Answer:
a)calculated molarity of NaOH would be lower
b) calculated molarity of NaOH would be lower
c) calculated molarity of NaOH would be lower
d) calculated molarity of NaOH would be unaffected
Explanation:
Let us recall that the reaction of NaOH and HCl is as follows;
NaOH(aq) + HCl(aq) ----> NaCl(aq) + H2O(l)
Since the reaction is 1:1, when the number of moles of HCl reacting with NaOH is low due to dilution, the calculated molarity of NaOH also becomes less than it's accurate value.
When 40mL of water is added to the titration flask rather than 25ml of water, the acid is more dilute hence less number of moles of acid than necessary reacts with the base thereby yielding a less than accurate value of the molarity of NaOH.
If the burette wet with water is not rinsed with NaOH solution, the concentration of the NaOH in the burette decreases due to dilution with water and a less than accuracy value is calculated for the molarity of NaOH.
If five drops of phenolphthalein is used instead of one or two drops, there is no qualms since enough phenolphthalein may be added to ensure that a sharp end point is obtained.
Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points from highest to lowest. CoCl3, NH4Cl, Li2SO4
Answer:
NH4Cl > Li2SO4 > CoCl3
Explanation:
Let us recall that the freezing point depression depends on the molality of the solution and the number of particles present.
Let us also recall that freezing point depression is a colligative property. It depends on the number of particles present in solution.
Usually, the more the number of particles present, the lower the freezing point. Hence, NH4Cl which has only two particles will have the highest freezing point while CoCl3 which has four particles will have the lowest freezing point.
11) Methane and oxygen react to form carbon dioxide and water. What mass of water is formed if 0.80 g of methane reacts with 3.2 g of oxygen to produce 2.2 g of carbon dioxide
A chemistry student needs 90.0mL of carbon tetrachloride for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of carbon tetrachloride is 1.59g*cm3- Calculate the mass of carbon tetrachloride the student should weigh out. Be sure your answer has the correct number of significant digits.
Answer:
volume = mass/density
Here, volume = 80g/1.59gcm-3 → 50.314 cm3
Explanation:
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 solution of hydrochloric acid had a hydrogen ion concentration of 1.0 mol/dm3
Water was added to hydrochloric acid until the ph increased by 1
What was the hydrogen ion concentration of the hydrochloric acid after had been added?
Answer:
pH = -log[H+]
Where [H+] = Hydrogen ion concentration
In this case,
[H+] = 1 × 10^(-2) = 10^(-2)
log{10^(-2)} = -2
-log{10^(-2)} = -(-2) = 2
pH = -log{10^(-2)} = 2
and hi.!!!
Answer:
0.1
Explanation:
Hydrogen ion concentration can be calculated using the formula [H+] = 10^-pH
pH can be concentrated using ph = -log[H+]
let's calculate the initial pH before anything was added: pH = -log(1) = 0
it increased by 1 so the final pH is 1.
Now we'll find the [H+] of a solution with a pH of 1:
concentration = 10^(-1) = 0.1
Rank the following compounds in order of decreasing boiling point: sodium chloride (NaCl), methane (CH4), and iodomethane (CH3I). Rank from highest to lowest boiling point.
Answer:
CH4< CH4I< NaCl
Explanation
NaCl has the boiling point of 1,413°C ( 2,575°F )
CH3I has a boiling point of 42°C ( 107°F )
CH4 has the boiling point of -161.6°C ( -258.9°F )
How many carbon atoms are there in 15 lbs of sugar, C12H22O11?
Answer:
A molecule of sucrose (C12H22O11) has 12 carbon atoms, 22 hydrogen atoms and 11 oxygen atoms.
Explanation:
if this does not help let me know :)
There are 1.4376 × 10²⁶ carbon atoms in 15 lbs of sugar (C12H22O11).
From the given information,
Using the standard conversion method;
1 lbs = 453.592 gram
∴
15 lbs = (453.592 gram × 15 lbs/1 lbs)
= 6803.88 grams
Now, we will need to determine the molar mass of the sugar compound C12H22O11.
Molar mass of C12H22O11 = (12 × 12) +(1 × 22) + (16 × 11)
Molar mass of C12H22O11 = 144 + 22 + 176
Molar mass of C12H22O11 = 342 g/mol
Using the relation:
[tex]\mathbf{Number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]
Number of moles = [tex]\dfrac{6803.88 }{ 342}[/tex]
Number of moles of C12H22O11 = 19.894 moles
Since we've known the number of moles present in C12H22O11, the next thing to do is determine the number of molecules of sugar by using Avogadro's constant:
i.e.
number of moles of sugar = [tex]19.894 moles \times \dfrac{6.023 \times 10^{23}}{mol}[/tex]
= 1.198 × 10²⁵ molecules of C12H22O11
Now to determine the number of carbon atoms in 15 lbs, we have:
= number of carbon atoms × amount of molecules
= 12 × 1.198 × 10²⁵ carbon atoms
= 1.4376 × 10²⁶ carbon atoms
Therefore, we can conclude that there are 1.4376 × 10²⁶ carbon atoms present in 15 lbs of sugar, C12H22O11
Learn more about atoms here:
https://brainly.com/question/657632?referrer=searchResults
Draw 2,3-dichloro octane
Answer:
Hi friend
I hope this image will help you if not I'm sorry
if this help you please mark me as brinalist or vote me.
Thankyou
Explain carefully what happen to the propanol-water system if approximately
50% of propanol by mass is fractionally distilled. What will be the distillate and
the residue?
Answer:
In the case of mixtures of ethanol and water, this minimum occurs with 95.6% by mass of ethanol in the mixture. The boiling point of this mixture is 78.2°C, compared with the boiling point of pure ethanol at 78.5°C, and water at 100°C. You might think that this 0.3°C doesn't matter much, but it has huge implications for the separation of ethanol / water mixtures. The next diagram shows the boiling point / composition curve for ethanol / water mixtures. I've also included on the same diagram a vapor composition curve in exactly the same way as we looked at on the previous pages about phase diagrams for ideal mixtures.
Write an equilibrium expression for each chemical equation involving one or more solid or liquid reactants or products.
Answer:
a.
[tex]Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}[/tex]
b.
[tex]Keq=[O_2]^3[/tex]
c.
[tex]Keq=\frac{[H_3O^+][F^-]}{[HF]}[/tex]
d.
[tex]Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}[/tex]
Explanation:
Hello there!
In this case, for the attached reactions, it turns out possible for us to write the equilibrium expressions by knowing any liquid or solid would be not-included in the equilibrium expression as shown below, with the general form products/reactants:
a.
[tex]Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}[/tex]
b.
[tex]Keq=[O_2]^3[/tex]
c.
[tex]Keq=\frac{[H_3O^+][F^-]}{[HF]}[/tex]
d.
[tex]Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}[/tex]
Regards!
Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:
element & mass %
phosphorus & 39.18%
sulfur & 60.82%
Write the molecular formula of X.
Answer:
Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:
element & mass %
phosphorus & 39.18%
sulfur & 60.82%
Write the molecular formula of X.
Explanation:
The given molecule of phosphorus and sulfur has molar mass --- 316.25 g.
Empirical formula calculation:
element: phosphorus sulfur
co9mposition: 39.185% 60.82%
divide with
atomic mass: 39.185/31.0 g/mol 60.82/32.0g/mol
=1.26mol 1.90mol
smallest mole ratio: 1.26mol/1.26mol =1 1.90mol/1.26 mol =1.50
multiply with 2: 2 3
Hence, the empirical formula is:
P2S3.
Mass of empirical formula is:
158.0g/mol
Given, molecule has molar mass --- 316.25 g/mol
Hence, the ratio is:
316.25g/mol/158.0 =2
Hence, the molecular formula of the compound is :
2 x (P2S3)
=[tex]P_4S_6[/tex]
a sample of salt has 1.74 moles of sodium chloride. how many formula units of the ionic compound are in the sample?
Answer:
A sample of salt has 1.74 moles of sodium chloride. how many formula units of the ionic compound are in the sample?
Explanation:
Given, 1.74 moles of NaCl.
Since one mole of NaCl consists of --- [tex]6.023 * 10^2^3[/tex] formula units.
Then, 1.74mol of NaCl contains how many formula units of NaCl?
[tex]1.74 mol x \frac{6.023x10^2^3}{1 mol} \\=10.5x10^2^3[/tex]formula units.
Hence, the given sample has 10.5x10^23 formula units.
A cylinder contains 3.1 L of oxygen at 300 K and 2.7 atm. The gas is heated, causing a piston in the cylinder to move outward. The heating causes the temperature to rise to 610 K and the volume of the cylinder to increase to 9.4 L.
How many moles of gas are in the cylinder?
Express your answer using two significant figures.
Answer: The moles of gas present in the cylinder is 0.34 moles.
Explanation:
Given: [tex]P_{1}[/tex] = 2.7 atm, [tex]V_{1}[/tex] = 3.1 L, [tex]T_{1}[/tex] = 300 K
[tex]P_{2}[/tex] = ?, [tex]V_{2}[/tex] = 9.4 L, [tex]T_{2}[/tex] = 610 K
Formula used to calculate the final temperature is as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{2.7 atm \times 3.1 L}{300 K} = \frac{P_{2} \times 9.4 L}{610 K}\\P_{2} = \frac{5105.7}{2820} atm\\= 1.81 atm[/tex]
Now, moles present upon heating the cylinder are as follows.
[tex]P_{2}V_{2} = n_{2}RT_{2}\\1.81 atm \times 9.4 L = n_{2} \times 0.0821 L atm/mol K \times 610 K\\n_{2} = \frac{17.014}{50.081} mol\\= 0.34 mol[/tex]
Thus, we can conclude that moles of gas present in the cylinder is 0.34 moles.
The enthalpy of Sodium is 235 calories. The enthalpy Chlorine is 435 calories. The enthalpy of Sodium chloride 670 joules, what is the change in enthalpy for this reaction?
Answer:
ΔH = -2446J
Explanation:
Based on the reaction:
2 Na(s) + Cl2(g) → 2NaCl
We can find the enthalpy of this reaction using Hess's law:
The enthalpy of a reaction is equal to the sum of the enthalpy of products times their reaction quotient subtracting the enthalpy of reactants times their reaction quotient. For the reaction of the problem:
ΔH = 2ΔH(NaCl) - [2ΔH(Na) + ΔHCl2)]
ΔH(NaCl) = 670J
ΔH(Na) = 235cal * (4.184J/1cal) = 983J
ΔHCl2 = 435cal * (4.184J/1cal) = 1820J
ΔH = 2*670J - [2*983J + 1820J]
ΔH = 1340J - [3786J]
ΔH = -2446JAnswer:
the heat content of a system at constant pressure
Explanation:
For the following reaction, 11.6 grams of sulfur are allowed to react with 23.8 grams of carbon monoxide .
sulfur(s) + carbon monoxide(g) sulfur dioxide(g) + carbon(s)
What is the maximum amount of sulfur dioxide that can be formed?
What is the formula for the limiting reagent?
What amount of the excess reagent remains after the reaction is complete?
Answer:
S + 2CO = SO2 + 2C
First, look for the amount of substance of sulfur:
n(S) = m / M
n(S) = 14.8 g/32 g / mol = 0.4625 mol
n(CO) = m (CO) / M (CO)
M(CO) = 12 + 16 = 28 g/mol
n(CO) = 19.9 g/28 g/mol = 0.71 mol
S in excess, so for calculating we take CO:
n(SO2) = n(CO)/2 = 0.71 mol/2 = 0.355 mol
m(SO2) = M(SO2)*n(SO2)
M(SO2) = 32 + 16*2 = 64 g/mol
m(SO2) = 64 g/mol * 0.355 mol = 22.74 g
Determine the molarity of the sodium ions when 78.0 g Na2S is dissolved in water for a final volume of 1.0 L.
Answer:
[Na⁺] = 1.99 M
Explanation:
Na₂S is a ionic salt that can be dissociated.
Dissociation equation is:
Na₂S → 2Na⁺ + S⁻²
1 mol of sodium sulfide can give 2 moles of sodium cation.
We convert moles of salt: 78 g . 1mol / 78.06 g = 0.999 moles
As ratio is 1:2, after dissociation we have (0.999 . 2) = 1.998 moles of Na⁺
Molarity is a type of concentration.
It indicates moles of solute in 1 L of solution and in this case, we have 1 L as final voulme.
Moles of Na⁺ are 1.998 moles. Then molarity (mol/L) is:
M =1.99 mol/L
What types of house are found in very cold regions
Answer:
the house found in a very cold are Igloos or the ice house
the pressure of a sample of helium in a 0.150 L container is 1520 torr. if the helium is compressed to the volume of 0.012 L without changing the temperature what would be the pressure of the gas
Answer:
19000 torr
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 0.150 L
Initial pressure (P₁) = 1520 torr
Final volume (V₂) = 0.012 L
Temperature = constant
Final pressure (P₂) =?
The final pressure of the gas can be obtained by using the Boyle's law equation as illustrated below:
P₁V₁ = P₂V₂
1520 × 0.150 = P₂ × 0.012
228 = P₂ × 0.012
Divide both side by 0.012
P₂ = 228 / 0.012
P₂ = 19000 torr
Thus, the final pressure of the gas is 19000 torr
In the reactionpyruvate lactatethat is catalyzed by the enzyme lactate dehydrogenase, the compound on which the enzyme works, pyruvate, is called the _______.
Answer:
Substrate
Explanation:
In biochemical sciences, a substrate is a substance that is acted upon by an enzyme to yield a product. Enzymes are known for catalyzing biochemical reactions. The substances that are usually worked with during this catalytic process are termed as SUBSTRATES.
Substrates, which are usually changed during the process, binds to the active site on the enzyme and form an enzyme-substrate complex.
According to this question, pyruvate is converted to lactate in a reaction that is catalyzed by the enzyme lactate dehydrogenase. This means that the compound on which the enzyme works, pyruvate, is called the SUBSTRATE.
According to the following pKa values listed for a set of acids, which would lead to the strongest conjugate base?
a. -2.
b. 1.
c. 7.
d. 25.
e. 50.
Predict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Be sure the equation you submit is balanced. (You can edit both sides of the equation to balance it, if you need to.)
______________ → BaBr2 + H2O
Answer:
Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O
Explanation:
We have the products of a reaction and we have to predict the reactants. Since the products are binary salt and water, this must be a neutralization reaction. In neutralizations, acids react with bases. The acid that gives place to Br⁻ is HBr, while the base the gives place to Ba²⁺ is Ba(OH)₂. The balanced chemical equation is:
Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O
this was in my science test just want to see if anyone knows?? the wording is so confusing
the relative atomic mass of an element compares the mass of an atom of an element with the mass of an atom of.......
Answer:
the molar mass of the element
6. Which compound contains no ionic character?
Answer:
The compound which doesn't contains ionic character is HC, H-atom and CL- atom shares 1 electron a to form covalent bond....
Draw the skeletal structure for: (E)-hept-5-en-2-one
Answer:
Draw the skeletal structure for: (E)-hept-5-en-2-one
Explanation:
The root word hept indicates that the given compound has seven carons in its longest chain.
-en- primary suffix indicates that the compound has one double bond in it.
2-one indicates that the compound has -C=O bond in the second carbon.
The prefix (E) indicates that the highest priority groups are on the opposite direction of the double bond.
The structure of the given molecule is:
The size of an atomic orbital is associated with:______________
a. the magnetic quantum number (ml).
b. the spin quantum number (ms).
c. the angular momentum quantum number (l).
d. the angular momentum and magnetic quantum numbers, together.
e. the principal quantum number (n).
Answer:
e. the principal quantum number (n).
Explanation:
The size of the orbital is governed and decided by the principal quantum number n, which is dependent on the overall average distance between the number of electrons as well as the nucleus. The orbital's shape is explained by the angular quantum number. The magnetic quantum number is concerned with the orbital's orientation in space. The quantum number's spin explains the spin of the electrons.
How much carbon dioxide is released when it is fully combusted with 4Kg of ethanol with more than enough oxygen? How do you work it out?
Answer:
7.640 kg
Explanation:
Step 1: Write the balanced complete combustion equation for ethanol
C₂H₆O + 3 O₂ ⇒ 2 CO₂ + 3 H₂O
Step 2: Calculate the moles corresponding to 4 kg (4000 g) of C₂H₆O
The molar mass of C₂H₆O is 46.07 g/mol.
4000 g × 1 mol/46.07 g = 86.82 mol
Step 3: Calculate the moles of CO₂ released
86.82 mol C₂H₆O × 2 mol CO₂/1 mol C₂H₆O = 173.6 mol CO₂
Step 4: Calculate the mass corresponding to 173.6 moles of CO₂
The molar mass of CO₂ is 44.01 g/mol.
173.6 mol × 44.01 g/mol = 7640 g = 7.640 kg
1. Calculate the percent recovery of benzoic acid, naphthalene and 3-nitroaniline if you were able to collect 10.75 g of benzoic acid, 5.41 g of naphthalene, and 7.81 g of 3-nitroaniline from a set of extractions. The starting mass of the mixture was 25.04 g. (0.6 pt) 2. Describe why it is important to use sodium hydroxide and hydrochloride acid in this experiment. Why was it necessary to initially start off with a 5% solution of the acid or base for this experiment
Explanation:
1.)
mass = 25.04
percentage recovery
[tex]benzoic acid = \frac{10.75}{25.04} = 0.4293*100 = 42.93%\\[/tex]
[tex]naphtalene =\frac{5.41}{25.04} = 0.2160*100 = 21.60\\3-nitroaniline=\frac{7.81}{25.04} =0.3119*100=31.19[/tex]
2. This experiment has these compounds, benzoic acid (which is an acid), naphthalene (this is neutral) and 3-nitroaniline (this is a base).
to extract, 5 percent of NaOH has to be used in order for benzoic acid to become with sodium hydroxide. the salt would then dissolve in H2O, the other two remaining are going to dissolve in organic layer. and this would make benzoic acid to leave the mix.
we make use of 5 percent of HCl so that the 3-nitroaniline will turn into ammonium salt with the hcl, then the ammonium salt would dissolve in water and naphtalene would become soluble in organic layer. when this happens we would then have the three compounds separated.
Diethyl ether (C2H5 )2O vaporizes at room temperature. If the vapor exerts a pressure of 233 mm Hg in a flask at 25 °C, what is the density of the vapor?
Answer: The density of the given vapor is 0.939 g/L.
Explanation:
Given: Pressure = 233 mm Hg (1 mm Hg = 0.00131579 atm) = 0.31 atm
Temperature = [tex]25^{o}C[/tex] = (25 + 273) K = 298 K
According to the ideal gas equation,
[tex]PV = \frac{m}{M}RT[/tex]
where,
P = pressure
V = volume
m = mass
M = molar mass
R = gas constant = 0.0821 L atm/mol K
T = temperature
This formula can be re-written as follows.
[tex]PM = \frac{m}{V}RT[/tex] (where, [tex]Density = \frac{mass (m)}{Volume (V)}[/tex] )
Hence, formula used to calculate density of diethy ether (molar mass = 74.12 g/mol) vapor is as follows.
[tex]d = \frac{PM}{RT}[/tex]
Substitute values into the above formula as follows.
[tex]d = \frac{PM}{RT}\\= \frac{0.31 atm \times 74.12 g/mol}{0.0821 L atm/mol K \times 298 K}\\= \frac{22.9772}{24.4658}\\= 0.939 g/L[/tex]
Thus, we can conclude that the density of the given vapor is 0.939 g/L.
What is the name for CH 3 CH 2 COCHCH 3 CH(CH 3 ) 2 ?
Answer:
Butanoic acid
Explanation:
The IUPAC name of CH3CH2CH2COOH is:
The IUPAC name for a given compound is Butanoic acid.
3. A 3.455-g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed. If 0.2815 g of barium sulfate was obtained, what was the mass percentage of barium in the sample
Answer:
[tex]Ba\ percentage\ in\ Mass=4.8\%[/tex]
Explanation:
From the question we are told that:
Mass of mixture [tex]m=3.455g[/tex]
Mass of Barium [tex]m_b=0.2815g[/tex]
Equation of Reaction is given as
[tex]Ba2+ + H2SO4 => BaSO4 + 2 H+[/tex]
Generally the equation for Moles of Barium is mathematically given by
Since
[tex]Moles of Ba^{2+} = Moles of BaSO_4[/tex]
Therefore
[tex]Moles of Ba^{2+} = \frac{mass}{molar mass of BaSO4}[/tex]
[tex]Moles of Ba^{2+} = \frac{0.2815}{233.39}= 0.0012061 mol[/tex]
Generally the equation for Mass of Barium is mathematically given by
[tex]Mass\ of\ Ba^{2+} = Moles * Molar mass of Ba^{2+}[/tex]
[tex]Mass\ of\ Ba^{2+} = 0.0012061 * 137.33 = 0.1656 g[/tex]
Therefore
[tex]Ba\ percentage\ in\ Mass = mass of Ba^{2+}/mass of sample * 100%[/tex]
[tex]Ba\ percentage\ in\ Mass= \frac{0.1656}{ 3.455 }* 100%[/tex]
[tex]Ba\ percentage\ in\ Mass=4.8\%[/tex]