.934[/tex]
Putting values in equation 3, we get:
[tex]122.7torr=0.066\times p_T\\\\p_T=\frac{122.7torr}{0.934}=131.37torr[/tex]
Hence, the total partial pressure of the solution is 131.37 torr.
Each of the following sets of quantum numbers is supposed to specify an orbital. Choose the one set of quantum numbers that does NOT contain an error.
a. n = 4, l = 3, ml =-4
b. n = 2, l = 2, ml =0
c. n = 3, l = 2, ml =-2
d. n = 2, l = 2, ml =+1
Answer:
n = 3, l = 2, ml =-2
Explanation:
Quantum numbers are a set of values which can be used to describe the energy and position of an electron in space.
There are four sets of quantum numbers;
1) principal quantum number
2) orbital quantum number
3) spin quantum number
4) magnetic quantum number.
The values of orbital quantum number include; -l to +l;
The set of quantum numbers without error is ; n = 3, l = 2, ml =-2
Fill in the blanks with each titration term with its definition.
a. Solution of an unknown concentration that has another solution slowly added to it ________________
b. Process of slowly adding a solution to react with another solution and determine the concentration of one of the solutions based on the reaction between them ______________
c. A reagent added to the analyte solution that changes color when the reaction is complete ______________
d. Glassware that allows a solution to be precisely and slowly added to another solution _____________
e. Solution of known concentration that is slowly added to a solution of unknown concentration ________________
f. When the required amount of one solution has been added to the second solution to complete the reaction ____________
Answer:
Fill in the blanks with each titration term with its definition.
a. Solution of an unknown concentration that has another solution slowly added to it ________________
b. Process of slowly adding a solution to react with another solution and determine the concentration of one of the solutions based on the reaction between them ______________
c. A reagent added to the analyte solution that changes color when the reaction is complete ______________
d. Glassware that allows a solution to be precisely and slowly added to another solution _____________
e. Solution of known concentration that is slowly added to a solution of unknown concentration ________________
f. When the required amount of one solution has been added to the second solution to complete the reaction ____________
Explanation:
a. Solution of an unknown concentration that has another solution slowly added to it is called analyte.
b. Process of slowly adding a solution to react with another solution and determine the concentration of one of the solutions based on the reaction between them is called titration.
c. A reagent added to the analyte solution that changes color when the reaction is complete is called an indicator.
d. Glassware that allows a solution to be precisely and slowly added to another solution is called a pipette.
e. Solution of known concentration that is slowly added to a solution of unknown concentration is called titrant.
f. When the required amount of one solution has been added to the second solution to complete the reaction is called neutralization.
What is bond energy
A sample of gas is placed into an enclosed cylinder and fitted with a movable piston. Calculate the work (in joules) done by the gas if it expands from 5.33 L to 11.05 L against a pressure of 1.50 atm.
Explanation:
here is the answer. Feel free to ask for more chem help
Which substrate is used in the last step of glycolysis
Pyruvate Kinase
Pyruvate Kinase performs a substrate level phosphorylation on ADP to generate an ATP and pyruvate, the final product of glycolysis.
PK dificiency is transmitted in an autosomal recessive disorder in which both alleles must contain the mutated gene, PK-LR.
Hope it helps you! ＼(^ᴥ^)／
if a bottle of vinegar has 4.78g of acetic acid (CH3COOH) per 100.0 g of solution (mixed with water, what is the molarity of the vinegar? Density of the solution is 1.00g/mL.
Answer:
0.796 M
Explanation:
Step 1: Given data
Gravimetric concentration (Cg): 4.78 g%g
Density of the solution (ρ): 1.00 g/mL
Step 2: Calculate the volumetric concentration of the solution (Cv)
We will use the following expression.
Cv = Cg × ρ
Cv = 4.78 g%g × 1.00 g/mL = 4.78 g%mL
Step 3: Calculate the molarity of the solution (M)
The volumetric concentration is 4.78 g%mL, that is, there are 4.78 g of acetic acid per 100 mL of solution. We can calculate the molarity using the following expression.
M = mass solute / molar mass solute × liters of solution
M = 4.78 g / 60.05 g/mol × 0.1 L = 0.796 M
The mass of a single tantalum atom is 3.01×10-22 grams. How many tantalum atoms would there be in 37.1 milligrams of tantalum?
Answer: There are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.
Explanation:
Given: Mass of single tantalum atom = [tex]3.01 \times 10^{-22} g[/tex]
Mass of tantalum atoms = 37.1 mg (1 mg = 0.001 g) = 0.0371 g
Therefore, number of tantalum atoms present in 0.0371 grams is calculated as follows.
[tex]No. of atoms = \frac{0.0371 g}{3.01 \times 10^{-22}}\\= 1.23 \times 10^{20}[/tex]
Thus, we can conclude that there are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.
There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.
Explanation:
Given:
Mass of single atom of tantalum =[tex]3.01\times 10^{-22} g[/tex]
To find:
The number of atoms of tantalum in 37.1 milligrams.
Solution:
Mass of tantalum = 37.1 mg
[tex]1 mg = 0.001 g\\37.1 mg=37.1\times 0.001 g=0.0371 g[/tex]
The number of atoms in 0.0371 grams of tantalum = N
Mass of a single atom of tantalum = [tex]3.01\times 10^{-22} g[/tex]
Then a mass of N atoms of tantalum will be:
[tex]0.0371 g=N\times 3.01\times 10^{-22} g\\N=\frac{0.0371 g}{ 3.01\times 10^{-22} g}\\=1.23\times 10^{20}[/tex]
There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.
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A student conducted an experiment 4 times. His results were very close to each other each time he ran the experiment and
were very close to the true or actual value. His results showed
A. None of these answers are correct
B. poor accuracy and poor precision
C. good accuracy and good precision
D. poor accuracy and good precision
E. good accuracy and poor precision
Answer:
d is the answer have a good one
Chloride ion is a strong nucleophile and bromide is a good leaving group. The major product of treating (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) is _________. (1S,2R)-1-chloro-2-bromobutane cis-2-butene (1R,2S)-1-chloro-2-bromobutane (S)-2-chlorobutane trans-2-butene (R)-2-chlorobutane
Answer:
Chloride ion is a strong nucleophile and bromide is a good leaving group. The major product of treating (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) is _________. (1S,2R)-1-chloro-2-bromobutane cis-2-butene (1R,2S)-1-chloro-2-bromobutane (S)-2-chlorobutane trans-2-butene (R)-2-chlorobutane
Explanation:
The reaction of (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) forms the following product:
The answer is (R)-2-chlorobutane.
The reaction take splace through [tex]S_{N} _2[/tex] mechansim and inversion in configuration happens.
b) What is the change in entropy of the reaction if ΔH° = -3.2 kJ mol-1?
howtocalculatethevolumeofcarbondioxideproducedwhen400gofmarblewereats.t.p
Answer:
so 0.15 moles X 22.4 dm3/mole=3.36 dm3. Next we find the moles of hexane combusted, and then the moles of CO2. Finally, we find the volume of CO2 using the fact that at STP, 1 mole of gas = 22.4 dm3.
If 11g of a gas occupies 5.6dm'3at s.t.p., calculate it's vapour density (1.0mol of a gas occupies 22.4dm'3at s.t.p.).
Answer:
[tex]{ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 = \frac{11}{m _{r}} \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\ \\ { \bf{vapour \: density = 2 \times m _{r}}} \\ = 2 \times 14.85 \\ = 29.7 \: { \tt{g {dm}^{ - 3} }}[/tex]
Mava=mbvb ma x5.0ml = 5.2ml x 0.10m
Answer:
[tex]M_{a}[/tex] = 0.104 m
Explanation:
This expression is used to determine either the mass or volume of an acid or a base used during titration process.
So that;
[tex]M_{a}[/tex][tex]V_{a}[/tex] = [tex]M_{b}[/tex][tex]V_{b}[/tex]
[tex]M_{a}[/tex] is the mass of the acid
[tex]V_{a}[/tex] is the volume of the acid
[tex]M_{b}[/tex] is the mass of the base
[tex]V_{b}[/tex] is the volume of the base
Given that:
[tex]M_{a}[/tex] x 5.0 m l = 5.2 ml x 0.10 m
[tex]M_{a}[/tex] x 5.0 = 0.52 m
[tex]M_{a}[/tex] = [tex]\frac{0.52}{5.0}[/tex]
= 0.104
[tex]M_{a}[/tex] = 0.104 m
The mass of the acid used is 0.104 m.
The doctor has ordered Claforan 1 g in 100 ml D5W to run IV piggyback for 30 minutes twice daily. The pharmacy sends Claforn 2 g in a powdered form, which when reconstituted has a concentration of 180 mg Claforan per ml. How much Claforn will you add to the bag of D5W
Answer:
0.111 g
Explanation:
1 g = 1000 mg
Doctor ordered the following concentration of Claforan:
C = 1 g/100 mL x 1000 mg/1 g = 10 mg/mL
If we add 2 g iof Claforan, we obtain:
2 g Claforn ---- 180 mg/mL Claforan
To reach a concentration equal to C (10 mg/mL), we need:
10 mg/mL Claforan x 2 g Claforn/(180 mg/mL Claforan) = 0.111 g Claforn
Therefore, we have to add 0,111 g (111 mg) of Claforn to the bag of 100 ml D5W to obtain the ordered concentration of 10 mg/mL Claforan.
Suppose a 48. L reaction vessel is filled with 1.6 mol of Br2 and 1.6 mol of OCl2. What can you say about the composition of the mixture in the vessel at equilibrium
Answer:
There will be very little of BrOCl BrCl
Explanation:
Based on the equilibrium:
Br2(g) + OCl2(g) ⇄ BrOCl(g) + BrCl(g)
The equilibrium constant, Kc, is:
Kc = 1.58x10⁻⁵ = [BrOCl] [BrCl] / [Br2] [OCl2]
As Kc is <<< 1, in equilibrium, the concentration of products will remain lower regard to the concentration of the reactants. That means, right answer is;
There will be very little of BrOCl BrClI need help solving this!
For the reaction C + 2H2 → CH4, how many moles of hydrogen are needed to make 146.6 grams of methane, CH4 ?
Round your answer to the nearest tenth. If you answer is a whole number like 4, report the answer as 4.0
Use the following molar masses. If you do not use these masses, the computer will mark your answer incorrect.:
Element
Molar Mass
Hydrogen
1
Carbon
12
Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, [tex]CH_{4}[/tex].
Explanation:
Given: Mass of methane = 146.6 g
As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol[/tex]
The given reaction equation is as follows.
[tex]C + 2H_{2} \rightarrow CH_{4}[/tex]
This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.
[tex]Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol[/tex]
Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, [tex]CH_{4}[/tex].
is -2 degrees Fahrenheit warmer or is -17 degrees fahrenheit?
A 1.5-m 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa. Now paddle-wheel work is done on the system until the pressure in the tank rises to 150 kPa. Determine the entropy change of carbon dioxide during this process. Assume constant specific heats
Answer:
The entropy change of carbon dioxide = 0.719 kJ/k
Explanation:
Given:
1.5 m - 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa
The objective is to determine the entropy change of carbon dioxide
Formula used:
ΔS=
Solution:
On considering,
[tex]C_{P} =0.846 kJ/kg K\\C_V=0.657 kJ/kg k\\[/tex]
ΔS=[tex]mc_{v} lu\frac{p_{2} }{P_{1} }[/tex]
On substituting the values,
ΔS=[tex]2.7*0.657lu\frac{150}{100}[/tex]
ΔS=0.719 kJ/k
The entropy change is "0.719 kJ/K".
Given values are:
Mass of tank,
m = 2.7 kgPressure,
P₁ = 100 kPaRised pressure,
P₂ = 150 kPaAssumption of constant specific heat is,
[tex]C_v = 0.657 \ kJ/kgK[/tex]As we know the formula,
→ [tex]\Delta S = mC_v \ ln(\frac{P_2}{P_1} )[/tex]
[tex]= (2.7)(0.657) \ ln (\frac{150}{100} )[/tex]
[tex]= 1.7739\times 0.4055[/tex]
[tex]= 0.7193 \ kJ/K[/tex]
Thus above answer is right.
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Based on the equations below, which metal is the least active? Pb(NO3)2(aq) + Ni (s) --> Ni(NO3)2 (aq)+ Pb(s) Pb(NO3)2(aq) + Ag(s) --> No reaction Cu(
Answer:
Ni
Explanation:
An active metal is a highly reactive metal. Active metals are found high up in the activity series.
Active metals react with other metals that are lower than them in the activity thereby displacing the lower metals from a solution of their salts. This is what may have happened in the other two reactions.
Ni is the most active metal listed in the question since it can react a compounds with Pb(NO3)2(aq) to liberate Pb metal.
Which substance has nonpolar covalent bonds?
CO
NO2
H2
NaBr
Answer:
H2
Explanation:
Answer:
[tex]{ \bf{H _{2} }} \\ { \tt{hydrogen \: gas}}[/tex]
An unidentified gas is determined to be 24.0% carbon and 76% fluorine by mass. What is the empirical formula of this gas
Answer:
CF₂
Explanation:
Let's assume we have 100 g of the gas. If that were the case we'd have
24 g of C76 g of FNow we convert both masses into moles, using their respective molar mass:
24 g C ÷ 12 g/mol = 2 mol C76 g F ÷ 19 g/mol = 4 mol FWe can express those results as C₂F₄.
To determine the empirical formula we reduce those coefficients to the lowest possible integers, leaving us with CF₂.
4019K Express your answer as a nuclear equation.
Answer:
[tex]_{19} ^{40} K[/tex] → [tex]_{20} ^{40} Ca[/tex] + [tex]_{-1} ^{0} e[/tex] + ¯v
Explanation:
[tex]_{19} ^{40} K[/tex] is an unstable potassium isotope which has the capacity to undergo beta decay. During the decay process, it transforms to calcium atom with a release of beta particle and an antineutrino.
The required nuclear equation is:
[tex]_{19} ^{40} K[/tex] → [tex]_{20} ^{40} Ca[/tex] + [tex]_{-1} ^{0} e[/tex] + ¯v
The calcium formed due to this process is a stable atom.
Platinum is one of the most dense elements (d = 21.5 g/cm3). What is the volume of a 10.0 g sample of the metal?
Answer:
0.465
Explanation:
To find the volume of a substance, divide the mass by the density.
M/D = V
10.0 / 21.5 = 0.4651163
Then round to 3 significant figures: and the density is 0.465
A compound was analyzed and found to contain 76.57% carbon, 6.43% hydrogen, and 17.00% oxygen by mass. Calculate the empirical formula of the compound. If the molar mass of the compound is 94.11 g/mol, what is the molecular formula of the compound?
A second compound is composed of 53.30% Carbon 11.19% Hydrogen and 35.51% Oxygen by mass.Please Calculate the empirical formula of the compound of the molar mass of the compound is 90.12g/mol, what is the molecular formula for that compound?
Answer:
See explanation
Explanation:
First we divide the percentage by mass of each element by it's relative atomic mass then we divide the quotients obtained by the lowest ratio obtained in the first step.
C- 76.57/12, H= 6.43/1, O = 17.00/16
C- 6.38/1.06, H= 6.43/1.06, O= 1.06/1.06
C- 6, H- 6, O- 1
Empirical formula: C6H6O
[(12 ×6) + (6 × 1) + (16 × 1)]n=94.11
[72 + 6 +16]n = 94.11
n = 94.11/94
n= 1
Molecular formula = C6H6O
2)
C- 53.30/12, H- 11.19/1, O- 35.51/16
C- 4.44/2.22, H- 11.19/2.22, O- 2.22/2.22
C- 2, H- 5, O- 1
Empirical formula: C2H5O
[(2×12) + (5× 1) + (1×16)]n = 90.12
[24 + 5 + 16] n = 90.12
n= 90.12/45
n= 2
Molecular formula = C4H10O2
A frozen TV dinner contains 21 g of protein, 59 g of carbohydrate, and 18 g of fat. What is the total number of kilojoules (kJ) of potential energy within this TV
dinner? The accepted values for potential energy are 17 kJ per gram of protein, 17 kJ per gramof carbohydrate, and 38 kJ per gram of fat.
Round your answer to the nearest tens place and with the appropriate units.
Answer:
2040 kJ
Explanation:
Step 1: Calculate the energy provided by 21 g of protein
17 kJ are provided per gram of protein.
21 g × 17 kJ/g = 357 kJ
Step 2: Calculate the energy provided by 59 g of carbohydrate
17 kJ are provided per gram of carbohydrate.
59 g × 17 kJ/g = 1003 kJ
Step 3: Calculate the energy provided by 18 g of fat
38 kJ are provided per gram of fat.
18 g × 38 kJ/g = 684 kJ
Step 4: Calculate the total energy provided by the dinner
357 kJ + 1003 kJ + 684 kJ = 2044 kJ ≈ 2040 kJ
an emerald can be described as...
Answer:
green gemstone
Explanation:
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