In finding the molarity of a solution, we use the following formula:
What is Molarity?
The number of moles of the solute is calculated by dividing the mass of the solute by its molar mass.
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The molar mass of NH4NO3 and (NH4)3PO4 are 80.043 g/mol and 149.0867 g/mol, respectively.
![[NH+4]=0.1596 mol20.0 L=7.98×10−3 M NH+4](https://tex.z-dn.net/?f=%5BNH%2B4%5D%3D0.1596%20mol20.0%20L%3D7.98%C3%9710%E2%88%923%20M%20NH%2B4)
![[PO3−4]=0.0296 mol20.0 L=1.48×10−3 M PO3−4](https://tex.z-dn.net/?f=%5BPO3%E2%88%924%5D%3D0.0296%20mol20.0%20L%3D1.48%C3%9710%E2%88%923%20M%20PO3%E2%88%924)
Therefore, ![[PO3−4]\\](https://tex.z-dn.net/?f=%5BPO3%E2%88%924%5D%5C%5C)
has a molarity of 
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Answer:
Empirical formula is Cr₂O₃.
Explanation:
Given data:
Percentage of Cr = 68.4%
Percentage of O = 31.6%
Empirical formula = ?
Solution:
Number of gram atoms of Cr = 68.4 / 52 = 1.3
2
Number of gram atoms of O = 31.6 / 16 = 1.98
Atomic ratio:
Cr : O
1.32/1.32 : 1.98/1.32
1 : 1.5
Cr : O = 1 : 1.5
Cr : O = 2(1 : 1.5)
Empirical formula is Cr₂O₃.
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Answer:</h3>
2.125 g
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Explanation:</h3>
We have;
- Mass of NaBr sample is 11.97 g
- % composition by mass of Na in the sample is 22.34%
We are required to determine the mass of 9.51 g of a NaBr sample.
- Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.
In this case,
- A sample of 11.97 g of NaBr contains 22.34% of Na by mass
A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass
% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100
Mass of the element = (% composition of an element × mass of the compound) ÷ 100
Therefore;
Mass of sodium = (22.34% × 9.51 g) ÷ 100
= 2.125 g
Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g
Answer:
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Explanation:
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