Answer:
KBr dissolved in water.
Explanation:
A substance conducts electricity as a result of the presence of mobile ions in the substance.
An ionic substance such as KBr when dissolved in water releases free ions which become charge carriers in solution hence the solution conducts electricity. Solid ionic substances such as solid KBr and solid baking soda do not conduct electricity because the ions are strongly bound to each other in the crystal lattice.
Molecular substances such as sugar and alcohol do not conduct electricity even in solution.
Answer:
-0.93 °C
Explanation:
Hello,
The freezing-point depression is given by:
Whereas 
is the freezing temperature of the solution, 
is the freezing temperature of the pure solvent (0 °C since it is water), 
the Van't Hoff factor (1 since the solute is covalent), 
the solvent's freezing point depression point constant (in this case 
) and 
the molality of the glucose.
As long as the unknown is , solving for it:
Best regards.
The mass in grams of NH₃ produced from the reaction is 3.4 g
<h3>Balanced equation</h3>
We'll begin by writing the balanced equation for the reaction. This illustrated below:
N₂ + 3H₂ -> 2NH₃
From the balanced equation above,
1 dm³ of N₂ reacted to produced 2 dm³ NH₃
<h3>How to determine the volume of NH₃ produced</h3>
From the balanced equation above,
1 dm³ of N₂ reacted to produced 2 dm³ NH₃
Therefore,
2.24 dm³ of N₂ will react to produce = 2.24 × 2 = 4.48 dm³ of NH₃
<h3>How to determine the mass of NH₃ produced</h3>
We'll begin by obtained the mole of 4.48 dm³ of NH₃. Details below:
22.4 dm³ = 1 mole NH₃
Therefore,
4.48 dm³ = 4.48 / 22.4
4.48 dm³ = 0.2 mole of NH₃
Finally, we shall determine the mass of NH₃ as follow:
- Molar mass of NH₃ = 17 g/mol
- Mole of NH₃ = 0.2 mole
- Mass of NH₃ =?
Mass = mole × molar mass
Mass of NH₃ = 0.2 × 17
Mass of NH₃ = 3.4 g
Learn more about stoichiometry:
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Answer:
Itś about 27 days so just pick 28!
Explanation:
Hope this helped!
Answer:
D / 15.0 g
Explanation:
3 % volume thus shows that there are 3 g of an solute in every 100mL of solutions
.. there will be 3 × 5000÷ 100 of H2O2 in a 500 mL bottle
