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natta225 [31]
2 years ago
14

A florist prepares a solution of nitrogen-phosphorus fertilizer by dissolving 5.66 g of NH₄NO₃ and 4.42 g of (NH₄)₃ PO₄ in enoug

h water to make 20.0 L of solution. What are the molarities of NH₄⁺ and of PO₄³⁻ in the solution?
Chemistry
1 answer:
Alexandra [31]2 years ago
5 0

In finding the molarity of a solution, we use the following formula:

M=moles soluteL solution

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.

molesNH+4inNH4NO3=5.66 g80.043 g/mol×1molNH+41molNH4NO3=0.0707 mol

molesNH+4in(NH4)3PO4=4.42 g149.0867 g/mol×3molNH+41mol(NH4)3PO4=0.0889 mol

total molesNH+4=0.0707 mol+0.0889 mol=0.1596 mol

molesPO3−4in(NH4)3PO4=4.42 g149.0867 g/mol×1molPO3−41mol(NH4)3PO4=0.0296 mol

[NH+4]=0.1596 mol20.0 L=7.98×10−3 M NH+4

[PO3−4]=0.0296 mol20.0 L=1.48×10−3 M PO3−4

Therefore, [PO3−4]\\ has a molarity of  1.48×10−3 M PO3−

To learn more about Molarity click on the link below:

brainly.com/question/19943363

#SPJ4

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lys-0071 [83]

Answer:

Explanation:

Given parameters:

           pH = 3.50

Unknown:

    concentration of [H₃0⁺] = ?

    concentration of [OH⁻] = ?

Solution:

In order to find the unknown, we use some simple expressions which best explains the pH scale and the equilibrium systems of aqueous solutions.

         pH = -log₁₀[H₃O⁺]

         [H₃O⁺] = inverse log₁₀ (-pH) = 10^{-pH} = 10^{-3.5}

          [H₃O⁺] = 3.2 x 10⁻⁴moldm⁻³

       

For the  [OH⁻]:

       we use : pOH = -log₁₀ [OH⁻]

     Recall: pOH + pH = 14

                  pOH = 14 - pH = 14 - 3.5 = 10.5

  Now we plug the value of pOH into pOH = -log₁₀ [OH⁻]

                                   [OH⁻] = 10^{-pOH}

                        [OH⁻] = 10^{-10.5} = 3.2 x 10⁻¹¹moldm⁻³

The solution is acidic as the concentration of H₃0⁺ is more than that of the OH⁻ ions.

                   

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Why are koalas so crusty?
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An intravenous saline drip has 9.8 g of sodium chloride per liter of water. by definition, 1 ml = 1 cm3.
ELEN [110]
Missing question: Express the salt concentration in kg/m³.
Answer is: the salt concentration is 9.8 kg/m³.
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V(solution) = 1 L = 1 dm³.
V(solution) = 1 dm³ ÷ 1000 dm³/m³.
V(solution) = 0.001 m³.
d(solution) = m(NaCl) ÷ V(solution).
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4 0
3 years ago
How much heat is required to convert 20.0 g of ice at 50.0⁰C to liquid water at 0.0⁰C? The specific heat of ice is 2.06 J/(g∙⁰C)
Alex777 [14]

Answer:

8740 joules are required to convert 20 grams of ice to liquid water.

Explanation:

The amount of heat required (Q), measured in joules, to convert ice at -50.0 ºC to liquid water at 0.0 ºC is the sum of sensible heat associated with ice and latent heat of fussion. That is:

Q = m\cdot [c\cdot (T_{f}-T_{o})+L_{f}] (1)

Where:

m - Mass, measured in grams.

c - Specific heat of ice, measured in joules per gram-degree Celsius.

T_{o}, T_{f} - Temperature, measured in degrees Celsius.

L_{f} - Latent heat of fussion, measured in joules per gram.

If we know that m = 20\,g, c = 2.06\,\frac{J}{g\cdot ^{\circ}C}, T_{f} = 0\,^{\circ}C, T_{o} = -50\,^{\circ}C and L_{f} = 334\,\frac{J}{g }, then the amount of heat is:

Q = (20\,g)\cdot \left\{\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot [0\,^{\circ}C-(-50\,^{\circ}C)]+334\,\frac{J}{g} \right\}

Q = 8740\,J

8740 joules are required to convert 20 grams of ice to liquid water.

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A student sets up the following equation to convert a measurement. (The ? stands for a number the student is going to calculate.
kvasek [131]

Answer:

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Explanation:

From the question given above,

the student is trying to convert 0.010 m/s to a number in cm/s.

Thus, we can convert 0.010 m/s to cm/s as illustrated below:

Recall:

1 m/s = 100 cm/s

Therefore,

0.010 m/s = 0.010 m/s × 100 cm/s ÷ 1 m/s

0.010 m/s = 1 cm/s

Therefore, 0.010 m/s is equivalent to 1 cm/s

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