Answer:
441.28 g Oxygen
Explanation:
- The combustion of hydrogen gives water as the product.
- The equation for the reaction is;
2H₂(g) + O₂(g) → 2H₂O(l)
Mass of hydrogen = 55.6 g
Number of moles of hydrogen
Moles = Mass/Molar mass
= 55.6 g ÷ 2.016 g/mol
= 27.8 moles
The mole ratio of Hydrogen to Oxygen is 2:1
Therefore;
Number of moles of oxygen = 27.5794 moles ÷ 2
= 13.790 moles
Mass of oxygen gas will therefore be;
Mass = Number of moles × Molar mass
Molar mass of oxygen gas is 32 g/mol
Mass = 13.790 moles × 32 g/mol
<h3> = 441.28 g</h3><h3>Alternatively:</h3>
Mass of hydrogen + mass of oxygen = Mass of water
Therefore;
Mass of oxygen = Mass of water - mass of hydrogen
= 497 g - 55.6 g
<h3> = 441.4 g </h3>
Answer:
0.286 moles
Explanation:
i hope this is helpful for you
I remember coming across this question and the options were:
KOH, HCN, NH₃, HI, Sr(OH)₂
Now, a substance with a low pH is one that dissociates completely in water to release hydrogen ions, while basic substances dissociate completely to release hydroxide ions. Therefore, in the order of increasing pH:
HI, HCN, NH₃, Sr(OH)₂, KOH
Answer-The correct option is option d with says all of the above.
Explanation- All three acids that are given combined together to form acid rain in which nitric and sulphuric acid are stronger acids present while carbonic acid is a weaker one.
The carbon dioxide admitted in air combines with water to form carbonic acid and gives a weak acidic nature to rainwater. Pollution in nature makes sulphur and nitrogen present in air react to form the stronger acids responsible for acid rain.
Half life is the time that it takes for half of the original value of some amount of a radioactive element to decay.
We have the following equation representing the half-life decay:
A is the resulting amount after t time
Ao is the initial amount = 50 mg
t= Elapsed time
t half is the half-life of the substance = 14.3 days
We replace the know values into the equation to have an exponential decay function for a 50mg sample
That would be the answer for a)
To know the P-32 remaining after 84 days we have to replace this value in the equation:
So, after 84 days the P-32 remaining will be 0.85 mg
