<span>To
solve this we assume that the gas is an ideal gas. Then, we can use the ideal
gas equation which is expressed as PV = nRT. At number of moles the value of PV/T is equal to some constant. At another
set of condition of temperature, the constant is still the same. Calculations
are as follows:</span>
P1V1/T1 = P2V2/T2
P2 = P1 (V1) (T2) / (T1) (V2)
P2 = 475 kPa (4 m^3) (277 K) / (290 K) (6.5 m^3)
P2 = 279.20 kPa
Therefore, the changes in the temperature and the volume lead to a change in the pressure of the system which is from 475 kPa to 279.20 kPa. So, there is a decrease in the pressure.
Answer:
Mass of NaF in 250 g tube of toothpaste = 100 mg of NaF
Explanation:
Mass of fluoride per gram of toothpaste = 1 mg
From the given data, the fluoride compound that is most often used in toothpaste is sodium fluoride, NaF, which is 45% fluoride by mass, mass of sodium fluoride per gram of toothpaste can be determined as follows:
1 mg of fluoride per gram of toothpaste = 45% total mass of NaF per gram of toothpaste
100% mass of NaF in toothpaste = 100/45 * 1 mg = 2.22 mg of NaF.
Therefore, mass of NaF per gram of toothpaste is 2.22 mg.
The mass of NaF in 250 g of toothpaste is then calculated below:
In 250 g of toothpaste, mass of NaF = 250 * 2.22 mg = 99.9 mg of NaF
Answers:
- a.) 10.0 mL of 0.0500 M HCl: 5.00 ml
- b.) 25.0 mL of 0.126 M HNO₃: 31.5 ml
- c.) 50.0 mL of 0.215 M H₂SO4: 215. ml
Explanation:
All the reactions are the neutralization of strong acids with the same strong base.
At the neutralization point you have:
- number of equivalents of the base = number of equivalent of the acid
And the number of equivalents (#EQ) may be calculated using the normality (N) concentration and the volume (V)
Then, at the neutralization point:
- # EQ acid = N acid × V acid
- # EQ base = N base × V base
- N acid × V acid = N base × V base
Also, you can use the formula that relates normality with molarity
- N = M × number of hydrogen or hydroxide ions
<u>a.) 10.0 mL of 0.0500 M HCl</u>
- The number of hydrogen ions for HCl is 1 and the number of hydroxide ions for NaOH is 1.
- 10.0 ml × 0.0500 M × 1 = V base × 0.100 M × 1
⇒ V base = 10.0 ml ×0.0500 M / 0.100 M = 5.00 ml
<u>b.) 25.0 mL of 0.126 M HNO₃</u>
- The number of hydrogen ions for HNO₃ is 1 and the number of hydroxide ions for NaOH is 1.
- 25.0 ml × 0.126 M × 1 = V base × 0.100 M × 1
⇒ V base = 25.0 ml ×0.126 M / 0.100 M = 31.5 ml
<u>c.) 50.0 mL of 0.215 M H₂SO4</u>
- The number of hydrogen ions for H₂SO4 is 2 and the number of hydroxide ions for NaOH is 1.
- 50.0 ml × 0.215 M × 2 = V base × 0.100 M × 1
⇒ V base = 50.0 ml ×0.215 M × 2 / 0.100 M = 215. ml
C. F, Br, Cl, At
Explanation: you can see the periodic table
