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frez
11 days ago
12

Describe how you would prepare approximately 2 l of 0.050 0 m boric acid, b(oh)3.

Chemistry
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How much volume (in cm3) is gained by a person who gains 11.1 lbs of pure fat?
Anarel [2989]

The density of human fat is 0.918 g/cm^{3}. The mass of the pure fat is 11.1 lbs.

First, convert the mass from pounds to grams as follows:

1 lb=453.592 g

Thus,

11.1 lb\rightarrow 11.1\times 453.592 g=5034.875 g

Density is defined as mass per unit volume, meaning volume can be calculated as:

V=\frac{m}{d}

By substituting the values,

V=\frac{5034.875 g}{0.918 g/cm^{3}}=5484.61 cm^{3}

Consequently, the volume gained by the individual will be 5484.61 cm^{3}.

6 0
2 months ago
All faculty members are happy to see students help each other. Dumbledore is particularly pleased with Hermione. Though, it shou
VMariaS [2998]

Answer:

m_{Mg}=30.8mgMg

Explanation:

Greetings,

According to the provided chemical equation, the production of 31.2 mL of hydrogen allows one to calculate its moles using the ideal gas equation as detailed below:

n_{H_2}=\frac{PV}{RT}=\frac{754torr*\frac{1atm}{760torr}*0.0312L}{0.082 \frac{atm*L}{mol*K}*298.15K}=1.27x10^{-3}molH_2

Since the ratio of hydrogen to magnesium is 1:1, its milligrams are derived through the following proportional factor calculation:

m_{Mg}=1.27x10^{-3}molH_2*\frac{1molMg}{1molH_2}*\frac{24.305gMg}{1molMg}*\frac{1000mgMg}{1gMg}\\m_{Mg}=30.8mgMg

Regards.

3 0
2 months ago
0.036549 round to three sig figs
Alekssandra [3086]
3 first significant figure
6 second significant figure
5 third significant figure
4 cannot exceed 5, so retain 5 instead of increasing it to 6

0.0365
6 0
2 months ago
Read 2 more answers
Equimolar samples of CH3OH(l) and C2H5OH(l) are placed in separate, previously evacuated, rigid 2.0 L vessels. Each vessel is at
Alekssandra [3086]

Answer:

Complete Question:  

Equimolar quantities of CH3OH(l) and C2H5OH(l) are placed in separate 2.0 L containers that have been evacuated beforehand. Pressure gauges are attached to each container, and the temperature is maintained at 300 K. In both containers, liquid is consistently visible at the bottom. The varying pressure within the vessel that contains CH3OH(l) is illustrated below.

In comparison to the equilibrium vapor pressure of CH3OH(l) at 300 K, the equilibrium vapor pressure of C2H5OH(l) at 300 K is

ANSWER : lower, since the London dispersion forces among C2H5OH molecules surpass those among CH3OH molecules.

Explanation:

To clarify the answer provided, let’s begin by defining some concepts.

The London dispersion force is the least strong type of intermolecular force. It is a temporary force that arises when the electron arrangement in two neighboring atoms creates transient dipoles.  

The vapor pressure of a liquid reflects the equilibrium pressure of its vapor above the liquid (or solid); specifically, it represents the pressure associated with the evaporation of a liquid (or solid) in a sealed environment above the substance.

The pressure will be lower due to the stronger London dispersion forces acting between C2H5OH molecules compared to those between CH3OH molecules. This implies that when intermolecular forces are stronger, they intensify the interactions binding the substance together, thereby reducing the liquid's vapor pressure at any given temperature and making it more difficult to vaporize the substance.

Note: The London dispersion force for C2H5OH is more substantial than for CH3OH because C2H5OH has more electrons than CH3OH.

3 0
3 months ago
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