Answer:
pH = 13.1
Explanation:
Hello there!
In this case, according to the given information, we can set up the following equation:
[tex]HCl+KOH\rightarrow KCl+H_2O[/tex]
Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:
[tex]n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol[/tex]
Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:
[tex]n_{KOH}=0.00768mol-0.00192mol=0.00576mol[/tex]
And the resulting concentration of KOH and OH ions as this is a strong base:
[tex][KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M[/tex]
And the resulting pH is:
[tex]pH=14+log(0.131)\\\\pH=13.1[/tex]
Regards!
What type of reaction is 3Mg + N2 → Mg3N2 ?
I’m confused and I don’t understand this, can someone help?
Wilma- tt
Wilbur- TT
c) 100% chance
d) 100% chance
it has been a while since I've done this so it could be wrong but im pretty sure its correct
Wilbur's family claims it is pure so they have TT
and his wife simply has big eyes so its tt
so the chances of their children having either eye shape is 100%
I'm sorry- im not too good at explaining it ><
A percent is a ratio expressed per 100 units. If you have 20 balls, and 1 is blue, you can say that 5% (or 5 out of 100 balls) is blue. Similarly, if a solution is 5% KI by mass, you can say that 5 g per 100 g solution is KI. When mass percent is used as a conversion factor, it can be expressed in the units required to solve the problem provided that the ratio remains constant. If the numerator is divided by 100, the denominator must also be divided by 100. Two examples are: express the grams of solute per 100 grams of solution, or express the centigrams (cg) of solute per 1 g of solution The latter example works because 100 cg is equivalent to 1 g. Complete the following statements regarding how many grams of KI are found in either 100 g or 1 kg of a 2.5% solution of KI.
a. A 2.5% (by mass) solution concentration signifies that there is ______ of solute in every 100 g
b. Therefore, when 2.5% is expressed as a ratio of solute mass over solution mass, that mass ratio would be ________
c. A solution mass of 1 kg is _______ times greater than 100 g, thus one kilogram (1 kg) of a 2.5% KI solution would contain ____________of
Answer:
a. 2.5 g
b. 0.025
c. i. 10 times ii 25 g of KI
Explanation:
a. A 2.5% (by mass) solution concentration signifies that there is ______ of solute in every 100 g
Since % by mass = mass of solute, m/mass of solvent,M × 100 %
2.5 % = m/M × 100%
since M = 100 g,
2.5 % = m/100 × 100 %
m/100 = 2.5/100
m = 2.5 g
b. Therefore, when 2.5% is expressed as a ratio of solute mass over solution mass, that mass ratio would be ________
The mass ratio = mass of solute/mass of solvent
= m/M
= 2.5 g/100 g
= 0.025
c A solution mass of 1 kg is _______ times greater than 100 g, thus one kilogram (1 kg) of a 2.5% KI solution would contain ____________of
i. A solution mass of 1 kg is _______ times greater than 100 g
Since 1 kg = 1000 g, 1 kg/100 g = 1000 g/100 g = 10
So,1 kg is 10 times greater than 100 g.
ii. Thus one kilogram (1 kg) of a 2.5% KI solution would contain ____________of KI
So, if M = 1 kg = 1000 g, we find m
Since % by mass = mass of solute, m/mass of solvent,M × 100 %
2.5 % = m/M × 100%
since M = 1000 g,
2.5 % = m/1000 × 100 %
m/1000 = 2.5/100
m = 2.5/100 × 1000 g
m = 2.5 × 10 g
m = 25 g
3. Suppose you had titrated your vinegar sample with barium hydroxide instead of sodium hydroxide:
Ba(OH)2(aq)+2CH3COOH(aq)⟶Ba(CH3COO)2(aq)+2H2O(l)
Consider a 0.586 M aqueous solution of barium hydroxide,
What volume (in mL) of 0.586 M Ba(OH)2 solution are required to neutralize 10 ml of vinegar containing 2.78 g of acetic acid?
Answer:
39.6 mL
Explanation:
Step 1: Write the balanced neutralization reaction
Ba(OH)₂(aq) + 2 CH₃COOH(aq) ⟶ Ba(CH₃COO)₂(aq) + 2 H₂O(l)
Step 2: Calculate the moles corresponding to 2.78 g of CH₃COOH
The molar mass of CH₃COOH is 60.05 g/mol.
2.78 g × 1 mol/60.05 g = 0.0463 mol
Step 3: Calculate the moles of Ba(OH)₂ needed to react with 0.0463 moles of CH₃COOH
The molar ratio of Ba(OH)₂ to CH₃COOH is 1:2. The moles of Ba(OH)₂ needed are 1/2 × 0.0463 mol = 0.0232 mol.
Step 4: Calculate the volume of 0.586 M solution that contains 0.0232 moles of Ba(OH)₂
0.0232 mol × 1 L/0.586 mol = 0.0396 L = 39.6 mL
What could have caused the size of the moon jelly population to increase? Claim 1: A change to the zoo plankton population caused births to increase in the moon jelly population. Claim 2: A change to the sea turtle population caused deaths to decrease in the Moon jellies population
Answer:
i think is claim 1
Explanation:
moon jellies eat zooplankton and the zooplankton population increased
Answer:It’s important to know WHY. Great explanations are what makes Brainly awesomeExplain everything. Imagine that the person who was asking, knows nothing at all so you need to start from the very beginningAlways include the steps and/or background required to get to the final answer. Let’s help other people understand and solve future problems on their ownShow all the calculation if it is a numerical question.
Explanation:
The diagram above shows repeating groups of atoms that make up to samples. Will the properties of the two samples likely be the same or different? (Examples of properties are smell, color, and the temperature at which a substance melts)
Answer:
I would need to see the diagram in order to answer properly
A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the next few minutes?
-Molecules in both the metal and the surrounding air will start moving at lower speeds. -Molecules in both the metal and the surrounding air will start moving at higher speeds. -The air molecules that are surrounding the metal will slow down, and the molecules in the metal will speed up.
-The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Answer:
until the next harvest, and seed must be held for the next season's ... successful grain storage is the moisture content of the crop. ... or both. If ambient temperatures are low, then air alone may cool the ... allow some of the drying to take place naturally while the crop ... employed to cool grain that has been placed in storage.
The exhaust from car engines spreads throughout our atmosphere and adds to pollution. What is this an example of?
A.
An increase in enthalpy
B.
An increase in thrust
C.
An increase in thermodynamics
D.
An increase in entropy
(D is the correct answer!)
Answer: When exhaust from car engines spreads throughout our atmosphere and adds to pollution then it is an example of increase in entropy.
Explanation:
A degree of randomness in the molecules of a substance is called entropy.
When exhaust from car engines spreads throughout our atmosphere then it means the molecules are moving at a faster speed due to which they spread into the atmosphere rapidly.
This means that the entropy of exhaust is increasing.
Thus, we can conclude that when exhaust from car engines spreads throughout our atmosphere and adds to pollution then it is an example of increase in entropy.
An increase in entropy
Calculate the mass of sucrose needed to prepare a 2000 grams of 2.5% sucrose solution.
Answer:
mass of sucrose = 17.115 grams
Explanation:
Given that:
mass = 2000 grams
molar mass of sucrose = 342.3 g/mol
2.5% of the sucrose soltion.
Let assume we are given 1 M of the sucrose solution;
2.5% of 1 M = 0.025 M
∴
Molarity = ( mass/molar mass )(1000/V)
mass = (0.025 * 342.3 * 2000)/(1000)
mass of sucrose = 17.115 grams
The chemical equation below is unbalanced.
NaOH + H3PO4 -> Na3PO4 + H2O
When the equation is correctly balanced, what will be the coefficient of H2O?
Answer:
3 NaOH (aq) + H3PO4 (aq) → Na3PO4 (aq) + 3 H2O (l)
Explanation:
This is an acid-base reaction (neutralization): NaOH is a base, H3PO4 is an acid.
When the equation is balanced, and the coefficient of H₂O is 3.
To balance the chemical equation:
NaOH + H₃PO₄ ⇒ Na₃PO₄ + H₂O
The number of atoms of each element is the same on both sides of the equation.
Balance the hydrogen atoms. There are three hydrogen atoms on the left side and two hydrogen atoms on the right side. To balance the hydrogen atoms, multiply H₂O by a coefficient of 3:
NaOH + H₃PO₄ ⇒ Na₃PO₄ + 3H₂O
If the oxygen (O) atoms are balanced. There are four oxygen atoms on the right side, but only two oxygen atoms on the left side. To balance the oxygen atoms, we need to multiply NaOH by a coefficient of 2.
To learn more about the chemical equation, follow the link:
https://brainly.com/question/28792948
#SPJ6
The masses of carbon and hydrogen in samples of four pure hydrocarbons are given above. The hydrocarbon in which sample has the same empirical formula as propene, C3H6
Sample Mass of carbon Mass of hydrogen
A 60g 12g
B 72g 12g
C 84g 10g
D 90g 10g
a. Sample A
b. Sample B
c. Sample C
d. Sample D
Answer:
Sample B
Explanation:
In this case, we need to determine the empirical formula for each sample. The one that match the formula of the propene would be the sample.
Let's do Sample A:
C: 60 g; H: 12 g
1. Calculate moles:
We need the atomic weights of carbon (12 g/mol) and hydrogen (1 g/mol):
C: 60 / 12 = 5
H: 12 / 1 = 12
2. Determine number of atoms in the formula
In this case, we just divide the lowest moles obtained in the previous part, by all the moles:
C: 5 / 5 = 1
H: 12 / 5 = 2.4 or rounded to two
3. Write the empirical formula:
Now, the prior results, represent the number of atoms in the empirical formula for each element, so, we put them with the symbol and the atoms as subscripted:
C₁H₂ = CH₂
Therefore, sample A is not the same as propene.
Sample B:
C: 72 g H: 12 g
Following the same steps, let's determine the empirical formula for this sample
C: 72 / 12 = 6 ---> 6 / 6 = 1
H: 12 / 1 = 12 ----> 12 / 6 = 2
EF: CH₂
Sample C:
C: 84 g H: 10 g
C: 84 / 12 = 7 ----> 7 / 7 = 1
H: 10 / 1 = 10 ----> 10 / 7 = 1.4 or just 1
EF: CH
Sample D
C: 90 g H: 10 g
C: 90 / 12 = 7.5 -----> 7.5 / 7.5 = 1
H: 10 / 1 = 10 -------> 10 / 7.5 = 1.33 or just 1
EF: CH
Neither compound has the same empirical formula as C3H6, but C3H6 is a molecular formula, so, if we just simplify the formula we have:
C3H6 -----> CH₂
Therefore, sample B is the one that match completely. Sample B would be the one.
Hope this helps
. Apply a coat of ____ to keep a wooden ladder structurally sound and functioning well. A. metallic paint B. oil-based paint C. linseed oil D.
Answer:
oil based paint
Explanation:
1.the volume of air in a mountain bicycle tire is constant. the initial pressure and temperature are 3.14 atm and 26 degrees celsius.
a) as the bike is ridden friction with the road causes the tire to heat up. at what kelvin temperature will the gas in the tire have a pressure of 3.96 atm
b) explain why your answer makes sense at the particle level. use the words collisions, force, and frequency in your responde.
Answer:
Explanation:
Givens
P1 = 3.14 atm
T1 = 26 + 273 = 299
P2 = 3.96
T2 = ?
Equation
P1 / T1 = P2 / T2
Solution
3.14 / 299 = 3.96 / x Cross multiply
3.14 x = 299 * 3.96 Divide by 3.14
x = 299*3.96 / 3.14
x = 1184.04/3.14
x = 377.08 oK Take off 273 to get the degrees Centigrade.
x = 104.08 oC
Part B
If you want the temperature to increase, the relationship is a direct one. That means as the temperature goes up, the pressure will as well. So for a bicycle, it makes sense that the volume remains constant. That makes the formula above without volume because the volume would cancel out on both sides of the equation.
So as the pressure increases, the particles making up the air begin to respond by the molecules moving faster. Temperature really is the a measurement of how fast molecules are moving. So the momentum is increasing. When the particles hit something, it is with a much greater force and that increases the apparent pressure.
How many moles of iron are needed to produce 10 moles of Fe2O3
Answer:
20 moles of Iron are needed
Explanation:
In 1 mole of Iron (III) oxide, Fe₂O₃, there are 2 moles of Iron, Fe. The equation that we can write is:
2 moles Fe = 1 mole Fe₂O₃
To find the moles of iron that are needed to produce 10 moles of Fe₂O₃ we must use this equation as follows:
10 moles of Fe₂O₃ * (2 moles Fe / 1mole Fe₂O₃) =
20 moles of Iron are neededQuestion 6
4 pts
6) Chromium crystallizes in a body-centered cubic unit cell. If the length of an edge of the unit cell
is 289 pm, what is the density (in g/cm3) of chromium?
Show your work on a paper and upload it here.
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Choose a File
Answer:
7.15g/cm³
Explanation:
To solve this question we must know that a body-centered cubic unit cell contains 2 atoms.
The volume of the unit cell is:
289pm = (289x10⁻¹²m)³ =
2.414x10⁻²⁹m³ * (1cm³ / 1x10⁻⁶m³) = 2.414x10⁻²³cm³
And the mass is -Molar mass Chromium = 51.9961g/mol:
2atoms * (1mol / 6.022x10²³atoms) * (51.9961g / mol) =
1.727x10⁻²²g
The density is:
1.727x10⁻²²g / 2.414x10⁻²³cm³ =
7.15g/cm³If 56.8 g of FeCl, are formed, how many grams of Fe reacted?
(Explain or show work plz)
Answer:
the molar mass for Fe and Cl
The following data were obtained by gas-liquid chromatography on a 40-cm packed column:
Compound tR, min W, min
Air 1.9 —
Methylcyclohexane 10 0.76
Methylcyclohexene 10.9 0.82
Toluene 13.4 1.06
Calculate:
a. an average number of plates from the data.
b. the standard deviation for the average in (a).
c. an average plate height for the column.
Answer:
a)
For Methylcyclohexane N₁ = 2770
For Methylcyclohexene N₂ = 2827
For Toluene N₃ = 2557
b) the standard deviation for the average in (a) is 143.31
c)
For Methylcyclohexane; H₁ = 0.01444 cm
For Methylcyclohexene; H₂ = 0.01415 cm
For Toluene; H₃ = 0.01564 cm
Explanation:
Given the data in the question;
Gas-liquid chromatography on a 40-cm packed column:
Compound tR, min W, min
Air 1.9 —
Methylcyclohexane 10.0 0.76
Methylcyclohexene 10.9 0.82
Toluene 13.4 1.06
a)
an average number of plates from the data;
To get the Number of plates N, we use the following expression;
N = 16( tR / W )², we use it for Methylcyclohexane, Methylcyclohexene and Toluene
-
For Methylcyclohexane N₁ = 16( 10 / 0.76 )² = 16( 173.13 ) = 2770.08 ≈ 2770
-
For Methylcyclohexene N₂ = 16( 10.9 / 0.82 )² = 16( 176.7 ) = 2827.2 ≈ 2827
-
For Toluene N₃ = 16( 13.4 / 1.06 )² = 16( 159.8078 ) = 2556.9 ≈ 2557
b) standard deviation for the average in (a).
First we get the mean;
Mean N" = ( N₁ + N₂ + N₃ ) / 3 = ( 2770 + 2827 + 2557 ) / 3 = 8154 / 3 = 2718
Next we determine the deviation
d₁² = (N₁ - N")² = (2770 - 2718)² = (52)² = 2704
d₂² = (N₂ - N")² = (2827 - 2718)² = (109)² = 11881
d₃² = (N₃ - N")² = (2557 - 2718)² = (-161)² = 25,921
∴ ∑d²[tex]_i[/tex] = 40506
Standard Deviation S = √( ∑d²[tex]_i[/tex] / ( n-1 ) )
Standard Deviation S = √( 40506 / ( 3-1 )
Standard Deviation S = √( 40506 / 2 )
Standard Deviation S = √( 20253 )
Standard Deviation S = 143.31
Therefore, the standard deviation for the average in (a) is 143.31
c)
an average plate height for the column
Given that; Gas-liquid chromatography on a 40-cm packed column, L = 40 cm
-
For Methylcyclohexane; H₁ = L/N₁ = 40 / 2770 = 0.01444 cm
-
For Methylcyclohexene; H₂ = L/N₂ = 40 / 2827 = 0.01415 cm
-
For Toluene; H₃ = L/N₃ = 40 / 2557 = 0.01564 cm
In each pair of compounds, pick the one with the higher vapor pressure at a given temperature. Explain your reasoning.
a. CH4 or CH3CI
b. CH3CH2CH2OH or CH3OH
c. CH3OH or H2CO
Answer:
The answer is:
[tex]PART \ A: CH_4\\\\PART \ B: CH_30H\\\\PART \ C: H_2CO[/tex]
Explanation:
Parts A:
The vapor pressure is higher in [tex]CH_4[/tex] because it is non-polar, while [tex]CH_3Cl[/tex] is polar. [tex]CH_4[/tex] has a lower molar weight as well.
Part B:
Although hydrogen bonding is found in both commodities, the vapor pressure is higher because of the smaller molar mass of [tex]CH_30H[/tex].
Part C:
[tex]H_2CO[/tex] does not show hydrogen due to the increased vapor pressure
[tex]CH_3OH[/tex] bonding.
Using stoichiometry, you predict that you should be able to use 314.0 g of Al to produce 1551 g of AlCi3. In your lab
exercise you actually produced 1400.0 g of aluminum chloride. What is your % yield for this reaction?
CORRECT ANSWER IS: 90.26% but what are the steps on how to get this answer ?
Answer:
90.26%
Explanation:
From the question given above, the following data were obtained:
Theoretical yield of AlCl₃ = 1551 g
Actual yield of AlCl₃ = 1400 g
Percentage yield =?
The percentage yield of the reaction can be obtained as follow:
Percentage yield = Actual yield / Theoretical yield × 100
Percentage yield = 1400 / 1551 × 100
Percentage yield = 140000 / 1551
Percentage yield = 90.26%
Thus, the percentage yield of the reaction is 90.26%
A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the temperature changed to:
(a) -5 ºC
(b) 95 ºF
(c) 1095 K
Answer: The new volume at different given temperatures are as follows.
(a) 109.81 mL
(b) 768.65 mL
(c) 18052.38 mL
Explanation:
Given: [tex]V_{1}[/tex] = 571 mL, [tex]T_{1} = 26^{o}C[/tex]
(a) [tex]T_{2} = 5^{o}C[/tex]
The new volume is calculated as follows.
[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL[/tex]
(b) [tex]T_{2} = 95^{o}F[/tex]
Convert degree Fahrenheit into degree Cesius as follows.
[tex](1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C[/tex]
The new volume is calculated as follows.
[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL[/tex]
(c) [tex]T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C[/tex]
The new volume is calculated as follows.
[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL[/tex]
How many sigma bond and pi bond are present
Sample A: 300 mL of 1M sodium chloride
Sample B: 145 mL of 1.5 M sodium chloride
Which sample contains the larger concentration of sodium chloride?
Answer:
Sample A
Explanation:
Molarity of a solution can be calculated using the formula:
Molarity = number of moles (mol) ÷ volume (vol)
For Sample A:
V = 300ml = 300/1000 = 0.3 L
Molarity = 1M
n = number of moles (mol)
1 = n/0.3
n = 0.3moles
For Sample B:
V = 145 mL = 145/1000 = 0.145L
Molarity = 1.5 M
n = number of moles
1.5 = n/0.145
n = 1.5 × 0.145
n = 0.22 moles
Based on the above results (moles), sample A with 0.3 moles contains the larger concentration of sodium chloride.
30. Which of the following atoms would form
a trivalent anion?
A. Aluminium
B. Oxygen
C. Lithium
D. Nitrogen
Aluminium and Nitrogen!
What is the molarity of a solution that contains 152 g NaCl in 4.00 L solution?
Answer:
200.0lg
Explanation:
please give a brainliest
Boyle's Law represents the relationship of volume as pressure is changed. The constants of Boyle's Law are
temperature and number of molecules of gas
Answer:
True.
Explanation:
Hello!
In this case, according to the ideal gas equation:
[tex]PV=nRT[/tex]
It turns out possible for us to obtain the Boyle's law, as a directly proportional relationship between pressure and volume, considering that both moles and temperature remain constant; in such a way, we write the following:
[tex]PV=k[/tex]
Therefore, for an initial (1) and final state (2) we are able to write:
[tex]P_2V_2=P_1V_1[/tex]
Regards!
When liquid water freezes into solid ice in the freezer,
Answer:
What about it? I don't get the question
A solution is prepared by dissolving 16.90 g of ordinary sugar (sucrose, C12H22O11, 342.3 g/mol) in 40.90 g of water. Calculate the boiling point of the solution. Sucrose is a nonvolatile nonelectrolyte.
Answer:
Explanation:
The boiling point will increase due to dissolution of sugar in water . Increase in boiling point ΔT
ΔT = Kb x m , where Kb is molal elevation constant water , m is molality of solution
Kb for water = .51°C /m
moles of sugar = 16.90 / 342.3
= .04937 moles
m = moles of sugar / kg of water
= .04937 / .04090
= 1.207
ΔT = Kb x m
= .51 x 1.207
= .62°C .
So , boiling point of water = 100.62°C .
4. Eggs are present in the ovaries when people are born, and after they go
through puberty. *
OTrue
OFalse
What is the usable form of nitrogen that is found in the ground?
4-ethylbenzoyl chloride structure
Answer:
Ethylbenzoyl chloride | C9H9ClO