We need to first come up with a balanced equation:
→ 
We know that the molar ratio of hydrogen to oxygen to water now is 4:1:2.
Converting the amount of grams given to moles is as follows:
Hydrogen: 
Oxygen: 
We know now that the limiting reactant is oxygen. We can then know that the number of moles of water are produced are double the number of moles of oxygen used due to the ratio that we established at the beginning - 4:1:2.
So we now can use 6.25 moles of water as the amount produced.
Then we convert moles of water to grams:
Now we know that there are 112.59g of water produced when we start with 50g of hydrogen and 50g of water.
Answer:
Mass of original sample = 100 g
Explanation:
Half life of cesium-137 = 30.17 years
Where, k is rate constant
So,
The rate constant, k = 0.02297 year⁻¹
Time = 90.6 years
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Initial concentration = ?
Final concentration = 12.5 grams
Applying in the above equation, we get that:-
![12.5\ g=[A_0]e^{-0.02297\times 90.6}](https://tex.z-dn.net/?f=12.5%5C%20g%3D%5BA_0%5De%5E%7B-0.02297%5Ctimes%2090.6%7D)
![[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g](https://tex.z-dn.net/?f=%5BA_0%5D%3D%5Cfrac%7B12.5%7D%7Be%5E%7B-0.02297%5Ctimes%2090.6%7D%7D%5C%20g%3D100%5C%20g)
<u>Mass of original sample = 100 g</u>
The molecular formula of sucrose is - C₁₂H₂₂O₁₁
molecular mass of sucrose - 342 g/mol
molarity of sucrose solution is 0.758 M
In 1 L solution the number of sucrose moles are - 0.758 mol
Therefore in 1.55 L solution, sucrose moles are - 0.758 mol/L x 1.55 L
= 1.17 mol
The mass of 1.17 mol of sucrose is - 1.17 mol x 342 g/mol = 4.00 x 10² g
Answer : The mass of 
needed are, 1.515 grams.
Explanation :
First we have to calculate the mole of 
.
Now we have to calculate the moles of .
The balanced chemical reaction will be,
![PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4](https://tex.z-dn.net/?f=PbSO_4%2BNa_2CrO_4%5Crightarrow%20PbCrO_4%2BNa_2SO_4%5Btex%5D%3C%2Fp%3E%3Cp%3EFrom%20the%20balanced%20chemical%20reaction%2C%20we%20conclude%20that%3C%2Fp%3E%3Cp%3EAs%2C%201%20mole%20of%20%5Btex%5DPbCrO_4)
produced from 1 mole of
So, 0.005 mole of produced from 0.005 mole of
Now we have to calculate the mass of
Therefore, the mass of needed are, 1.515 grams.
