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Anna [14]
3 years ago
13

. ¿Qué cantidad de HNO3 concentrado y de solvente se utilizará en la preparación de 500 mLal 5% (v/v)?

Chemistry
1 answer:
Viefleur [7K]3 years ago
6 0

Answer:

464.29mL de solvente y 35.71mL de ácido nítrico concentrado deben agregarse.

Explanation:

El ácido nítrico concentrado viene al 70% v/v por temas de estabilidad. El volumen de ácido nítrico que se debe agregar si se quieren hace 500mL al 5% de HNO3 es:

500mL * (5mL / 100mL) = 25mL de ácido nítrico se deben agregar.

Como el ácido nítrico está al 70%:

25mL ácido nítrico * (100mL / 70mL ácido nítrico) = 35.71mL de ácido nítrico concentrado deben agregarse.

Y el volumen de solvente debe ser:

500mL - 35.71mL = 464.29mL

You might be interested in
Several properties of water are shown. Classify each property as a physical property or a chemical property.
max2010maxim [7]

can be split into hydrogen and oxygen             chemical property

is liquid at room temperature                              physical property

has a density of 1.0 g/cm³                                    physical property

reacts with certain metals                                  chemical property

Explanation:

The physical properties of matter are the properties that occur when no change is occurring to a matter. They are usually observed with our senses or some simple laboratory experiments.

Some of the physical properties are boiling point, density, state of matter, e.t.c

The chemical properties are the properties that occurs when a matter undergoes some chemical changes.

For example rusting, decomposition, reactivity e.t.c

Learn more:

Chemical properties brainly.com/question/2376134

#learnwithBrainly

8 0
3 years ago
In each row check off the boxes that apply to the highlighted reactant. reaction The highlighted reactant acts as a... (check al
tekilochka [14]

The given question is incomplete. The complete question is :

In each row check off the boxes that apply to the underlined reactant. The underlined reactant acts as a... (check all that apply)

1. HCH_3CO_2(aq)+NH_3(aq)\rightarrow CH_3COO^-(aq)+NH_4^+(aq)

here underlined is HCH_3CO_2

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

2. BH_3(aq)+NH_3(aq)\rightarrow BH_3NH_3(aq)

Here underlined is NH_3

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

3. HNO_2(aq)+C_2H_5NH_2(aq)\rightarrow NO_2^-(aq) + C_2H_5NH_3^+(aq)

Here underlined is C_2H_5NH_2

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

Answer: 1. Brønsted-Lowry acid

2. Lewis base

3. Brønsted-Lowry base

Explanation:

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

1.  HCH_3CO_2(aq)+NH_3(aq)\rightarrow CH_3CO^{2-}(aq)+NH_4^+aq)

As HCH_3CO_2(aq) is donating a proton , it acts as a bronsted acid.

2. BH_3(aq)+NH_3(aq)\rightarrow BH_3NH_3(aq)

As NH_3 contains a lone pair of electron on nitrogen , it can easily donate electrons to BH_3 and act as lewi base.

3.  HNO_2(aq)+C_2H_5NH_2(aq)\rightarrow NO_2^-(aq) + C_2H_5NH_3^+(aq)

As C_2H_5NH_2(aq) is accepting a proton , it acts as a bronsted base.

7 0
3 years ago
A 26.08 g mixture of zinc and sodium is reacted with a stoichiometric amount of sulfuric acid. The reaction mixture is then reac
Over [174]

Answer:

Molar percent of sodium in original mixture is 88,50%

Explanation:

The last reaction is:

BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl

The moles of BaCl₂ are:

0,132L × 3,80M = 0,502 moles of BaCl₂

As the amount of BaCl₂ is the maximum possible to produce BaSO₄, the moles of BaCl₂ must be the same than moles of Na₂SO₄.

0,502 moles of BaCl₂ ≡ 0,502 moles of Na₂SO₄

These moles of Na₂SO₄ comes from:

2 Na + H₂SO₄ → Na₂SO₄ + H₂

As the reaction is in stoichiometric amounts, moles of Na are twice the moles of Na₂SO₄

0,502 moles of Na₂SO₄ ×\frac{2molesNa}{1moleNa_{2}SO_{4}}× 22,99 g/mole = 23,08 g of Na

Molar percent of sodium in original mixture is:

\frac{23,08g}{26,08g}*100 = <em>88,50% </em>

I hope it helps

4 0
3 years ago
The specific heat of copper metal is 0. 385 J/(g °C). How much energy must be added to a 35. 0-gram sample of copper to change t
Rus_ich [418]

The amount of heat required for changing the temperature of copper has been 606 J. Thus, option B is correct.

Specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The heat required to raise the temperature has been expressed as:

\rm Heat=mass\;\times\;specific\;heat\;\times\;Change\;in\;temperature

<h3>Computation for the heat energy required</h3>

The given specific heat of copper has been \rm 0.385\;J/g^\circ C

The mass of copper has been, \rm 35\;g

The initial temperature of copper has been, \rm 20^\circ C

The final temperature of copper has been, \rm 65^\circ C

The change in temperature has been, \Delta T

\Delta T=\text{Final\;temperature-Initial\;temperature}\\\Delta T =65^\circ \text C-20^\circ \text C\\\Delta T=45^\circ \text C

Substituting the values for the heat required as:

\rm Heat=35\;g\;\times\;0.385\;J/g^\circ C\;\times\;45^\circ C\\Heat=606\;J

The amount of heat required for changing the temperature of copper has been 606 J. Thus, option B is correct.

Learn more about specific heat, here:

brainly.com/question/2094845

7 0
2 years ago
A 0.216 g sample of an aluminium compound X reacts with an excess of water to produce a single hydrocarbon gas. This gas burns c
makkiz [27]
0.216g of aluminium compound X  react with an excess of water water to produce gas. this gas burn completely  in O2  to form H2O and 108cm^3of CO2 only . the volume of CO2 was measured at room temperature and pressure

0.108 / n  =  24 / 1 
n = 0.0045 mole ( CO2 >>0.0045 mole 
0.216 - 0.0045 = 0.2115
so Al =   0.2115 / 27  =>  0.0078 mole 
C = 0.0045 * 1000 => 4.5    and Al  = 0.0078 * 1000 = 7.8 

7 0
3 years ago
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