The salt NaC2H3O2 would form from the combination of:

Moles of phosphorous in 15-35-15 fertilizer in 10g

guapka [62]

<span>The composition of a fertilizer is usually express in NPK number. NPK number is in terms of Percent by mass of the said element which are Nitrogen, Phosphorus and Potassium. A 15-35-15 fertilizer has 15% Nitrogen, 35% Phosphorous, and 15% Potassium by mass. If you have 10 g of this fertilizer, to get the number of moles of phosphorus, you multiply the mass by 35%, which is equal to 10*0.35 or 3.5 g phosphorus. Then you divide the calculated mass of phosphorous by its molar mass which is 30.97 g/mol. Therefore, you have 3.5/30.97 which is equal to 0.1130 mol Phosphorus. This is the amount of Phosphorus in moles in the fertilizer.</span>

6 0

3 years ago

Question 6 (4 points)

pishuonlain [190]

Answer:

20.9%

Explanation:

I took the test i hope this helps:)

8 0

3 years ago

Read 2 more answers

The question is in the picture below

Rus_ich [418]

Answer:

$\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3$

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

$\bf{\text{A} \rightarrow 2\text{B}}$ (Δ $\text{H}_1$ ) -----(1)

Since we have 2B, multiply the whole of II. by 2:

$\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}$ (2Δ $\text{H}_2$ ) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is $\text{A} \rightarrow 2\text{C} +2\text{D}$ .

Reversing III. gives us a negative enthalpy change as such:

$\bf{2\text{D} \rightarrow \text{E}}$ (-Δ $\text{H}_3$ ) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of $\text{A} \rightarrow 2\text{C} +\text{E}$ , which is also the equation of interest.