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seropon [69]
2 years ago
12

Ammonium phosphate is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric acid with

liquid ammonia. Calculate the moles of ammonia needed to produce 0.060 mol of ammonium phosphate. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Chemistry
1 answer:
ankoles [38]2 years ago
8 0

Answer:

n_{NH_3}=0.18molNH_3

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

H_3PO_4(aq)+3NH_3(l)\rightarrow (NH_4)_3PO_4

In such a way, considering the 3 to 1 molar relationship between ammonia and ammonium phosphate, the moles of ammonia result:

n_{NH_3}=0.060mol(NH_4)_3PO_4*\frac{3molNH_3}{1mol(NH_4)_3PO_4}\\n_{NH_3}=0.18molNH_3

Best regards.

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saveliy_v [14]

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A. Cesium sulfide

Explanation:

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7 0
2 years ago
What is FeCl3+NaOH=Fe(OH)3+NaCl
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This type of formula is use in chemistry it has  Iron ,Sodium, Hydroxyl and  Chloride unit s in the equation.
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3 0
3 years ago
I’m very confused on how to solve this
Bond [772]
Recall that density is Mass/Volume. We are given the mL of liquid which is volume so all we need is mass now. We are given the mass of the granulated cylinder both with and without the liquid, so if we subtract them, we can get the mass of the liquid by itself. So, 136.08-105.56= 30.52g. This is the mass of the liquid. We now have all we need to find the density. So, let’s plug these into the density formula. 30.52g/45.4mL= 0.672 g/mL. This is our final answer since the problem requests the answer in g/mL, but be careful, because some problems in the future may ask for g/L requiring unit conversions. Also note that 30.52 was 4 sigfigs and 45.4 was 3 sigfigs, and so dividing them required an answer that was 3 sigfigs as well, hence why the answer is in the thousandths place
4 0
3 years ago
What is the value of delta Hrxn for this equation:
Ratling [72]

Answer:

The ΔHrxn for the above equation = 179 kJ/mol

Explanation:

The reaction bond enthalpies are for the reactant;

3 × N-H = 3 × 390 = 1,170 kJ/mol

2 × O=O = 2 × 502 = 1004 kJ/mol

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6 0
2 years ago
A solution has a poh of 7. 1 at 10∘c. what is the ph of the solution given that kw=2. 93×10−15 at this temperature? remember to
Gnom [1K]

A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .

It is given that,

pOH of solution = 7.1

Kw =2.93×10^(-15)

Firstly, we will calculate the value of pKw

The expression which we used to calculate the pKw is,

pKw=-log [Kw]

Now by putting the value of Kw in this expression,

pKw =−log{2.93×10^(-15)}

pKw =15log(2.93)

pKw=14.5

Now we have to calculate the pH of the solution.

As we know that,

pH+pOH=pKw

Now put all the given values in this formula,

pH+7.1=14.5

pH=7.4

Therefore, we find the value of pH of the solution is, 7.4.

learn more about pH value:

brainly.com/question/12942138

#SPJ4

7 0
1 year ago
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