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GrogVix [38]
2 years ago
7

Unit Conversion Help Thank you

Chemistry
1 answer:
AlekseyPX2 years ago
5 0

Answer : 1721.72 g/qt are in 18.2 g/cL

Explanation :

As we are given: 18.2 g/cL

Now we have to convert 18.2 g/cL to g/qt.

Conversions used are:

(1) 1 L = 100 cL

(2) 1 L = 1000 mL

(3) 1 qt = 946 qt

The conversion expression will be:

\frac{18.2g}{1cL}\times \frac{100cL}{1L}\times \frac{1L}{1000mL}\times \frac{946mL}{1qt}

=1721.72\text{ g/qt}

Therefore, 1721.72 g/qt are in 18.2 g/cL

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Zinc has a specific heat capacity of 0.390 J/goC. What is its molar heat capacity? Enter your answer numerically to three signif
ch4aika [34]

Answer:

The answer to your questions is  Cm = 25.5 J/mol°C  

Explanation:

Data

Heat capacity = 0.390 J/g°C

Molar heat capacity = ?

Process

1.- Look for the atomic number of Zinc

     Z = 65.4 g/mol

2.- Convert heat capacity to molar heat capacity

       (0.390 J/g°C)(65.4 g/mol)

- Simplify and result

   Cm = 25.5 J/mol°C  

3 0
1 year ago
Assuming equal concentrations of conjugate base and acid, which one of the following mixtures is suitable for making a buffer so
BartSMP [9]

Answer:

NH₃/NH₄Cl

Explanation:

We can calculate the pH of a buffer using the Henderson-Hasselbalch's equation.

pH=pKa+log\frac{[base]}{[acid]}

If the concentration of the acid is equal to that of the base, the pH will be equal to the pKa of the buffer. The optimum range of work of pH is pKa ± 1.

Let's consider the following buffers and their pKa.

  • CH₃COONa/CH3COOH (pKa = 4.74)
  • NH₃/NH₄Cl (pKa = 9.25)
  • NaOCl/HOCl (pKa = 7.49)
  • NaNO₂/HNO₂ (pKa = 3.35)
  • NaCl/HCl Not a buffer

The optimum buffer is NH₃/NH₄Cl.

4 0
1 year ago
The dissociation of calcium carbonate has an equilibrium constant of Kp = 1.16 at 1073 K . CaCO3(s) ⇄ CaO(s) + CO2(g) If you pla
tino4ka555 [31]

Answer:

<h3>Pressure of CO_2 in the container=1.6 atm</h3>

Explanation:

First balance the chemical equation:

CaCO_3(s) ⇄  CaO(s) + CO_2(g)

two components are solid so these two will not exert any kind of pressure in the container so at equilibrium only CO2 will apply pressure on the container

Therefore only partial pressure of CO2 will be taken for the calculation of equilibrium pressure constant i.e. Kp

K_p=[CO_2]

[CO_2]=p

K_p=p

p=K_p = 1.16atm

Pressure of CO_2 in the container=1.6 atm

8 0
2 years ago
How are NGC 1427A and U different? How are they the same?
agasfer [191]

Answer:

that looks pretty and also well NGC 1427A has no general shape, so it is an irregular galaxy. U has a bulge in the center and arms, so it is a spiral galaxy. They are similar in the both certain plenty of dust and gas. Both also have active star-forming sites.

3 0
1 year ago
Read 2 more answers
Tag all the carbon atoms with pi bonds in this molecule. If there are none, please check the box.
snow_tiger [21]

Answer:

Pi bonds (π bonds) are covalent chemical bonds where two lobes of an orbital involved in the bond overlap with two lobes of the other orbital involved. These orbitals share a nodal plane that passes through the nuclei involved. Are generally weaker than sigma links, because their negatively charged electronic density is further from the positive charge of the atomic nucleus, which requires more energy.

They are frequent components of multiple bonds, as is the molecule indicated in our exercise.

The characteristics that distinguish pi bonds from other kinds of interactions between atomic species are described below, beginning with the fact that this union does not allow the free rotation movement of atoms, such as carbon. For this reason, if there is rotation of the atoms, the bond is broken.

Explanation:

In order to describe the formation of the pi bond, first we must talk about the hybridization process, as this is involved in some important links.

Hybridization is a process where hybrid electronic orbitals are formed; that is, where orbitals of atomic sub-levels s and p can get mixed. This causes the formation of sp, sp2 and sp3 orbitals, which are called hybrids.

In this sense, the formation of pi bonds occurs thanks to the overlapping of a pair of lobes belonging to an atomic orbital over another pair of lobes that are in an orbital that is part of another atom.

This orbital overlap occurs laterally, so the electronic distribution is mostly concentrated above and below the plane formed by the linked atomic nuclei, and causes the pi bonds to be weaker than the sigma bonds.

When talking about the orbital symmetry of this type of junction, it should be mentioned that it is equal to that of the p-type orbitals as long as it is observed through the axis formed by the bond. In addition, these junctions are mostly made up of p orbitals.

Since pi bonds are always accompanied by one or two more links (one sigma or another pi and one sigma), it is relevant to know that the double bond that is formed between two carbon atoms has less bond energy than that corresponding to two Sometimes the sigma link between them.

4 0
2 years ago
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