In formation of a Type II Binary Compound, the metal atom present is<span>
NOT</span> found in either Group 1 or Group 2 on the periodic table. For the choices, Ba is under Group 2 on the periodic table, which makes it the atom not involved in formation of type II compounds. The answer is B.
Answer:
Explanation:
- State of benzene at RTP = liquid
- State of chloroform at RTP = liquid
- Boiling point of benzene = 80.1 °C
- Boiling point of chloroform = 61.2 °C
Since, both of the chemicals are liquids, we can separate it by the process of distillation.
<u>Distillation:</u>
- is the process in which we separate two liquids on the basis of their difference in boiling points.
<u>How it works:</u>
Since chloroform has less boiling point, it will evaporate and collected first and benzene will follow it after sometime.
- Apparatus of distillation is in the attached file.
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Yes , the chemist can answer if the compound in K2O or K2O2
The chemical formula and composition of both the compounds is entirely different. The compound K2O2 has an additional molecule of oxygen than K2O and hence will have have higher molecular mass.
In the compound K2O
molecular mass= 2x 39+16 =94
mass ratio of K in compound= 78/94 = 0.830
In the compound K2O2
molecular mass= 2x 39+16X2 =110
mass ratio of K in compound= 78/110 = 0.710
and hence by the required ratio while extracting K , the chemist may know if the compound is K20 or K2O2
If the ratio is anything different from 0.830 and 0.710 then the compund will be something different
#SPJ9
Answer:
Explanation:
The heat received by water is equal to the heat rejected by the piece of metal. That is to say:
The initial temperature of the piece of metal is:
Answer:
Total pressure = 16.42× 10⁻⁹atm
Explanation:
Given data:
Moles of H₂ = 2.50 × 10⁻³ mol
Moles of He = 1.00 × 10⁻³ mol
Mass of Ne = 3 × 10⁻⁴ mol
Volume = 10 L
Temperature = 35°C
Total pressure = ?
Solution:
Pressure of hydrogen:
P = nRT / V
P = 2.50 × 10⁻³ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L
p = 63.22× 10⁻³ atm. L /10 L
P = 6.3 × 10⁻³atm
Pressure of helium:
P = nRT / V
P = 1.00 × 10⁻³ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L
p = 25.29 × 10⁻³ atm. L /10 L
P = 2.53× 10⁻³ atm
Pressure of neon:
P = nRT / V
P = 3 × 10⁻⁴ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L
p = 75.86× 10⁻³ atm. L /10 L
P = 7.59× 10⁻³ atm
Total pressure of mixture:
P(mixture) = pressure of hydrogen + pressure of helium+ pressure of neon
P(mixture) = 6.3 × 10⁻³atm + 2.53× 10⁻³ atm + 7.59× 10⁻³ atm
P(mixture) = 16.42× 10⁻⁹atm
