A 0.506 g sample of solid calcium hydroxide was dissolved in some water and then titrated with 28.85 mL of hydrochloric acid to neutralize, then the molarity of HCl is 0.471 M.
How do we calculate molarity?Molarity will be calculated by using the below equation as:
M = n/V, where
V = volume of solvent
n is the moles of solute and it will be calculated by using the below equation:
n = W/M, where
W = given mass
M = molar mass
Given chemical reaction is:
Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
Moles of Ca(OH)₂ = 0.506g / 74.093g/mol = 0.0068 mol
From the stoichiometry of the reaction:
0.0068 moles of Ca(OH)₂ = reacts with 2×0.0068 = 0.0136 moles of HCl
Given volume of HCl = 28.85mL = 0.02885 L
On putting values on the molarity equation, we get
M = 0.0136 / 0.02885 = 0.471 M
Hence required molarity of HCl is 0.471 M.
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Water is a _____ molecule, which gives it many of its unique properties, including its ability as a universal solvent.
Answer:
The answer is Polar.
Explanation:
Because polar has many unique properties.
The blood carries nutrients to where they are needed. True False
Answer:
True
Explanation:
Answer:
true
Explanation:
Select all the correct answers.
Which two generalizations can be made based on what you know about cycles of matter in a closed system?
o New matter is added, and old matter is destroyed.
O Matter changes its physical form, allowing it to return to its original state.
O The amount of matter within the system remains the same.
O Matter and energy can cross the boundaries of the system.
The cycle has a well-defined starting and stopping point.
PLSS HELP
Answer:
3) The amount of matter within the system remains the same
5) The cycle has a well-defined starting and stopping point.
Explanation:
A closed system is a system that the boundaries of a system cannot be passed by either matter or energy.
Hence, generalisations can be made based on a closed system and the law of conservation of mass which states that matter can neither be created nor destroyed but can only change form.
sodium carbonate (+heat) →
Answer:
heat carbunate
Explanation:
thats true im good in here like and heart byebye
What state of matter has the most kinetic energy?
A gas
B liquid
C solid
Answer:
gaseous state.
Explanation:
that's the answer
Best way to deal with calculations in chemistry?
Answer: Look up a video, and they will tell/show you.
Which equation shows how frequency is related to velocity and wavelength?
The equation that shows how frequency is related to velocity and wavelength is f = v / λ (Option A)
What is frequency?This is defined as the number of oscillations made in one second. It is expressed as
Frequency (f) = 1 / period (T)
f = 1 / T
Relationships between frequency, wavelength and velocityThe velocity, frequency and wavelength of a wave are related according to the following equation
v = λf
Making f the subject, we have
f = v / λ
Thus, we can conclude that the correct answer to the question is:
f = v / λ (Option A)
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The correct IUPAC name for 2-isopropylbutane is: a) 2-ethylpentane b) 2,3-dimethylpentane c) 2-methylhexane d) 3-methylhexane
Answer:
a
Explanation:
On drawing the structure, the most Carbon atoms in a row is 5
⇒ pentaneThe remaining two carbon atoms are connected to the 2nd carbon atom
2-ethylIUPAC name is :
2-ethyl + pentane2-ethylpentaneHow many atoms of oxygen are present: 3Al2(SO4)3
a. 36
b. 12
c. 10
d. 24
Answer:
the answer is 12
Explanation:
In one formula unit of Al2(SO4)3 A l 2 ( S O 4 ) 3 , there would be twelve atoms of oxygen.
PROBLEM SETS: %BY MASS, % BY VOLUME, MOLARITY, MOLALITY (show your solution)
1. What is the molarity of a solution in which 0.850 grams of ammonium nitrate are dissolved in 345 mL of solution?
2. Calculate the molality of a solution of 13.5g of KF dissolved in 250. g of water.
3. Calculate the molality of a solution containing 16.5 g of dissolved naphthalene (C10H8) in 54.3 g benzene (C6H6).
4. What is the mass percent of each component in the mixture formed by adding 12 g of calcium sulfate, 18 g of sodium nitrate, and 25 g of potassium chloride to 500 g of water?
5. A solution is made by dissolving 125 g of sodium chloride in 1.5 kg of water. What is the percent by mass?
6. What is the percent by volume of a solution formed by added 15 L of acetone to 28 L of water?
The concentration of a substance can be expressed using molarity, molality or percent.
What is concentration?The term concentration has to do with the amount of substance present in solution. Now let us solve the problems individually;
a) Number of moles = 0.850 grams/80 g/mol = 0.011 moles
molarity = 0.011 moles/345 * 10^-3 L = 0.032 M
b) Number of moles = 13.5g /58 g/mol = 0.23 moles
molality = 0.23 moles/250 * 10^-3 Kg = 0.92 m
c) Number of moles = 16.5 g/128 g/mol = 0.13 moles
molality = 0.13 moles/54.3 * 10^-3 Kg =2.39 m
d) Total mass present = 12 g + 18 g + 25 g + 500 g = 555 g
mass percent of calcium sulfate = 12 g/555 g * 100/1 = 2.2 %
mass percent of sodium nitrate = 18 g/ 555 g * 100/1 = 3.2 %
mass percent of potassium chloride = 25 g / 555 g * 100/1 = 4.5%
mass percent of water = 500 g / 555 g * 100/1 =90.1%
e) Total mass present = 125 g + 1500g = 1625 g
Mass percent of NaCl = 125 g/1625 g * 100/1 = 7.7%
f) Total volume of solution = 15 L + 28 L = 43 L
percent by volume of acetone = 15 L/43 L * 100/1 = 34.9%
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2AI(NO3)3 + 3Na2CO3 → Al2(CO3)3(s) + 6NaNO3
What is the ratio of moles of AI(NO3)3 to moles Na2CO3?
The ratio of moles of AI(NO3)3 to moles Na2CO3 in the reaction is 2:3
In the reaction 2 moles of AI(NO3)3 reacted with 3 moles Na2CO3 so the ratio of the moles is 2 is to 3 represented as 2:3
hope it helps
Can you think of an specific adaptations plants have made to survive in unique conditions?
Answer:
Plant Adaptations is a unique feature a plant has that allows it to live and survive in its own particular habitat (the place that it lives). Desert Plant Adaptations – Plant Adaptation is really a unique have a plant has that enables it to reside and survive in the own particular habitat (the area it lives).
Explanation:
If a watershed suddenly receives a great deal of rainfall, what will happen to its wetlands and rivers?
Answer:
Depending on the actual amount of water, it's going to overfill.
Explanation:
If your watershed is already one that has a lot of water and is always filled, it's likely that your wetlands and rivers will start to overfill, causing a form of flooding.
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I hope this helps!
-No one
- If you have bottle filled with Sulfur Hexafluoride and there are 1.00 L of the liquid with a density of 1.60 g/mL, how many moles of the liquid are present?
Answer:
About 11.0 moles.
Explanation:
We are given a bottle filled with 1.00 L of sulfur hexafluoride (SF₆) and we want to determine the number of moles of the liquid that is present.
First, determine its mass with the given density:
[tex]\displaystyle \begin{aligned}\rho & = \frac{m}{V} \\ \\ m & = \rho V \\ \\ & = \left(\frac{1.60\text{ g}}{\text{mL}}\right)(1.00\text{ L})\left(\frac{1000\text{ mL}}{1\text{ L}}\right) \\ \\ & = 1.60\times 10^3 \text{ g}\end{aligned}[/tex]
The molecular weight of SF₆ is 146.07 g/mol. Hence:
[tex]\displaystyle 1.60\times 10^3\text{ g SF$_6$} \cdot \frac{1\text{ mol SF$_6$}}{146.07\text{ g SF$_6$}} = 11.0\text{ mol SF$_6$}[/tex]
Therefore, about 11.0 moles of sulfur hexafluoride is present.
Calculate the specific heat in J/(g·ºC) of an unknown substance if a 2.50-g sample releases 12.0 cal as its temperature changes from 25.0ºC to 20.0ºC. ________J/(g·°C)
The specific heat of an unknown substance is 0.96 J/(g·ºC).
What is the specific heat capacity?The specific heat capacity of a substance can be described as the quantity of heat needed to raise the temperature in one unit of substance by 1 degree Celcius.
Whenever the heat is lost or absorbed there is a change in the temperature of the substance:
Q = mCΔT
Given, the amount of energy released, Q = 12 cal
The initial temperature of the substance = 25.0ºC
The final temperature of the substance = 20.0ºC
The change in the temperature of a substance, ΔT = 25 -20 = 5°C
The mass of the substance, m = 2.50 g
We have to find the specific heat capacity of an unknown substance.
12 cal = 2.50 g × C × 5°C
C = 0.96 J/(g·ºC)
Therefore, the specific heat capacity of an unknown substance is equal to 0.96 J/(g·ºC).
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The amount of heat needed to raise a substance's temperature by one degree Celsius per gram is known as specific heat. The specific heat in J/(gºC) of an unknown substance is 0.96 J/gºC.
The amount of heat needed to increase the temperature of the entire substance by one degree is the heat capacity of a substance. Specific heat capacity is what is used when the substance's mass is equal to one.
The hydrogen bonds in water, which account for its high specific heat.
The expression used to calculate the specific heat capacity is:
Q = mcΔT
Here Q = 12 Cal
m = 2.50 g
T₁ = 25.0ºC
T₂ = 20.0ºC
ΔT = 5°C
12 cal = 2.50 g × c × 5°C
c = 0.96 J/gºC
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What is the specific heat of a mystery substance that requires 79 joules of energy to heat 4.52
grams of substance from 23 degrees Celsius to 54 degrees Celsius?
The specific heat of a mystery substance that requires 79 joules of energy to heat 4.52 grams of substance from 23 degrees Celsius to 54 degrees Celsius is 0.56 J/g°C.
How to calculate specific heat capacity?The specific heat capacity of a substance can be calculated using the following formula:
Q = mc∆T
Where;
Q = quantity of heat absorbed or released (J)m = mass of substancec = specific heat capacity∆T = change in temperature (°C)c = Q/m∆T
c = 79 ÷ 4.52 × (54 - 23)
c = 79 ÷ 140.12
c = 0.56 J/g°C
Therefore, the specific heat of a mystery substance that requires 79 joules of energy to heat 4.52 grams of substance from 23 degrees Celsius to 54 degrees Celsius is 0.56 J/g°C.
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Which of the following UV rays has the highest energy that damages your skin the most?
A. UVA
B. UVB
C. X-ray
D. Infrared (IR)
H2S what species are present at 10-6 mol/L or greater when dissolved in water
Answer:
the answer has been given below have a good day
Explanation:
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The species are present at 10-6 mole/L or greater when dissolved in water is when in solutions, weak acids and bases partially ionize.
What is ionization?Ionization is defined as anything that causes electrically neutral atoms or molecules to gain or lose electrons in order to become electrically charged atoms or molecules (ions). One of the main mechanisms by which radiation, including charged particles and X-rays, transmits their energy to matter is ionization. An anion is created when an atom or molecule picks up an electron; a cation is created when they lose an electron.
Weak Acids and Weak Bases Ionizing. Numerous acids and bases are weak, meaning they do not completely ionize in aqueous solution. The nonionized acid, hydronium ion, and conjugate base of a weak acid are all mixed together to form a solution in water, with the nonionized acid being present in the highest concentration.
Thus, the species are present at 10-6 mole/L or greater when dissolved in water is when in solutions, weak acids and bases partially ionize.
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The partial pressures of gases A, B, and C in a mixture are
0.75 atmosphere, 0.25 atmosphere, and 1.25 atmospheres,
respectively. What is the total pressure of the gas mixture
in millimeters of Hg?
A. 1710 mm of Hg
B.
1140 mm of Hg
C. 760.0 mm of Hg
D. 570.0 mm of Hg
Answer:
A
Explanation:
.75 + .25 + 1.25 = 2.25 atm
1 atm is 760 mm hg
2.25 * 760 = 1710 mm HG
Answer:
[tex]\huge\boxed{\sf 2.25\ atm = 1710\ mm\ of\ Hg}[/tex]
Explanation:
Partial pressure of gas A = 0.75 atm
Partial pressure of gas B = 0.25 atm
Partial pressure of gas C = 1.25 atm
Total partial pressure = 0.75 atm + 0.25 atm + 1.25 atm
= 2.25 atm
We know that:
1 atm = 760 mm of Hg
Multiply 2.25 to both sides
2.25 atm = 760 × 2.25 mm of Hg
2.25 atm = 1710 mm of Hg
[tex]\rule[225]{225}{2}[/tex]
Decomposition of hydrogen peroxide: 2H2O2 --> O2(g) + 2H2O(l) How many molecules of water are produced from the decomposition of 3.4g of hydrogen peroxide, H2O2?
Answer:
Explanation:
You have the equation. Now change the 3.4 g H2 to moles. moles = grams/molar mass
3.4 g/2.016 = 1.686 moles.
Now using the coefficients in the balanced equation, convert moles H2O2 to moles H2O.
1.686 moles H2 x (2 moles H2O/2 moles H2O2) = 1.686 x (2/2) = 1.686 x (1/1) = 1.686 moles H2O.
Now you know that 1 mole of water is composed of 6.022 x 10^23 molecules. So
1.686 moles H2O x (6.022 x 10^23 molecules H2O/1 mole H2O) = ?? molecules.
Please Help! Chemistry questions below!
So
Moles of Oxygen:-
44.7/321.396mol3 mol of O_2 produces 2 mol SO_2
1 mol of ao_2 produces 2/3=0.6mol So_2
1.396 mol produces 0.8376mol SO_2
Mass of SO_2
0.8376(64)53.6gAn unidentified covalent molecular compound contains only carbon, hydrogen, and oxygen. When 7.40 mg of this compound is burned, 17.80 mg of CO2 and 2.08 mg of H2O are produced.
The freezing point of camphor is lowered by 26.4°C when 3.044 g of the compound is dissolved in 18.00 g of camphor (Kf = 40.0°C kg/mol).
What is the molecular formula of the unidentified compound?
Choose appropriate coefficients in the molecular formula below.
C H O
The molecular formular shows all the atoms in the compound. The molecular formula of the compound is C15H10O5.
What is molecular formula?The molecular formula is the formula of the compound that shows the total number of atoms in a compound.
We know that;
ΔT = K m i
ΔT = 26.4°C
K = 40.0°C kg/mol
m = ?
i = 1
26.4°C = 40.0°C kg/mol * m * 1
m = 26.4°C/40.0°C kg/mol = 0.66 m
We know that
m = mass/molar mass ÷ mass of solvent(in Kg)
Let the molar mass be MM
m = 3.044 g/MM ÷ (18 * 10^-3 Kg)
0.66 m = 3.044 g/0.018MM
0.66 m * 0.018MM = 3.044 g
MM = 3.044 g/0.66 m * 0.018
MM = 256 g/mol
Now;
Mass of C = 17.80 * 10^-3/44 * 12 = 0.00485 g
Moles of C = 0.00485 g/12 g/mol = 0.0004 moles
Mass of H = 2.08 * 10^-3/18 * 2 = 0.0002 g
Moles of H = 0.0002 g/1 g/mol = 0.0002 moles
Mass of O = [7.40 * 10^-3 - (0.00485 + 0.0002)] = 0.00235 g
Moles of O = 0.00235 g/16 g/mol = 0.000147 moles
Divide through by the lowest number of moles;
C - 0.0004/0.000147 H - 0.0002/0.000147 O - 0.000147 /0.000147
C - 3 H - 2 O - 1
The empirical formula is C3H2O
So;
[3(12) + 2(1) + 1(16)]n = 256
[36 + 2 + 16]n = 256
n = 5
Molecular formula = C15H10O5
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During photosynthesis, a green plant produces 122 mL of oxygen gas at STP. What mass of glucose
(C6H12O6) is produced during this reaction?
6CO2+ 6H2O > C6H12O6+ 602
Show all steps (Answer: 0.164g)
Answer:
About 0.164 g of glucose.
Explanation:
We can determine the mass of glucose produced given the volume of oxygen gas produced with stoichiometry.
Recall that at STP, a mole of any gas occupies a volume of 22.4 L.
From the reaction, six moles of oxygen gas is produced for every one mole of glucose.
Lastly, the molecular weight of glucose is 180.18 g/mol.
Therefore:
[tex]\displaystyle \begin{aligned} 122\text{ mL O$_2$} & \cdot \frac{1\text{ L O$_2$}}{1000\text{ mL O$_2$}}\cdot \frac{1\text{ mol O$_2$}}{22.4\text{ L O$_2$}} \cdot \frac{1\text{ mol C$_6$H$_{12}$O$_6$}}{6\text{ mol O$_2$}} \cdot \frac{180.18\text{ g C$_6$H$_{12}$O$_6$}}{1\text{ mol C$_6$H$_{12}$O$_6$}} \\ \\ & = 0.164\text{ g C$_6$H$_{12}$O$_6$}} \end{aligned}[/tex]
Therefore, about 0.164 g of glucose is produced.
There are 7 named classes of hazardous materials.
O True
O False
False, there are 9 named classes of hazardous materials.
What are hazardous materials?Hazardous materials are substances or chemicals that pose a health hazard, a physical hazard, or harm to the environment.
There are 9 hazardous substances symbols you need to know: flammable, oxidising, explosives, gas under pressure, toxic, serious health hazard, health hazard, corrosive and environmental hazard.
Hence, the statement is false.
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A sample of calcium fluoride was decomposed into the constituent elements. If the sample produced 625 mg of calcium, how many grams of fluorine were formed?
The mass of fluorine that were formed is 592 mg
StoichiometryFrom the question, we are to determine the mass of fluorine formed
First, we will write the balanced chemical equation for the decomposition reaction
The balanced chemical equation for the reaction is
CaF₂ → Ca + F₂
This means
1 mole of calcium fluoride decomposes to give 1 mole of calcium and 1 mole of fluorine
Now, we will determine the number of moles of calcium produced
From the given information,
Mass of calcium produced = 625 mg = 0.625 g
Using the formula,
[tex]Number\ of \ moles =\frac{Mass}{Atomic\ mass}[/tex]
Atomic mass of calcium = 40.078 g/mol
Then,
Number of moles of calcium produced = [tex]\frac{0.625}{40.078}[/tex]
Number of moles of calcium produced = 0.01559 mole
Since
1 mole of calcium fluoride decomposes to give 1 mole of calcium and 1 mole of fluorine
Then,
0.01559 mole of calcium fluoride will decompose to give 0.01559 mole of calcium and 0.01559 mole of fluorine
∴ Number of mole of fluorine formed was 0.01559 mole
Now, for the mass of fluorine formed
Using the formula,
Mass = Number of moles × Molar mass
Molar mass of fluorine = 38 g/mol
Then,
Mass of fluorine formed = 0.01559 × 38
Mass of fluorine formed = 0.59242 g
Mass of fluorine formed = 592.42 mg
Mass of fluorine formed ≅ 592 mg
Hence, the mass of fluorine that were formed is 592 mg.
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21.10g of NaOH and Ba3(OH)2 mixture is dissolved water to prepare 1.0dm³ Solution. To neutralize 25.OO mL of this solution needs 0.5 moldm-³ HCl 15.00mL. calculate the percentage of NaOH by mass in the mixture.
From the equation of te reaction, we know that the mass percent of NaOH in the mixture is 1.4%.
What is neutralization?Neutralization is a reaction that occurs between an acid and a base to yield salt and water only.
In tis case, the reaction of the NaOH and HCl occurs as follows; NaOH + HCl ----> NaCl + H2O
Number of moles of HCl reacted = 15/1000 * 0.5 moldm-³ = 0.0075 moles
Since the reaction is 1:1, 0.0075 moles of NaOH reacted.
Mass of NaOH = 0.0075 moles of NaOH * 40 g/mol = 0.3 g
Percent of NaOH = 0.3 g/21.10g * 100/1 = 1.4%
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Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute.
a. 0.100 m K2S
b. 21.5 g of CuCl2 in 4.50 * 102 g water
c. 5.5% NaNO3 by mass (in water)
The freezing points of each of the solutions are as follows;
0.100 m K2S - - 0.558oC
21.5 g of CuCl2 in 4.50 * 102 g water - -2oC
5.5% NaNO3 by mass (in water) - - 2.6oC
What is freezing point?The freezing point is the point at which liquid changes to solid. Let us now look at the freezing point of each solution.
a)
Since;
ΔT = K m i
K = 1.86 oC m-1
m = 0.100 m
i = 3
ΔT = 1.86 oC m-1 * 0.100 m * 3 = 0.558oC
Freezing point = 0oC - 0.558oC = - 0.558oC
b) Number of moles of CuCl2 = 21.5 g/134.45 g/mol = 0.16 moles
molality = 0.16 moles/0.45 Kg = 0.36 m
ΔT = K m i
ΔT = 1.86 oC m-1 * 0.36 m * 3 = 2oC
Freezing point = 0oC - 2 = -2oC
c) Number of moles of NaNO3 = 5.5g/85 g/mol = 0.065 moles
molality of the solution = 0.065 moles/0.0945 Kg = 0.69 m
ΔT = 1.86 oC m-1 * 0.69 m * 2 = 2.6oC
Freezing point = 0oC - 2.6oC = - 2.6oC
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what is galvanization
A sample of methane gas contains 3.62x10^29 atoms of hydrogen. What is the mass of the sample?
The mass of the sample to methane that contains 3.62×10²⁹ atoms of hydrogen is 4810631.24 g
Avogadro's hypothesis6.02×10²³ atoms = 1 mole of Hydrogen
How to determine the mass of Hydrogen1 mole of Hydrogen = 2 g
Thus,
6.02×10²³ atoms = 2 g of Hydrogen
Therefore,
3.62×10²⁹ atoms = (3.62×10²⁹ × 2) / 6.02×10²³
3.62×10²⁹ atoms = 1202657.81 g of Hydrogen
How to determine the mass of methane 1 mole of methane, CH₄ = 12 + (4×1) = 16 gMass of H in 16 g of CH₄ = 4 × 1 = 4 gThus,
4 g of Hydrogen is present in 16 g of methane.
Therefore,
1202657.81 g of Hydrogen will be present in = (1202657.81 × 16) / 4 = 4810631.24 g of methane
Thus, we can conclude that the mass of the sample of methane is 4810631.24 g
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How many grams of aluminum will be deposited by 0.1F? (Al=27) a.0.3g b. 0.9. c. 9.0g. d. 2.7g
Based on the charge on the aluminium ion, 0.9 g of aluminium are deposited by 0.1 F of electricity.
What is electrolysis?Electrolysis is the decomposition of a substance known as an electrolyte when electric current is passed through it.
The mass and hence moles an electrolyte deposited when current is passed through it depends on the charge on the ion.
Aluminium ion has a charge of +3 and requires 3F of electricity to deposit 1 mole or 27 g of aluminium
0.1 F will discharge = 0.1/3 × 27 g of aluminium
mass of aluminium deposited = 0.9 g of aluminium.
Therefore, 0.9 g of aluminium are deposited by 0.1 F of electricity.
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