Speed = distance/time
10m/2s = 5m/s
The speed is 5 m/s
The number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole
We'll begin by calculating the number of mole of NH₄NO₃ in the solution. This can be obtained as follow:
Volume = 125 mL = 125 / 1000 = 0.125 L
Molarity = 1.40 M
<h3>Mole of NH₄NO₃ =? </h3>
Mole = Molarity x Volume
Mole of NH₄NO₃ = 1.40 × 0.125
<h3>Mole of NH₄NO₃ = 0.175 mole</h3>
Finally, we shall determine the number of mole of ammonium ion, NH₄⁺ in the solution. This can be obtained as follow:
NH₄NO₃(aq) —> NH₄⁺(aq) + NO₃¯(aq)
From the balanced equation above,
1 mole of NH₄NO₃ contains 1 mole of NH₄⁺
Therefore,
0.175 mole of NH₄NO₃ will also contain 0.175 mole of NH₄⁺
Thus, the number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole
Learn more: brainly.com/question/25469095
Answer: 483 mL of the cleaning solution are used to clean hospital equipment
Explanation:
The question requires us to calculate the volume, in mL, of solution is used to clean hospital equipment, given that 415g of this solution are used and the specific gravity of the solution is 0.860.
Measurements > Density
Specific gravity is defined as the ratio between the density of a given substance to the density of a reference material, such as water:
The density of a substance is defined as the ratio between the mass and the volume of this substance:
Considering the reference substance as water and its density as 1.00 g/mL, we can determine the density of the substance which specific gravity is 0.860:
Thus, taking water as the reference substance, we can say that the density of the cleaning solution is 0.860 g/mL.
Now that we know the density of the cleaning solution (0.860 g/mL) and the mass of solution that is used to clean hospital equipment (415g), we can calculate the volume of solution that is used to clean the equipment:
Therefore, 483 mL of the cleaning solution are used to clean hospital equipment.
Answer:
= 72.73%
Explanation:
The percentage by mass of an element is given by;
% element = total mass of element in compounds/molar mass of compound × 100
The mass of oxygen in carbon dioxide = 32 g
Molar mass of CO2 = 44 g
Therefore;
% of O2 = 32/44 × 100%
<u>= 72.73%</u>