Answer:
2.52 atm
Explanation:
Change C to K.
Solve for P1=P2V2T2/V1T1
- Hope that helps! Please let me know if you need further explanation.
If electrons are negative, losing an electron makes the ion positive.
Therefore Cu1+ contains 1 unpaired electron.
The number of protons and the electron configuration of each
Answer:
- <u><em>9.01 g of fertilizer</em></u>
Explanation:
Assume initially that the amount in grams of the multinutrient fertilizer is 100 g, and calculate the amount of nitrogen supplied by every compound present in these 100 g of fertilizer.
<u>1) Urea (CH₄N₂O)</u>:
- Molar mass of urea: 60.06 g/mol
- Atomic mass of N: 14.007 g/mol
- Total mass of N in the formula: 2 × 14.007 g/mol = 28.014 g/mol
- Amount of N in 100 g of compound: 100 g × 54.8% × 28.014 g / 60.06 g = 25.56 g
<u>2) KNO₃</u>
- Molar mass of KNO₃: 101.1032 g/mol
- Atomic mass of N: 14.007 g/ mol
- Total mass of N in the formula: 14.007 g/mol
- Amount of N in 100 of the compound: 100 g × 26.3% × 14.007 / 101.1032 = 3.64 g
<u>3) (NH₄)₂PO₄</u>
- Molar mass of (NH₄)₂PO₄: 132.06 g/mol
- Atomic mass of N: 14.007 g/ mol
- Total mass of N in the formula: 2×14.007 g/mol = 28.014 g/mol
- Amount of N in 100 of the compound: 100 g × 14.1% × 14.007 / 132.06 = 2.99 g
4) <u>Total mass of N in 100 g of fertilizer</u>:
Add all the amounts of N obtained above
- 25.56g + 3.69 g + 2.99 g = 32.19 g of N
5) <u>Mass of fertilizer that should be applied to provide 2.90 g of N to a plant</u>.
Set a proportion:
- 32.19 g of N / 100 g of fertilizer = 2.90 g of N / X
Solve for X:
- X = 2.90 g of N × 100 g of fertilizer / 32.19 g of N = 9.01 g of fertilizer
That is the answer: 9.01 g of fertilizer should be applied to provide 2.90g of N to a plant.
