What is the concentration of a solution in which 15 grams of sugar is dissolved in 0.2 L of water?
Answer:
0.2 M
Explanation:
Step 1: Given data
Mass of sugar (sucrose): 15 gVolume of water: 0.2 L (we will assume it is the volume of the solution)There are different ways to express the concentration of a solution. We will calculate molarity, which is one of the most used.
Step 2: Calculate the moles of sucrose
The molar mass of sucrose is 342.3 g/mol.
15 g × 1 mol/342.3 g = 0.044 mol
Step 3: Calculate the molarity of the solution
Molarity is equal to the moles of solute divided by the liters of solution.
M = 0.044 mol/0.2 L = 0.2 M
Calculate the concentration of a solution with 0.8g of NaCl in 280mL of water.
Answer: The molarity of NaCl solution is 0.0489 M
Explanation:
Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:
[tex]\text{Molarity of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (mL)}}[/tex] .....(1)
We are given:
Given mass of NaCl = 0.8 g
Molar mass of NaCl = 58.44 g/mol
Volume of the solution = 280 mL
Putting values in equation 1, we get:
[tex]\text{Molarity of solution}=\frac{0.8\times 1000}{58.44\times 280}\\\\\text{Molarity of solution}=0.0489M[/tex]
Hence, the molarity of NaCl solution is 0.0489 M
g Ions B and C react to form the complex BC. If 35.0 mL of 1.00 M B is combined with 35.0 mL of 1.00 M C, 0.00500 mol of BC is formed. Determine the equilibrium constant for this reaction.
Answer:
Kf = 0.389.
Explanation:
Hello there!
In this case, it turns out possible for us to solve this problem by firstly writing the equilibrium chemical equation and equilibrium expression for the formation of this complex:
[tex]B+C\rightleftharpoons BC\\\\Kf=\frac{[BC]}{[B][C]}[/tex]
Thus, we firstly calculate the concentrations at equilibrium, knowing that the reaction extent in this case is 0.00500mol (same as the formed moles of BC):
[tex][B]=[C]=\frac{0.0350L*1.00mol/L-0.00500mol}{0.0700L} =0.429M[/tex]
[tex][BC]=\frac{0.00500mol}{0.0700L} =0.0714M[/tex]
And finally, the equilibrium constant:
[tex]Kf=\frac{0.0714}{[0.429][0.429]}\\\\Kf=0.389[/tex]
Regards!
In Denver, Colorado the elevation is about 5,280 feet above sea level. Explain what potential effects this may have on the solubility of a gaseous solute in a liquid solution.
Answer:
The solubility of the gaseous solute decreases
Explanation:
As we know, pressure decreases with altitude. This means that, at higher altitudes, the pressure is much lower than it is at sea level.
The solubility of a gas increases with increase in pressure and decreases with decrease in pressure.
Hence, in Denver, Colorado where the elevation is about 5,280 feet above sea level, a gaseous solute is less soluble than it is at sea level due to the lower pressure at such high altitude.
Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH.
pH= 2.89
Answer: The value of [tex][H_{3}O^{+}][/tex] is 0.0012 M and [tex][OH^{-}][/tex] is [tex]1.02 \times 10^{-14}[/tex].
Explanation:
pH is the negative logarithm of concentration of hydrogen ion.
It is given that pH is 2.89. So, the value of concentration of hydrogen ions is calculated as follows.
[tex]pH = - log [H^{+}]\\2.89 = - log [H^{+}]\\conc. H^{+} = 0.0012 M[/tex]
The relation between pH and pOH value is as follows.
pH + pOH = 14
0.0012 + pOH = 14
pOH = 14 - 0.0012 = 13.99
Now, pOH is the negative logarithm of concentration of hydroxide ions.
Hence, [tex][OH^{-}][/tex] is calculated as follows.
[tex]pOH = - log [OH^{-}]\\13.99 = - log [OH^{-}]\\conc. OH^{-} = 1.02 \times 10^{-14} M[/tex]
Thus, we can conclude that the value of [tex][H_{3}O^{+}][/tex] is 0.0012 M and [tex][OH^{-}][/tex] is [tex]1.02 \times 10^{-14}[/tex].
Select all the correct locations on the image.
Select the areas that would receive snowfall because of the lake effect.
(The right answer for this question, I got it right on the edmentrum)
Answer:
A:Itasca
A:ItascaB:Hubbard
A:ItascaB:HubbardC:Douglas
A:ItascaB:HubbardC:DouglasD:Grand Marais
E:Two harbors
F:Duluth
Answer:
Duluth, Twin Harbors and Grand Marais because they are on the coast of the lake.
Explanation:
Which of the following is an example of a nonrenewable resource?
a
cattle
b
uranium
c
cotton
d
trees
Answer:
b . uranium, It is not a renewable resource.
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What is the Ke of the 20 gram shoe as it falls to the ground at 6 m/s?
PLEASE HELP ITS 7TH GRADE SCIENCE!!!
Match the name of each gas law to the properties it compares. (3 points)
1. Boyle's law
2. Avogadro's law
3. Gay-Lussac's law
a. Volume and moles
b. Pressure and volume
c. Pressure and temperature
Answer:
1. B
2. A
3. C
Explanation:
1. Boyle's law is one of the gas laws that states that the pressure of a gas is inversely proportional to its volume at a constant temperature. PV = K. Hence, this gas law compares the properties of the pressure (P) and the volume (V)
2. Avogadro's law states that the volume of a gas is directly proportional to the number of molecules of that gas, at a constant temperature and pressure. K = Vn. Hence, this gas law compares the properties of the number of moles (n) and the volume (V).
3. Gay-Lussac's law states that the pressure of a gas is directly proportional with the temperature of the gas at a constant volume. K = PT. Hence, this law compares the properties of the pressure (P) and the temperature (T)
The properties compared by Boyle's law are pressure and volume, Avogadro's law is volume and moles, and Gay-Lussac's law is pressure and temperature.
What are the variations of ideal gas law?The ideal gas has the absence of interatomic collisions and follows the ideal gas equation.
The ideal gas law at constant pressure and temperature is termed Boyle's and Gay Lussac's law. The laws can be given as:
Boyle's law was the nature of ideal gas at a constant temperature and compares the pressure and volume of the gas.Avogadro's law was the ideal nature of gas, with the moles of gas and volume in specific value at constant temperature and pressure.Gay-Lussac's law compares the pressure and temperature of the gas at constant volume.Learn more about ideal gas laws, here:
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What is the pCu of the resulting solution if 20.00 mL of 0.08 M EDTA (H4Y) is added to 15.00 mL of 0.10 M CuSO4 and buffered at pH 10? The Kf’ for complex CuY2- is 2.21 x 1018
Answer:
The answer is "5.4".
Explanation:
[tex]BoH + HCL =BCL +H_2o \\\\At eq \\\\N_1V_1=N_2V_2 \\\\v_2=20 \ ml\\\\[BCL]=\frac{20 \times 0.08}{20+20}=0.04\\\\pH = \frac{1}{2} [pkw - pk_b - \log e]\\\\pk_b = 2 pH - Pkw + \Log C\\\\pK_b=5.4[/tex]
Draw the Lewis structure of NCl3NCl3 . Include lone pairs. Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Cl N
Answer:
See explanation and image attached
Explanation:
Nitrogen trichloride NCl3 contains one nitrogen and three chlorine atoms. Each chlorine atom has three lone pairs of electrons while nitrogen has one lone pair of electrons.
The molecule geometry of the molecule is trigonal pyramidal while it's electron domain geometry is tetrahedral. Nitrogen is the central atom and is found in sp3 hybridization.
There is no formal charge on the molecule.
the mixture of base and acid
Answer:
Mixture of a Strong Acid and a Strong Base
On mixing a strong acid and strong base neutralization (pH = 7) takes place. The resulting solution may be an acid or base depending on the Concentration. Say, N1, V1 is the strength and volume of the strong acid and N2, V2 is the strength and volume of the strong base
Explanation:
Most introductory chemistry books will teach that the reaction between an acid and a base is called neutralization, and the products formed are water and a salt
Can the properties of different substances in a mixture be used to separate them?
Answer: here you go
Explanation:
Physical properties of the substances in a mixture are different, so this allows the substances to be separated. Think about the example of a mixture of salt water.
Answer:
Mixtures can be classified on the basis of particle size into three different types: solutions, suspensions, and colloids. The components of a mixture retain their own physical properties. These properties can be used to separate the components by filtering, boiling, or other physical processes.
Explanation:
điện phân nóng chảy hoàn toàn 14,9 gam muối clorua của 1 kim loại kềm R thu đc 2,24 lít khí . R là kim loại gì ?
Answer:
d
Explanation:
nCl2 = 0,1 -> nRCI = 0,2
M muéi = R + 35,5 = 14,9/0,2
-> R = 39: R la K
Answer:
d
Explanation:
nCl2 = 0,1 -> nRCI = 0,2
M muéi = R + 35,5 = 14,9 / 0,2
-> R = 39: R la K
The seagulls on the beach -
table salt is 42.7% sodium how many grams of salt contain 76 g of sodium
Answer:
Số gam muối ăn cần là 76:42,7%=177,986 g
Explanation:
When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing point of pure . Calculate the mass of potassium bromide that must be dissolved in the same mass of to produce the same depression in freezing point. The van't Hoff factor for potassium bromide in .
The question is incomplete, the complete question is:
When 177. g of alanine [tex](C_3H_7NO_2)[/tex] are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is [tex]5.9^oC[/tex] lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is [tex]7.2^oC[/tex] lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.
Answer: The van't Hoff factor for potassium bromide in X is 1.63
Explanation:
Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.
The expression for the calculation of depression in freezing point is:
[tex]\Delta T_f=i\times K_f\times m[/tex]
OR
[tex]\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)
When alanine is dissolved in mystery liquid X:[tex]\Delta T_f=5.9^oC[/tex]
i = Vant Hoff factor = 1 (for non-electrolytes)
[tex]K_f[/tex] = freezing point depression constant
[tex]m_{solute}[/tex] = Given mass of solute (alanine) = 177. g
[tex]M_{solute}[/tex] = Molar mass of solute (alanine) = 89 g/mol
[tex]w_{solvent}[/tex] = Mass of solvent = 800.0 g
Putting values in equation 1, we get:
[tex]5.9=1\times K_f\times \frac{177\times 1000}{89\times 800}\\\\K_f=\frac{5.9\times 89\times 800}{1\times 177\times 1000}\\\\K_f=2.37^oC/m[/tex]
When KBr is dissolved in mystery liquid X:[tex]\Delta T_f=7.2^oC[/tex]
i = Vant Hoff factor = ?
[tex]K_f[/tex] = freezing point depression constant = [tex]2.37^oC/m[/tex]
[tex]m_{solute}[/tex] = Given mass of solute (KBr) = 177. g
[tex]M_{solute}[/tex] = Molar mass of solute (KBr) = 119 g/mol
[tex]w_{solvent}[/tex] = Mass of solvent = 800.0 g
Putting values in equation 1, we get:
[tex]7.2=i\times 2.37\times \frac{177\times 1000}{119\times 800}\\\\i=\frac{7.2\times 119\times 800}{2.37\times 177\times 1000}\\\\i=1.63[/tex]
Hence, the van't Hoff factor for potassium bromide in X is 1.63
why is it preferred to use calcium oxide rather than calcium chloride in preparation of iron (III) chloride
Answer:
Calcium Oxide is a drying agent, hence it dehydrates the reaction to give pure solid Iron ( III ) chloride, which cannot be done by calcium chloride.
It preferred to use calcium oxide rather than calcium chloride in preparation of iron (III) chloride because Calcium Oxide is a drying agent, hence it dehydrates the reaction to give pure solid Iron ( III ) chloride, which cannot be done by calcium chloride.
What is Dehydration ?A process such as a chemical reaction that removes water.The atoms which constitute the molecule of water that is removed.
Hence,It preferred to use calcium oxide rather than calcium chloride in preparation of iron (III) chloride because Calcium Oxide is a drying agent,
Thus, it dehydrates the reaction to give pure solid Iron ( III ) chloride, which cannot be done by calcium chloride.
Learn more about reactions here ;
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Which statement shows why mass is conserved for this reaction? 2Mg + O2 → 2MgO
Answer:
The same number of each atom are on both sides of the equation.
The Law of conservation of mass states that matter is neither destroyed nor created in a chemical reaction. Therefore, the mass of the reactants will be equal to the mass of the products in a chemical reaction. According to the law matter is neither created nor destroyed in a chemical reaction.
Which of the following is NOT likely to cause a change in average annual temperatures on Earth?
a. Human activity. b. Solar eclipses.
c. Photosynthesis by plants and algae. d. Strength of solar radiation.
Answer:
i think C is the answer
Explanation:
The change in average annual temperatures on earth will be due to "photosynthesis by plants and algae".
What is photosynthesis?Photosynthesis can be defined as a process in which plants, as well as other organisms, as well as other organisms, utilize to transform sunlight into chemical energy which can then be released to power the organism's activities using cellular respiration.
What is plants?
Plants seem to be mostly photosynthetic eukaryotes belonging to the plantae kingdom.
Therefore, photosynthesis cannot change in average annual temperature on Earth.
To know more about photosynthesis.
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The hydrogen fluoride molecule, HF, is more polar than a water molecule, H2O (for example, has a greater dipole moment), yet the molar enthalpy of vaporization for liquid hydrogen fluoride is lesser than that for water. Explain.
Answer:
Water forms more hydrogen bonds than HF
Explanation:
The answer to this question goes back to the idea of hydrogen bonding. Hydrogen bonding occurs when hydrogen is bonded to a highly electronegative atom such as fluorine or oxygen.
However, in HF, there are three lone pairs of electrons on fluorine atom and one hydrogen atom bonded to fluorine.
In H2O, there are two lone pairs of electrons on oxygen atom and two hydrogen atoms bonded to oxygen. This simply means that water can form four hydrogen bonds while HF only forms two hydrogen bonds.
This implies that H2O molecules possess more hydrogen bonding than HF molecules. Hence, the molar enthalpy of vaporization for liquid hydrogen fluoride is lesser than that for water.
g A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is __________ compared to a 300 mOsM solution and ___________ compared to a cell with 300 mOsM (non-penetrating solutes) interior.
Answer:
A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is hypotonic compared to a 300 mOsM solution and hypotonic compared to a cell with 300 mOsM (non-penetrating solutes) interior.
Explanation:
The osmolarity is calculated from the molar concentration of the active particles in the solution. We have a solution that is composed of NaCl, CaCl₂ and urea.
When they are dissolved in water, they dissociate into particles as follows:
NaCl → Na⁺ + Cl⁻ (2 particles per compound)
CaCl₂ → Ca²⁺ + 2 Cl⁻ (3 particles per compound)
urea: not dissociation (1 particle per compound)
Then, we have to calculate the osmolarity of the solution. We multiply the molarity of each compound by the number of particles produced by the compound in water:
Osm = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) + (20 mM urea x 1) = 280 mOsm
Compared with 300 mOsm, 280 mOsm has a lower osmolarity, so it is a hypotonic solution.
To compare with a cell's osmolarity, we have to consider only the non-penetrating solutes. Urea is considered a penetrating solute for mammalian cells. So, the osmolarity of non-penetrating solutes (NaCl and CaCl₂) is calculated as:
Osm (non-penetrating solutes) = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) = 260 mOsm
Therefore, we have:
Compared to 300 mOsm solution ⇒ 280 mOsm solution is a hypotonic solution
Compared to a cell with 300 mOsm ⇒ 260 mOsm solution is hypotonic
The specific rate constant, k, for radioactive beryllium–11 is 0.049 s–1. What mass of a 0.500 mg sample of beryllium–11 remains after 28 seconds? This reaction was found to be first order.
Answer: The mass of sample that remained is 0.127 mg
Explanation:
The integrated rate law equation for first-order kinetics:
[tex]k=\frac{2.303}{t}\log \frac{a}{a-x}[/tex] ......(1)
Given values:
a = initial concentration of reactant = 0.500 mg
a - x = concentration of reactant left after time 't' = ?mg
t = time period = 28 s
k = rate constant = [tex]0.049s^{-1}[/tex]
Putting values in equation 1:
[tex]0.049s^{-1}=\frac{2.303}{28s}\log (\frac{0.500}{(a-x)})\\\\\log (\frac{0.500}{(a-x)})=\frac{0.049\times 28}{2.303}\\\\\frac{0.500}{a-x}=10^{0.5957}\\\\frac{0.500}{a-x}=3.94\\\\a-x=\frac{0.500}{3.942}=0.127mg[/tex]
Hence, the mass of sample that remained is 0.127 mg
Give the IUPAC name for the following alkyl group, and classify it as primary, secondary, or tertiary.
CH3(CH2)9CH2
Answer:
Give the IUPAC name for the following alkyl group, and classify it as primary, secondary, or tertiary.
CH3(CH2)9CH2
Explanation:
In the given alkyl group there are eleven carbon atoms.
So, the alkyl group name is:
n-undecyl.
Pimary carbon is the one which is attached only one other carbon atom,group.
Secondary carbon is the one which is attached to two carbons.
Thertiary carbon is the one which is attached to three other carbons.
In the given alkyl group,
the primary,secondary alkyl groups are shown below:
There is no tertiary carbon atom in the given molecule.
help please and thank you!
Answer:
a) N2(g) + H20 (aq) --> HNO3 (aq) + NO (g)b) 40 NO2(g) + H20(aq) = 20 HNO3(aq) + 20 NO(g)Explanation:
what is a property of every mixture
Explanation:
can u post a picture of the question ?
convert 12nanometer to centimeter
Answer:
1x10^-6
Explanation:
5 compounds that has electrovalent and covalent bond
Answer:
electrovalent
NaCl
Lithium Carbonate
ammonium phosphate
aluminium floride
potassium hydride
covalent
methane
benzene
carbon iv oxide
hydro flouride
hydro chloride
phương pháp VI PHÂN ĐỒ THỊ để xác định bậc phản ứng
Answer:
mwlooy kagabi jal
64 JAHA VI PHÂN KAY