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4 years ago

Which statement would be the most useful for deriving the ideal gas law

Chemistry

2 answers:

Anvisha [2.4K]4 years ago

8 0

Volume is directly proportional to the number of moles

Brums [2.3K]4 years ago

3 0

Answer:

volume is directly proportional to the number of moles

Explanation:

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electric current is passed through water, it decomposes inti hydrogen and oxygen. write full balanced chemical equation for the

Tems11 [23]

Answer:

2H2O ------------------] 2H2 + O2

6 0

3 years ago

During the transamination process, the enzyme transaminases uses ________ as a cofactor.

s344n2d4d5 [400]

During the transamination process, the enzyme transaminases use Pyridoxal pyrophosphate as a cofactor.

All transamination reactions, as well as several amino acid oxylation and deamination processes, involve the coenzyme pyridoxal phosphate. The aminotransferase enzyme's epsilon-amino group of a particular lysine group forms a Schiff-base bond with the aldehyde group of pyridoxal phosphate.

The epsilon-amino group of the lysine residue in the active site is replaced by the alpha-amino group of the amino acid substrate. The ensuing intermediate, a quinoid, undergoes deprotonation to become an aldimine, which is then protonated to become a ketimine by accepting a proton in a different position. Ketamine undergoes hydrolysis, leaving the amino group on the protein complex intact.

Know more about Pyridoxal pyrophosphate at:

brainly.com/question/14117818

#SPJ4

5 0

1 year ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

3 years ago

Choose the compound below that contains at least one polar covalent bond, but is nonpolar.

Ghella [55]

Answer:

The correct choice is the letter <em>d. CF₄ is a compound that contains at least one polar covalent bond (four in fact), bust is nonpolar</em>.

Explanation:

For a compound to have <em>at least one polar covalent bond, but being a not polar compound</em>, the geometric configuration of the molecule must lead to a symmetrical distribution of the charges. This is because, in that way, the dipoles will cancel each other, producing a null net dipole, which makes a non-polar molecule.

Consider the four given molecules:

Since, Se has six valence electrons, and it is bonded to four Br atoms, there will be one lone electron pair which will make that the charge is not symmetrically distributed, leading to a polar molecule. so this is not correct choice.

HCN's structure is linear: H-C≡N.

Nitrogen is more electronegative than carbon an so it is pulling the electrons on in the triple bond with carbon (C≡N) more than the hydrogen on the other side of carbon (H-C)

Therefore, there is a net a dipole moment towards N and the molecule is polar.

This is quite evident non-symmtrical and so you quickly can predict that is is polar.

This molecule has four F atoms around a C atom.

Since, F is more electronegative than F, each C - F bond has a dipole moment (negative charge is closer to F atoms). Then, whe have four dipole moments (<em>covalent bonds</em>) on each CF₄ molecule.

Now, since the molecule is symmetric the four dipoles will cancel each other and the compund in nonpolar.

In conclusion, the correct choice is the letter <em>d. CF₄ is a compound that contains at least one polar covalent bond (four in fact), bust is nonpolar.</em>

8 0

3 years ago

The heat of vaporization for liquid zinc is 1.76 kj/g. How much heat is needed to boil 11.2 g of liquid zinc already at its boil

sladkih [1.3K]

<h3>Answer:</h3>

= 19.712 kJoules

<h3>Explanation:</h3>

Heat of vaporization refers to the amount of heat required to change a unit mass of a substance from liquid to gaseous state without change in temperature.

To calculate the amount of heat, we use,

Amount of heat = Mass × Heat of vaporization

Q = m×Lv

Given;

Mass of liquid Zinc = 11.2 g

Lv of liquid Zinc = 1.76 kJ/g

Therefore;

Q = 11.2 g × 1.76 kJ/g

= 19.712 kJ

Thus, the amount of heat needed to boil 11.2 g of zinc is 19.712 kilo-joules.

6 0

3 years ago