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bija089
2 days ago
8

A solution has a hydroxide-ion concentration of 0.0040 M. What is the pOH of the solution? A solution has a pH value of 3.66. Wh

at is the pOH for this solution? What is the pH of a 0.0072 M hydrochloric acid solution?
Chemistry
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A 2.950×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in w
Anarel [2989]

Answer:

The glycerol solution has a molality of 2.960×10^-2 mol/kg.

Explanation:

Calculating the moles of glycerol involves the formula: Moles = Molarity × Volume of solution = 2.950×10^-2 M × 1 L = 2.950×10^-2 moles.

To find the mass of water, use: Mass = Density × Volume = 0.9982 g/mL × 998.7 mL = 996.90 g, which converts to 0.9969 kg.

The formula for molality is: Molality = Moles of solute/Mass of solvent (in kg) = 2.950×10^-2/0.9969 = 2.960×10^-2 mol/kg.

7 0
3 months ago
(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak
KiRa [2933]

Response:

Here's my calculation

Clarification:

Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We need to determine the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's compile all the information in one location.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} =  5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}

3. Plot the initial values

The graph below visualizes the initial concentrations as plotted on the vertical axis.

7 0
4 months ago
A sample of a monoprotic acid (ha) weighing 0.384 g is dissolved in water and the solution is titrated with aqueous naoh. if 30.
Alekssandra [3086]
The formula for a monoprotic acid can be represented as HA, and its reaction with a base is shown as follows: HA + NaOH ---> NaA + H₂O. The stoichiometry between the acid and the base is 1:1. At the point of neutralization, the moles of HA equals the moles of the base. The moles of NaOH that reacted can be calculated as 0.100M / 1000 mL/L x 30.0 mL = 0.003 mol. Consequently, the moles of HA that reacted equal 0.003 mol. The mass of the acid is 0.384 g, yielding a molar mass of 0.384 g / 0.003 mol = 128 g/mol.
3 0
3 months ago
A 25.0-g sample of ice at -6.5oC is removed from the freezer and allowed to warm until it melts. Given the data below, select al
KiRa [2933]

Answer:

B, D

Explanation:

We need to recognize that the ice will rise in temperature from -6.5 ºC to 0 ºC for it to change into water.

Let's define q₁ as the heat needed to warm the ice to 0ºC, and q₂ as the heat for the transition from solid to liquid.

The calculation for q₁ is as follows:

q₁ = s x m x ΔT, where s represents the specific heat of ice (2.09 J/gºC), m is the mass, and ΔT is the temperature difference.

For q₂, the enthalpy of fusion is computed as:

q₂ = C x ΔT

with C indicating the specific heat for the phase transition, denoted as AH in kJ/mol.

All necessary data for computing q₁, q₂, and the total heat change (q₁ + q₂) is provided.

q₁ = 25.0 g x (2.09 J/gºC) x (0 - (-6.5 ºC))

q₁ = 339.6 J = 0.339 kJ

q₂ = (25 g/18 g/mol) x 6.02 kJ/mol = 1.39 x 6.02 kJ = 8.36 kJ

Combining these values gives us qtotal = 0.339 kJ + 8.36 kJ = 8.70 kJ.

Now we can answer the question:

(a) False, AH refers to the heat capacity during melting.

(b) True, as we concluded earlier.

(c) False, there’s only one phase transition from solid (ice) to liquid.

(d) True based on our calculations above.

(e) False, according to our findings.

7 0
3 months ago
When 1.00 g of boron is burned in o2(g) to form b2o3(s), enough heat is generated to raise the temperature of 733 g of water fro
VMariaS [2998]
To tackle this problem, one must first determine the specific heat of water, which is the energy required to raise the temperature of 1 g of water by 1 degree C. The relationship is given by the formula q = c X m X delta T, where q indicates the specific heat of water, m signifies the mass, and delta T denotes the temperature change. The specific heat of water is 4.184 J/(g X degree C). The temperature of the water increased by 20 degrees, therefore: 4.184 x 713 x 20.0 = 59700 J, rounded to 3 significant digits, equals 59.7 kJ. This value indicates the energy required to produce B2O3 from 1 gram of boron. To convert this to kJ/mole, additional calculations are required. The gram atomic mass of Boron is 10.811, so dividing 1 gram of boron by 10.811 results in.0925 moles of boron. Given that 2 moles of boron are needed for the formation of 1 mole of B2O3, dividing the moles of boron by two yields.0925/2 =.0462 moles. Consequently, dividing the energy in KJ by the number of moles provides KJ/mole: 59.7/.0462 = 1290 KJ/mole.
7 0
3 months ago
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