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weqwewe
2 months 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 [2.9K]2 months 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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The parent element decays by beta decay. which daughter element will form?
castortr0y [3046]
Beta decay occurs in two forms: β⁻ decay and β⁺ decay. In β⁻ decay, a neutron is transformed into a proton through the emission of an electron. If β⁻ decay occurs, the mass number of the daughter nucleus remains unchanged, but the number of protons increases by 1 and the number of neutrons decreases by 1, compared to the parent nucleus. On the other hand, β⁺ decay involves the conversion of a proton into a neutron via the emission of a positron. In this case, the mass number of the daughter nucleus stays the same, while the number of protons decreases by 1 and the number of neutrons increases by 1 compared to the parent nucleus.
7 0
1 month ago
The density of an alcohol is 0.788 g/mL. What volume in microliters, μL, will correspond to a mass of 20.500 mg?
lions [2927]

Answer:

B.  26.0 μL.

Explanation:

Hello,

Considering the provided mass and density, the volume calculates to be:

V=\frac{m}{\rho} =\frac{25.000mg}{0.788g/mL}*\frac{1g}{1000mg}*\frac{1000\mu L}{1mL} \\ \\V=26.0\mu L

Thus, the solution is B.  26.0 μL.

Best regards.

6 0
2 months ago
Which procedure cannot be performed on a hot plate, requiring a Bunsen burner instead
Tems11 [2777]

Answer: The process of heating a crucible to eliminate moisture from a hydrate.

Explanation:

The available choices are:

a. Heating a solvent to aid in the dissolution of a solute.

b. Heating a solid in isolation to remove moisture.

c. Bringing water to a boil for use in a water bath.

d. Heating a crucible to eliminate moisture from a hydrate.

Possible actions that can be done on a hot plate include:

a. Heating a solvent to assist a solute in dissolving.

b. Heating a solid in isolation to dry it.

c. Heating water to boiling for a water bath.

However, it's important to note that using a hot plate for heating a crucible to remove water from a hydrate is not advisable. Silica or ceramic materials are not meant to be heated on a hot plate.

Consequently, the correct procedure is heating a crucible to remove water from a hydrate.

4 0
3 months ago
An unknown element is found to have three naturally occurring isotopes with atomic masses of 35.9675 (0.337%), 37.9627 (0.063%)
Tems11 [2777]

Answer:

The correct choice for your inquiry is option A, Argon.

Explanation:

Isotope               Atomic mass                      Percent (%)

    1                       35.9675                              0.337

    2                      37.9627                              0.063

    3                      39.9624                            99.6

To calculate the average atomic mass: (Mass of isotope 1)(percent of 1) + (Mass of isotope 2)(percent of 2) + (Mass of isotope 3)(percent of 3)

Average atomic mass = (35.9675)(0.00337) + (37.9627)(0.00063) + (39.9624)(0.996)

Average atomic mass = 0.1212 + 0.0239 + 39.8025

Average atomic mass = 39.9476

                   Theoretical  Atomic mass

a) Ar                         39.95

b) K                          39.10

c) Cl                         35.45

d) Ca                       40.08

                 

5 0
2 months ago
Combustion analysis of an unknown compound containing only carbon and hydrogen produced 0.2845 g of co2 and 0.1451 g of h2o. wha
VMariaS [2998]
CxHy + (x+0.25)O₂ → xCO₂ + 0.5yH₂O

m(CO₂)/{xM(CO₂)}=m(H₂O)/{0.5yM(H₂O)}

0.2845/{44.01x}=0.1451/{9.01y}

x/y=0.4=2:5

The empirical formula is C₂H₅.
7 0
2 months ago
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