Answer: see below
Explanation: for 75 ml, take 45*75 mg of BSA and make up to 75 mL with distilled water.
Answer:
44 grams of CO₂ will be formed.
Explanation:
The balanced reaction is:
C + O₂ → CO₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- C: 1 mole
- O₂: 1 mole
- CO₂: 1 mole
Being the molar mass of each compound:
- C: 12 g/mole
- O₂: 32 g/mole
- CO₂: 44 g/mole
By stoichiometry the following mass quantities participate in the reaction:
- C: 1 mole* 12 g/mole= 12 g
- O₂: 1 mole* 32 g/mole= 32 g
- CO₂: 1 mole* 44 g/mole= 44 g
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
If 12 grams of C react, by stoichiometry 32 grams of O₂ react. But you have 40 grams of O₂. Since more mass of O₂ is available than is necessary to react with 12 grams of C, carbon C is the limiting reagent.
Then by stoichiometry of the reaction, you can see that 12 grams of C form 44 grams of CO₂.
<u><em>44 grams of CO₂ will be formed.</em></u>
Answer:
We use aluminum almost every day, aluminum can be used for juice boxes, chip bags, cans, foils, electronics, and even our cars have aluminum.
Explanation:
Answer: 27.09 ppm and 0.003 %.
First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.
Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.
So, according to the <em>law of ideal gases,</em>
PV = nRT
where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)
The moles of CO will be,
n = 35 mg x
x
→ n = 0.00125 mol
We clear V from the equation and substitute P = 0.92 atm and
T = -30 ° C + 273.15 K = 243.15 K
V = 
→ V = 0.0271 L
As 1000 cm³ = 1 L then,
V = 0.0271 L x
= 27.09 cm³
<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>
c = 27 cm³ / m³ = 27 ppm
<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:
c = 27.09
x
x 100%
c = 0.003 %
So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.