Molality is the measure of concentration of solute in 1 kg of solution. The molality of the solution is 3.05 mol/kg.
10% of HCl (by mass) means 10 g of HCl and in 90 g of water.
Molar mass of HCl = 36.5 g/mol
Molality:
It is the measure of concentration of solute in 1 kg of solution. It can be calculated by the formula.
[tex]\bold {m = \dfrac {n }{w}\times 1000}[/tex]
Where,
m- molality
n - number of moles
w - weight of solvent in grams
Number of moles of HCl
[tex]\bold {n = \dfrac w{m} = \dfrac {10}{36.5} = 0.274 g}[/tex]
put the value in molality formula,
[tex]\bold {m = \dfrac {0.274 }{90}\times 1000}\\\\\bold {m = 3.05\ g/mol}[/tex]
Therefore, the molality of the solution is 3.05 mol/kg.
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Final answer:
To calculate the molarity of a 10.0% HCl solution, convert 10 grams of HCl to moles, and then divide by the volume of the solution in liters. This results in a molarity of 2.74 M.
Explanation:
To calculate the molarity of a 10.0% (by mass) aqueous solution of hydrochloric acid (HCl), start by understanding that a 10.0% solution means there are 10 grams of HCl in 100 grams of the solution. First, since the solution is aqueous, we can assume the density is close to that of water, which is approximately 1 g/mL, so 100 grams of the solution is roughly equivalent to 100 mL (0.1 L) of solution.
Next, convert the mass of HCl to moles. The molar mass of HCl is about 36.46 g/mol:
10 grams HCl × (1 mol HCl / 36.46 grams) = 0.274 moles HCl
Then, divide the moles of HCl by the volume of the solution in liters to find the molarity:
Molarity = Moles of solute / Volume of solution in liters
Molarity = 0.274 moles HCl / 0.1 L = 2.74 M
The molarity of the 10.0% HCl solution is therefore 2.74 M.
Suppose a piston automatically adjusts to maintain a gas at a constant pressure of 5.80 atm . For the initial conditions, consider 0.04 mol of helium at a temperature of 240.00 K . This gas occupies a volume of 0.14 L under those conditions. What volume will the gas occupy if the number of moles is increased to 0.07 mol (n2) from the initial conditions?
Answer: the new volume will be 0.245L
Explanation:Please see attachment for explanation
Answer:
The volume will be 238 mL or 0.238 L
Explanation:
Step 1: Data given
Pressure = constant = 5.80 atm
The initial moles of helium = 0.04 moles
Temperature = 240.00 K
Volume = 0.14L
The number of moles increases to 0.07 moles
Step 2: Calculate the new volume
p*V = n*R*T
V = (n*R*T)/p
⇒ with n = the number of moles = 0.07 moles
⇒ with R = the gas constant = 0.08206 L*atm/mol*K
⇒ with T = the temperature = 240.00 K
⇒ with p = the pressure = 5.80 atm
V = (0.07 * 0.08206 * 240.00) / 5.80
V = 0.238 L = 238 mL
The volume will be 238 mL or 0.238 L
Identify the factors that govern the speed and direction of a reaction. Check all that apply.
a. Reaction rates increase when the products are more concentrated
b. Reaction rates increase when the reactants are more concentrated
c. Reaction rates increase as the temperaturenses
d. Reaction rates decrease when als we present
Answer: option B and option C
Explanation:
b. Imagine an unusual life form in which the N atom in an amino acid is changed to a C atom. Could a hydrogen bond in this unusual alpha helix occur? Why or why not?
Answer: No hydrogen bond cannot occur in this alpha helix structure.
Explanation: For hydrogen bond to form, the electronegativity difference should be more than 1.7. carbon has an electronegativity of 2.5 whereas hydrogen has 2.1 so their electronegativity difference is 0.3. So in this alpha helix structure cannot occur.
Phosphorous can form an ion called phosphide, which has the formula P3−.
This ion ______.
A. contains 18 electrons
B. forms when a phosphorus atom loses three protons
C. has properties very similar of P
D. is called a cation
Answer: A. contains 18 electrons
Explanation:
Atomic number is defined as the number of protons or number of electrons that are present in an atom. It is characteristic of an element.
Atomic number of phosphorous is 15.
The electronic configuration of phosphorous (P) will be,
[tex]P:15:1s^22s^22p^63s^23p^3[/tex]
Atomic number = Number of electrons = Number of protons = 15
As the phosphorous atom has gained 3 electrons, it will have 15+3= 18 electrons , the phosphorous anion will be having a charge of -3.
The electronic configuration of [tex]P^{3-}[/tex] will be,
[tex]P^{3-}:18:1s^22s^22p^63s^23p^6[/tex]
Thus the correct statement is this ion contains 18 electrons
The phosphide ion contains 18 electrons. The correct answer is Option A.
Explanation:The ion formed by phosphorus, called phosphide, has the formula P3−. To determine its properties, we can look at its electron configuration. Phosphorus has an atomic number of 15, meaning it has 15 electrons in its neutral state. When phosphorus forms the phosphide ion (P3−), it gains three extra electrons to achieve a stable electron configuration. So, the phosphide ion contains a total of 18 electrons.
Therefore, the correct answer is A. The ion contains 18 electrons.
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Calculate your experimentally determined percent mass of water in Manganese(II) sulfate monohydrate. Report your result to 2 or 3 significant figures, e. g. 9.8% or 10.2%.
Answer:
10.6%
Explanation:
The determined percent mass of water can be calculated from the formula of the hydrate by
dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and
multiplying this fraction by 100.
Manganese(ii) sulphate monohydrate is MnSO4 . H2O
1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of
hydration must be included.
1 Manganes 52.94 g = 63.55 g
1 Sulphur 32.07 g =
32.07 g 2 Hydrogen is = 2.02 g
4 Oygen =
64.00 g 1 Oxygen 16.00 = 16.00 g
151.01 g/mol 18.02 g/mol
Formula Mass = 151.01 + (18.02) = 169.03 g/mol
2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and
multiply this fraction by 100.
Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%
The final result is 10.6% after the two steps calculations
The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.
What is percentage mass?The percentage mass is the ratio of the mass of the element or molecule in the given compound.
The percentage can be given as:
[tex]\text{Percent Mass} = \frac{\text{Mass of molecule}}{\text{total mass of compound}} \times 100 \%[/tex]
The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.
Thus,
[tex]\text{Percent Mass} = \frac{\text{18.02}}{\text{169.03 }} \times 100 \%\\\\\text{Percent Mass} = 10.6 \%}[/tex]
Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.
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A chemist adds of a 0.0013 mM copper(II) fluoride solution to a reaction flask. Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Round your answer to significant digits.
The question is incomplete, here is the complete question:
A chemist adds 345.0 mL of a 0.0013 mM (MIllimolar) copper(II) fluoride [tex]CuF_2[/tex] solution to a reaction flask.
Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
Answer: The mass of copper (II) fluoride is 0.13 mg
Explanation:
We are given:
Millimolarity of copper (II) fluoride = 0.0013 mM
This means that 0.0013 millimoles of copper (II) fluoride is present in 1 L of solution
Converting millimoles into moles, we use the conversion factor:
1 moles = 1000 millimoles
So, [tex]0.0013mmol\times \frac{1mol}{1000mmol}=1.3\times 10^{-6}mol[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We are given:
Moles of copper (II) fluoride solution = [tex]1.3\times 10^{-6}mol[/tex]
Molar mass of copper (II) fluoride = 101.5 g/mol
Putting values in above equation, we get:
[tex]1.3\times 10^{-6}mol=\frac{\text{Mass of copper (II) fluoride}}{101.5g/mol}\\\\\text{Mass of copper (II) fluoride}=(1.3\times 10^{-6}mol\times 101.5g/mol)=1.32\times 10^{-4}g[/tex]
Converting this into milligrams, we use the conversion factor:
1 g = 1000 mg
So,
[tex]\Rightarrow 1.32\times 10^{-4}g\times (\frac{1000mg}{1g})=0.13mg[/tex]
Hence, the mass of copper (II) fluoride is 0.13 mg
Answer:
The mass of copper(II) fluoride is 45.54 micrograms
Explanation:
A chemist adds 345.0 mL of a 0.0013 mM copper(II) fluoride solution to a reaction flask. Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Round your answer to significant digits.
Step 1: Data given
Molarity of the copper(II) fluoride solution = 0.0013 mM =
Volume of the solution 345.0 mL = 0.345 L
Molar mass copper(II) fluoride = 101.54 g/mol
Step 2: Calculate moles of copper(II) fluoride
Moles CuF2 = molarity * volume
Moles CuF2 = 0.0000013 M * 0.345 L
Moles CuF2 = 0.0000004485 moles
Step 3: Calculate mass of CuF2
Mass CuF2 = moles * Molar mass
Mass CuF2 = 0.0000004485 moles * 101.54 g/mol
Mass CuF2 = 0.00004554 grams = 0.04554 miligrams = 45.54 micrograms
The mass of copper(II) fluoride is 45.54 micrograms