What functional feature(s) does the phosphate group contribute to the structure of a phospholipid? select all that apply. select

Given data:

CO concentration in air = 15 ppm

Volume of air inhaled per breath = 0.50 L

Breaths taken per minute = 20

CO density = 1.2 g/L

Objective:

milligrams of CO inhaled over 6 hours

Clarification:

A concentration of 15 ppm of CO means that there are 15 liters of CO per 10⁶ liters of air

. Consequently, the volume of CO inhaled through 0.50 L of air is

Next, considering there are 20 breaths in a minute,

the total number of breaths in 360 minutes (or 6 hours) will be

= 360 min * 20 breaths/1 min = 7200 breaths

Thus, the total volume of CO inhaled in that time frame is

= 7200 breaths * 7.50*10⁻⁶L/1 breath = 0.054 L

Given that the density of CO is 1.2 g/L

the mass of CO inhaled equals Density*Volume

= 0.054 * 1.2 = 0.0648 g = 64.8 mg

Thus, the mass of CO inhaled over 6 hours is 64.8 mg

1 atomic mass unit (amu) represents the mass of an atom or is used to measure mass on an atomic scale. It is also referred to as a dalton, abbreviated as Da, while atomic mass unit is indicated as amu.

1 amu can be translated into grams as follows:

For conversion into grams:

Therefore, the mass of Te is 204.16 * 10^-2^4 g

Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.

Explanation:

1) Molarity of 0.250 L HCl solution: 0.0328 M

The amount of HCl in the 0.250 L solution = 0.0082 moles

2) Molarity of 0.100 L NaOH solution: 0.0245 M

The amount of NaOH in the 0.100 L solution = 0.00245 moles

3) Determining the concentration of hydrochloric acid in the final solution.

0.00245 moles of NaOH will neutralize 0.00245 moles of HCl from the original 0.0082 moles of HCl.

The total volume of the mixture becomes 0.100 L + 0.250 L = 0.350 L

Remaining moles of unreacted HCl = 0.0082 moles - 0.00245 moles = 0.00575 moles

Concentration of the remaining HCl:

0.0164 molar concentration of hydrochloric acid in the resulting solution.

Clarification:

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Response:

The specific heat of the alloy

Clarification:

Weight of the alloy = 25 gm

Initial temperature = 100°c = 373 K

Weight of the water = 90 gm

Initial temperature of water = 25.32 °c = 298.32 K

Final temperature = 27.18 °c = 300.18 K

Using the energy balance equation,

Heat released by the alloy = Heat absorbed by the water

[[ - ] = ( - )

25 × × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

This gives us the specific heat of the alloy.