What is the mass of a cylinder of lead that is 1.80 in in diameter, and 4.12 in long. The density of lead is 11.4 g/mL.

Resting potential.

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

= 15 L CO * 0.50 L air/10⁶ L air = 7.5 *10⁻⁶ L CO/breath

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

The formula for Molarity is given by:

                                  M = moles / V
To isolate V,
                              V = moles / M ------------------(1)
Moles can also be calculated as:
                                  moles = mass / M.mass -------------(2)
Substituting the value of moles from equation 2 into equation 1 yields:
                                  V = (mass / M.mass) / M
Plugging in the numbers gives:
                                  V = (45 g / 164 g/mol) / 1.3 mol/dm³

                                       V = 0.21 dm³.

Response:

2200 mg of antibiotic

Explanation:

The prescribed antibiotic dosage is 40 mg/kg of body weight.

For a patient weighing 55 kg, we calculate the dose of antibiotic as follows:

If we analyze 40/1000000, we can determine antibiotic allocation in kg per kg of weight

= 0.00004 kg of antibiotic for each kilogram

0.00004 multiplied by 55 (to find out the required amount for a 55 kg individual)

= 0.0022 kg

This 0.0022 figure converts to milligrams as follows

0.0022*10^6

= 2200 mg of antibiotic is indicated for a patient weighing 55 kg.

The answer is - 0.138 M. The buffer pH can be determined using the Henderson equation. Here, acts as a weak acid and serves as its corresponding conjugate base. The weak acid has two protons, while the base contains one. The equation can therefore be expressed in terms of protons transferred. Phosphoric acid can donate protons in three stages; the equation we’ve referenced pertains to the second stage, as the acid then has only two protons available and the base only one. Given the concentration of the acid as 0.10 M, we need to calculate the concentration of the base necessary to form a buffer with a pH of exactly 7.0. Substituting the values into the equation leads us to the solution. Cross-multiplying, we find that [base] = 1.38(0.10), yielding [base] = 0.138. Therefore, the concentration of the base needed for the buffer is 0.138 M.