An oxide of niobium has a cubic unit cell in which there are oxide ions at the middle of each edge and niobum atoms at the cente

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

To determine the mass of the lead piece, we use the following equation: Q(heat) = mC delta T, where Q equals 78.0 j, M is the mass we want to find, C is the specific heat capacity (0.130 j/g/C), and delta T shows the temperature difference, set at 9.0 c. Rearranging the formula to solve for M gives us M = Q / c delta T. By substituting in the values, we conclude that M = 78.0 j / (0.130 j/g/C * 9.0 C), calculating this gives us a mass of 66.7 g of lead.

Response:

22.9 moles (Option 5)

Clarification:

A substance being 35.7% carbon by mass signifies that in every 100 g of the substance, there is 35.7 g of carbon.

Using a rule of three, we get:

100 g of substance ___ has __ 35.7 g of C

769 g of substance ___ has ___ ( 769.35.7) / 100 = 274.5 g of C

1 mol of C = 12 g/m

Mass / Molar mass = Moles

274.5 g / 12 g/m = 22.9 moles

Response:
             4.5 m³

Resolution:
              The statement indicates the presence of two blocks on a lid of a container with a volume of 9 m³. The lid's weight is equal to that of the two blocks. Thus, there were initially four blocks (or 4 atm pressure) acting on a volume of 9 m³.

After adding four additional blocks on the lid, the pressure rises from 4 atm to 8 atm (2 atm from the lid, 2 atm from the original blocks, and 4 atm from the new blocks).

Hence, The data established is,

                  P₁  =  4 atm

                  V₁  =  9 m³

                  P₂  =  8 atm

                  V₂  =?

Using Boyle's Law,

                               P₁ V₁  =  P₂ V₂

Resolving for V₂,
                               V₂  =  P₁ V₁ / P₂

Substituting values yields:
                               V₂  =  (4 atm × 9 m³) ÷ 8 atm

                               V₂  =  4.5 m³

Answer:

evaporated

Explanation:

Once the solution evaporates, only salt will remain, as the sole other component in the solution is water.