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Kaylis [27]
2 years ago
6

I added and multiplied i still don't know

Chemistry
1 answer:
hjlf2 years ago
8 0

Answer:

it'a answer number 2

Explanation: you divide the mass by volume and 32.2 divided by 4 is 8.05

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This type of chemical bond forms when electrons are either gained or lost by an atom.
Ket [755]
It's ionic because electrictrons are only lost or gained in ionic bonds
5 0
3 years ago
Qué tipo de reacciones pertenecen las combustiones?
navik [9.2K]

Answer:

Si una reacción química libera energía, se llama reacción exotérmica. El ejemplo más común es la combustión, en la cual la energía se manifiesta en forma de calor y luz. Si, por el contrario, la reacción química requiere energía del medio para efectuarse, recibe el nombre de reacción endotérmica.

5 0
2 years ago
Ammonia can be produced in the laboratory by heating ammonium chloride
AnnyKZ [126]

Answer:

Mass = 2.89 g

Explanation:

Given data:

Mass of NH₄Cl = 8.939 g

Mass of Ca(OH)₂ = 7.48 g

Mass of ammonia produced = ?

Solution:

2NH₄Cl   +  Ca(OH)₂     →    CaCl₂ + 2NH₃ + 2H₂O

Number of moles of NH₄Cl:

Number of moles = mass/molar mass

Number of moles = 8.939 g / 53.5 g/mol

Number of moles = 0.17 mol

Number of moles of Ca(OH)₂ :

Number of moles = mass/molar mass

Number of moles = 7.48 g / 74.1 g/mol

Number of moles = 0.10 mol

Now we will compare the moles of ammonia with both reactant.

                      NH₄Cl          :          NH₃

                          2              :           2

                         0.17          :          0.17

                   Ca(OH)₂         :          NH₃

                        1                :           2

                    0.10              :          2/1×0.10 = 0.2 mol

Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 0.17 mol × 17 g/mol

Mass = 2.89 g

6 0
2 years ago
Water has a boiling point of 100.0°C and a Kb of 0.512°C/m. What is the boiling
Reil [10]

Answer:

The boiling  point of a 8.5 m solution of Mg3(PO4)2 in water is<u> 394.91 K.</u>

Explanation:

The formula for molal boiling Point elevation is :

\Delta T_{b} = iK_{b}m

\Delta T_{b} = elevation in boiling Point

K_{b} = Boiling point constant( ebullioscopic constant)

m = molality of the solution

<em>i =</em> Van't Hoff Factor

Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .

In solution Mg3(PO4)2 dissociates as follow :

Mg_{3}(PO_{4})_{2}\rightarrow 3Mg^{2+} + 2 PO_{4}^{3-}

Total ions after dissociation in solution :

= 3 ions of Mg + 2 ions of phosphate

Total ions = 5

<em>i =</em> Van't Hoff Factor = 5

m = 8.5 m

K_{b} = 0.512 °C/m

Insert the values and calculate temperature change:

\Delta T_{b} = iK_{b}m

\Delta T_{b} = 5\times 0.512\times 8.5

\Delta T_{b} = 21.76 K

Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K

\Delta T_{b} = T_{b} - T_{b}_{pure}

T_{b}_{pure} = 373.15 K[/tex]

21.76 = T - 373.15

T = 373.15 + 21.76

T =394.91 K

8 0
3 years ago
How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that
tia_tia [17]

Answer:

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,  

= 2.50 L × 0.800 mol/L

= 2 mol

The given pH of a buffer is 8.53

pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,  

= -log (1.8 ×10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:  

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73= 5.37

[NH₄⁺ = 5.37 × 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,  

mass of NH₄Cl = 10.74 mol × 53.5 g/mol

= 574.59 grams.  

8 0
3 years ago
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