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3 years ago

What volume of 0.205 M K3PO4 solution is necessary to completely react with 134 mL of 0.0102 M NiCl2?

1 answer:

5 0

the balanced equation for the above reaction is as follows;
3NiCl₂ + 2K₃PO₄ ---> Ni₃(PO₄)₂ + 6KCl
stoichiometry of NiCl₂ to K₃PO₄ is 3:2
the number of NiCl₂ moles reacted - 0.0102 mol/L x 0.134 L = 0.00137 mol
according to molar ratio
if 3 mol of NiCl₂ reacts with 2 mol of K₃PO₄
then 0.00137 mol of NiCl₂ reacts with - 2/3 x 0.00137 = 0.000911 mol of K₃PO₄

molarity of given K₃PO₄ solution - 0.205 M
there are 0.205 mol in 1000 mL
therefore volume of 0.000911 mol - 0.000911 mol / 0.205 mol/L = 4.44 mL
volume of K₃PO₄ needed is 4.44 mL

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Which type of solar radiation is the most powerful?

QveST [7]

Gamma ray outbursts are an extremely dangerous sort of radiation.
Gamma rays have the most energy on the electromagnetic spectrum
They also have the shortest wave length.
If a gamma ray burst hit us, the rays would instantly destroy half of the ozone layer, stripping away our most important protective layer. This would cause mass extinction as the other uv rays from our sun would kill us

7 0

3 years ago

Read 2 more answers

Select the correct answer. A certain reaction has this form: aA bB. At a particular temperature and [A]0 = 2.00 x 10-2 Molar, co

Pie

Answer:

$\large \boxed{\text{D. 23.34 min}}$

Explanation:

1. Find the order of reaction

Use information from the graph to find the order.

If a plot of ln[A] vs time is linear, the reaction is first order and the slope = -k.

2. Find the half-life

$k = \dfrac{\ln2}{ t_{\frac{1}{2}}}\\\\k = \text{-slope} = -(-2.97 \times 10^{-2} \text{ min}^{-1}) =2.97 \times 10^{-2} \text{ min}^{-1} \\ t_{\frac{1}{2}} =\dfrac{\ln 2}{k} = \dfrac{\ln 2}{2.97 \times 10^{-2}\text{ min}^{-1}} =\textbf{23.34 min}\\\\\text{The half-life is $\large \boxed{\textbf{23.34 min}}$}$

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4 years ago

How many neutrons does an atom of Copper have in its nucleus?

klasskru [66]

Answer:34.55
explanation:number of neutrons=mass number-atomic number
the mass number for copper is 63.55 so you do 63.55-29 (29 is the atomic number) and you would get 34.55

4 0

3 years ago

Calculate the molar concentrations of H+ and OH− in solutions that have the following pH values:

Anastaziya [24]

Answer:

Explanation:

a )

pH = - log[ H⁺]

8.26 = - log[ H⁺]

[ H⁺] = 10⁻⁸°²⁶ mole / l

= 5.49 x 10⁻⁹ moles / l

[ H⁺] [OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 5.49 x 10⁻⁹

= .182 x 10⁻⁵ moles / l

b )

10.25 = - log[ H⁺]

[ H⁺] = 10⁻¹⁰°²⁵ mole / l

= 5.62 x 10⁻¹¹ moles / l

[ H⁺] [OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 5.62 x 10⁻¹¹

= .178 x 10⁻³ moles / l

c )

4.65 = - log[ H⁺]

[ H⁺] = 10⁻⁴°⁶⁵ mole / l

= 2.24 x 10⁻⁵ moles / l

[ H⁺] [OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 2.24 x 10⁻⁵

= .4464 x 10⁻⁹ moles / l

7 0

3 years ago

The extraction of aluminum metal from the aluminum hydroxide found in bauxite by the Hall-Héroult process is one of the most rem

son4ous [18]

Answer:

10 kg Al(OH)₃

Explanation:

There is some info missing. I think this is the original question.

The extraction of aluminum metal from the aluminum hydroxide found in bauxite by the Hall-Héroult process is one of the most remarkable success stories of 19th-century chemistry, turning aluminum from a rare and precious metal into the cheap commodity it is today.

In the first step, aluminum hydroxide reacts to form alumina (Al₂O₃) and water: 2 Al(OH)₃(s) → Al₂O₃(s) + 3H₂O(g). In the second step, alumina (Al₂O₃ and carbon react to form aluminum and carbon dioxide: 2Al₂O₃(s)+3C(s)→4Al(s)+3CO₂(g). Suppose the yield of the first step is 63% and the yield of the second step is 89%.

Calculate the mass of aluminum hydroxide required to make 2.0 kg of aluminum. Be sure your answer has a unit symbol, if needed, and is rounded to the correct number of significant digits.

Let's consider the 2 steps in the synthesis of Al.

Step 1: 2 Al(OH)₃(s) → Al₂O₃(s) + 3 H₂O(g)

Step 2: 2 Al₂O₃(s) + 3 C(s) → 4 Al(s) + 3 CO₂(g)

In Step 2, the percent yield of Al is 89% and the real yield is 2.0 kg. The theoretical yield is:

2.0 kg (R) × (100 kg (T) / 89 kg (R)) = 2.2 kg = 2.2 × 10³ g

In Step 2, the mass of Al is 4 × 26.98 g = 107.9 g and the mass of Al₂O₃ is 2 × 101.96 g = 203.92g. The mass of Al₂O₃ that produced 2.2 × 10³ g of Al is:

2.2 × 10³ g Al × (203.92g Al₂O₃ / 107.9 g Al) = 4.2 × 10³ g Al₂O₃

In Step 1, the percent yield of Al₂O₃ is 63% and the real yield is 4.2 × 10³ g. The theoretical yield is:

4.2 × 10³ g (R) × (100 g (T)/ 63 g (R)) = 6.7 × 10³ g

In Step 1, the mass of Al₂O₃ is 101.96 g and the mass of Al(OH)₃ is 2 × 78.00 g = 156.0 g. The mass of Al(OH)₃ that produced 6.7 × 10³ g of Al₂O₃ is:

6.7 × 10³ g Al₂O₃ × (156.0 g Al(OH)₃ / 101.96 g Al₂O₃) = 1.0 × 10⁴ g Al(OH)₃ = 10 kg Al(OH)₃

7 0

4 years ago