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weqwewe
26 days ago
10

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

ater and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 998.7 mL . The density of water at 20.0∘C is 0.9982 g/mL. Part A Calculate the molality of the glycerol solution.
Chemistry
1 answer:
Anarel [2.5K]26 days ago
7 0

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.

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How many minutes will it take for a car traveling 74 miles per hour to cover 6.50 kilometers?
lions [2633]

Answer: 3.28 mins

Explanation:

Here’s how it breaks down:

Conversions

74 mph = 33.08 m/s

6.5 km = 6500 m

(6500 m)/(33.08 m/s) = 196.5 seconds

196.5 seconds is equivalent to 3.28 minutes

8 0
11 days ago
What is the emperical formula for a compound containing 68.3% lead, 10.6% sulfur, and the remainder oxygen? a. Pb2SO4 b. PbSO3 c
Tems11 [2387]

Solution:

The molecular formula is PbSO₄, indicating lead sulfate

Option c.

Explanation:

The percentage makeup shows that in 100 g of this compound, there are:

68.3 g of Pb, 10.6 g of S, and (100 - 68.3 - 10.6) = 21.1 g of O

To find the moles of each element, we divide by their molar masses:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

Next, we find the mole ratio by dividing each by the smallest number of moles:

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Thus, the molecular formula is PbSO₄, representing lead sulfate.

8 0
20 days ago
Read 2 more answers
A tritium nucleus is formed by combining two neutrons and a proton. the mass of this nucleus is 9.106 × 10–3 universal mass unit
lions [2633]
E = mc²

where E = energy produced

m = mass of the nucleus

C = speed of light

m = 9.106 x 10⁻³ x 1.67 x 10⁻²⁷ kg

C = 3 x 10⁸ m/s, thus C² = 9 x 10¹⁶

E = 1.37 x 10⁻¹² J

7 0
1 month ago
Read 2 more answers
How many protons neutrons and electrons are there in a neutral atom of 43k (potassium-43)?
castortr0y [2720]

Answer:

             Protons: 19

             Neutrons: 25

             Electrons: 19

Explanation:

Protons:

                   The atomic number determines the number of protons in an atom. Consequently, with Potassium's atomic number being 19, it contains 19 protons.

Neutrons:

                   The formula to find neutrons is:

                           # of Neutrons  =  Atomic Mass - # of Protons

Given:

                                Atomic Mass  =  43

                                # of Protons  =  19

Thus,

                                # of Neutrons  =  43 - 19

                                # of Neutrons  =  24

Electrons:

                                   In a neutral atom, the quantity of electrons matches that of protons. Therefore, a neutral Potassium atom with 19 protons must equally have 19 electrons.

3 0
1 month ago
Sea water's density can be calculated as a function of the compressibility, B, where p = po exp[(p - Patm)/B]. Calculate the pre
lions [2633]

Answer:

A pressure of 137.14 MPa exists 10,000 m beneath the ocean surface.

At this same depth, the density measures 2039 kg/m3.

Explanation:

P0 and ρ0 symbolize the pressure and density at sea level (indicative of atmospheric conditions). With an increase in ocean depth, both pressure and density likewise rise.

The relationship between pressure and density can be expressed as:

\frac{dP}{dy}=\rho*g=\rho_0*g*e^{(P-P_0)/\beta\\\\

By rearranging

\frac{dP}{e^{(P-P_0)/\beta}}= \rho_0*g*dy\\\\\int\limits^{P}_{P_0} {e^{-(P-P_0)/\beta}}dP =\int\limits^y_0 {\rho_0*g*dy}\\\\(-\beta*e^{-(P-P_0)/\beta})-(\beta*e^0)=\rho_0*g*(y-0)\\\\-\beta*(e^{-(P-P_0)/\beta}-1)=\rho_0*g*y\\\\e^{-(P-P_0)/\beta}=1-\frac{\rho_0*g*y}{\beta}\\\\-\frac{P-P_0}{\beta} =ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\

This equation allows for computation of P at 10,000 m beneath the ocean's surface:

P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-200MPa*ln(1-\frac{1027kg/m^3*9.81m/s^2*10,000m}{200MPa})\\\\P-P_0=-200MPa*ln(1-\frac{1027*9.81*10,000Pa}{200*10^6Pa})\\\\P-P_0=-200MPa*ln(1-0.5037)\\\\P-P_0=-200MPa*(-0.6857)=137.14MPa

The density found at a depth of 10,000 m in the ocean is

\rho=\rho_0*e^{(P-P_0)/\beta}\\\rho=1027kg/m^3*e^{(137.14/200)}=1027*e^{0.686}kg/m^3\\\rho=1027*1.985 kg/m^3\\\rho=2039\,kg/m^3

4 0
1 month ago
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