1.) 0.875atm x 760.0 mmHg/atm
2.) 8I
7I 0
6I 0
5I 0
4I 0
3I 0
2I--------- 000000000 0I
1I-0------------------------ I---------------
0 50 100 150
Boiling point(degrees Celsius)
3.) The warmer the molecules are the faster they move like boiling water the gases are coming out of the water
4.)no clue
5.) A {solution} is always transparent, light passes through with no scattering from solute particles which are the molecule in size. The solution is homogeneous and does not settle out. A solution cannot be filtered but can be separated using the process of distillation.
A {suspension} is cloudy and heterogeneous. The particles are larger than 10,000 Angstroms which allows them to be filtered. If a suspension is allowed to stand the particles will separate out.
<span>A {colloid} is intermediate between a solution and a suspension. While a suspension will separate out a colloid will not. Colloids can be distinguished from solutions using the Tyndall effect. Light passing through a colloidal dispersion, such as smoky or foggy air, will be reflected by the larger particles and the light beam will be visible. A hydrocolloid can simply be defined as a substance that forms a gel when it comes in contact with water. Such substances include both polysaccharides and proteins.
6.)</span><span>The random movement of microscopic particles suspended in a liquid or gas, caused by collisions with molecules of the surrounding medium. Also called Brownian motion, molecular movement, pedesis.
hope that helps please mark me as brainly
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Answer:
The total mass of D-Glucose dissolved in a 2μL aliquot is 1 E-4 g
Explanation:
providing a solution to 5% weight-volume as found in commerce:
⇒ % 5 = (5g d-glucose/ 100 mL sln)×100
⇒ 0.05 = g C6H12O6/mL sln
⇒ g C6H12O6 = (2 μL sln)×(0.001 mL/μL)×(0.05 g C6H12O6/mL sln)
⇒ g C6H12O6 = 1 E-4 g C6H12O6
Its a strong electrolyte!.
Answer:
Explanation:
1. The reaction will proceed backward, shifting the equilibrium position to the left.
2. The reaction will proceed forward, shifting the equilibrium position to the right.
3. Either add more of the products ( H2O or Cl2) or remove the reactant (HCl or O2)
