Answer:
pHe = 3.2 × 10⁻³ atm
pNe = 2.5 × 10⁻³ atm
P = 5.7 × 10⁻³ atm
Explanation:
Given data
Volume = 1.00 L
Temperature = 25°C + 273 = 298 K
mHe = 0.52 mg = 0.52 × 10⁻³ g
mNe = 2.05 mg = 2.05 × 10⁻³ g
The molar mass of He is 4.00 g/mol. The moles of He are:
0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol
We can find the partial pressure of He using the ideal gas equation.
P × V = n × R × T
P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K
P = 3.2 × 10⁻³ atm
The molar mass of Ne is 20.18 g/mol. The moles of Ne are:
2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol
We can find the partial pressure of Ne using the ideal gas equation.
P × V = n × R × T
P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K
P = 2.5 × 10⁻³ atm
The total pressure is the sum of the partial pressures.
P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm
Answer:
D. The ionic number.
Explanation:
Whenever an element losses or gains an electron; it becomes charged, thus turns to an ion. The ion can either be positively charged when it losses an electron or negarively charged when it gains an electron. The number shows the required number of electrons gained (negative) or lost (positive).
The volume of acetone, in milliliters, has a mass of 44.2 g is 55.96mL.
<h3>HOW TO CALCULATE VOLUME?</h3>
The volume of a substance can be calculated by dividing the mass by its density. That is;
Volume (mL) = mass (g) ÷ density (g/mL)
According to this question, acetone is a solvent with density of 0.7899g/mL and mass of 44.2g. The volume is calculated as follows:
Volume = 44.2g ÷ 0.7899g/mL
Volume = 55.96mL
Therefore, the volume of acetone, in milliliters, has a mass of 44.2g is 55.96mL.
Learn more about volume at: brainly.com/question/1578538
A. Lowering the temperature.
I hope this helped!
Answer:
+5
Explanation:
Let's consider the following balanced equation.
Mn(NO₃)₂ + 2 HCl --> MnCl₂ + 2 HNO₃
We can determine the oxidation number of N in (NO₃)₂²⁻ taking into account that the sum of the oxidation numbers of the elements present in a chemical species is equal to this overall charge. For this compound we know:
- The oxidation number of O is -2.
- The overall charge is 2-.
- We have 2 atoms of N and 2× 3 = 6 atoms of O.
2 N + 6 O = 2-
2 N + 6 (-2) = 2-
2 N = +10
N = +5
