A solution in which the pH is 1.6 would be described asa.
very acidic
b.
slightly acidic
c.
very basic
d.
slightly basic
e.
neutral

Answers

Answer 1
Answer:

Answer:

a.  very acidic

Explanation:

A solution with a pH of 1.6 would be described as a very acidic solution.

The pH scale is a convenient scale for expressing the level of acidity or alkalinity of aqueous solutions.

The scale ranges from 1 to 14 in which case 1 is the lower limit and 14 the upper limit.  

  • An acidic solution has a pH value less than 7. As pH decreases, the acidity of the solution increases.
  • Neutral solutions have a pH of 7
  • An alkaline solution has a pH greater than 7 and the level of alkalinity increases as pH increases.

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A cube of aluminum measures 3 cm on each side and weighs 81 g. What is its density? (HINT: find the volume of the block first)D=

M=

V=

Answers

The density of the aluminum is 3 g/cm³

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance.

Density = (Mass )/(Volume )

To obtain the density of the aluminum, we'll begin by calculating the volume of the aluminum. This can be obtained as follow:

From the question given above, each side of the aluminum cube measures 3 cm. Thus,

Length (L) = 3 cm

Width (W) = 3 cm

Height (H) = 3 cm

Volume = Length × Width × Height

Volume = 3 × 3 × 3

Volume = 27 cm³

Therefore, the volume of the aluminum is 27 cm³

Finally, we shall determine the density of the aluminum. This can be obtained as follow:

Mass = 81 g

Volume = 27 cm³

Density =?

Density = (Mass )/(Volume )\n\nDensity = (81)/(27)

Density = 3 g/cm³

Therefore, the density of the aluminum is 3 g/cm³

Learn more: brainly.com/question/19239648

Answer:

27

Explanation: volume-3 times 6=18 /mass- 81 times 6= 486....  486 divided by 18 = 27.... a cube has 6 sides.

Part D 2ClO2(g)+2I−(aq)→2ClO−2(aq)+I2(s) Drag the appropriate labels to their respective targets. ResetHelp e−→e Superscript- rightarrow ←e−leftarrow e Superscript- Cathode Cathode Anode Anode II Superscript- ClO−2C l O Subscript 2 Superscript- Request Answer Part E Indicate the half-reaction occurring at Anode. Express your answer as a chemical equation. Identify all of the phases in your answer. nothing

Answers

The half reaction occurring at anode is:

2I^-(aq)---- > I_2(s)+2e^-

Half reaction for the cell:

The substance having highest positive  potential will always get reduced and will undergo reduction reaction.

Balanced chemical equation:

2ClO_2(g)+2I^-(aq)----- > 2ClO^(2-)(aq)+I_2(s)

The half reaction follows:

Oxidation half reaction:  2I^-(aq)---- > I_2(s)+2e^- , Reduction potential is 0.53V

Reduction half reaction:  ClO_2(g)+e^----- > ClO_2^-   ( × 2 ), Oxidation potential is +0.954 V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is :

2I^-(aq)---- > I_2(s)+2e^-

Find more information about Reduction potential here:

brainly.com/question/7484965

Answer: The half reaction occurring at anode is 2I^-(aq.)\rightarrow I_2(s)+2e^-

Explanation:

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

For the given chemical equation:

2ClO_2(g)+2I^-(aq)\rightarrow 2ClO^(-2)(aq.)+I_2(s)

The half reaction follows:

Oxidation half reaction:  2I^-(aq.)\rightarrow I_2(s)+2e^-;E^o_(I_2/I^-)=0.53V

Reduction half reaction:  ClO_2(g)+e^-\rightarrow ClO_2^-(aq.);E^o_(ClO_2/ClO_2^-)=+0.954V    ( × 2 )

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is 2I^-(aq.)\rightarrow I_2(s)+2e^-

A gas mixture with a total pressure of 770 mmHg contains each of the following gases at the indicated partial pressures: 128 mmHg CO2, 232 mmHg Ar, and 183 mmHg O2. The mixture also contains helium gas. What is the partial pressure of the helium gas?

Answers

The partial pressure of  Helium gas can be calculated by applying Daltons law of partial pressure and that comes out to be 227 mmHg.

What is  Dalton's law of partial pressure?

Dalton's law of partial pressure helps in calculating the total pressure exerted by all the gases that are present in a mixture that is equal to the sum of partial pressure of all individual gases.

Partial pressure is the pressure that is exerted by one gas in a mixture of gases on the walls of the container. This law is applicable to only mixture of gases that are ideal gases. Ideal gases are the one which do not have any forces of attraction or repulsion between the particles of thee gas.

In our question

p_(total) =p_{CO_(2) }+p_(Ar)+p_{O_(2) } +p_{{He} }

p_{{He} } =p_(total)-[p_{CO_(2) }+p_(Ar)+p_{O_(2) } ]

substituting all given values

p_{{He} } =770-(128+232+183)

      =227mmHg

Thus the partial pressure of Helium gas is 227mmHg

To learn more about Dalton's law of partial pressure, here:

brainly.com/question/14119417

#SPJ2

Answer:

227 mmHg

Explanation:

Ptotal= PCO2 +PAr+PO2+PHe

PHe= Ptotal-PCO2 -PAr-PO2

PHe= (770 mmHg)-(128 mmHg)-(232 mmHg)-(183 mmHg)

PHe=227 mmHg

Plants make their own food through a process called ____________.(6th grade science)

Answers

Answer:

Photosynthesis

Explanation:

Question:

Plants make their own food through a process called ____________.(6th grade science)

Answer:

The correct answer to the blank is photosynthesis. Photosynthesis is the process that enables plants to get energy from the sun. Light energy from the sun is converted into chemical energy by chlorophyll. Chlorophyll gives plants their green color.

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A solution of salt and water is 33.0% salt by mass and has a density of 1.50 g/ml. what mass of the salt in grams is in 5.00l of this solution?

Answers

To answer this item, we solve first for the mass of the solution by multiplying the density by the volume. That is,

      m = (density)(volume)

Substituting the known values,
    m = (1.50 g/mL)(5L)(1000 mL/1L)
      m = 7500 grams

To determine the mass of the salt in the solution, multiply the calculated mass of the solution by the decimal equivalent of the percent salt in the solution.

      m of salt = (7500 g)(0.33)
      m of salt = 2475 grams

Answer: 2475 grams

Microwave ovens use microwaves with a wavelength of 0.12 meters. How much energy does this wave have? 2 x 10-20 10 mm FI​

Answers

Answer:

E = 165.75×10⁻²⁶ J

Explanation:

Given data:

Wavelength = 0.12 m

Energy of wave = ?

Solution:

Formula:

E = h c/λ

c = 3×10⁸ m/s

h = 6.63×10⁻³⁴ Js

Now we will put the values in formula.

E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 0.12 m

E = 19.89×10⁻²⁶ J.m / 0.12 m

E = 165.75×10⁻²⁶ J