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3 years ago

Calculate the empirical formula for DCBN, given that the percent composition is 48.9%C, 1.76%H, 41.2%Cl, and 8.15%N.

Chemistry

1 answer:

sweet [91]3 years ago

5 0

Answer:

empirical formula is C7H3NCl2

Explanation:

too much work too explain and im lazy

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Is it possible for the equivalence point of a titration to not be at pH 7? Explain your answer.

lara31 [8.8K]

<span>The reason it will be 7 for some titrations is that when you titrates a strong acid with a strong base for example HCl and NaOH the salt formed is conjugate base of strong acid and will be a very weak base
That means that it cannot produce any OH^-1 and all the H+ has been converted to water.The only source of H+ or OH is water with a Ka of 10^-14 so the pH = -log [H+]=-log 10^-7 = 7
second reason is
When you titrates a weak acid with strong base at equivalence point
only a water solution of the conjugate base exists

CH3COOH + NaOH ----- Na+ CH3COO^-1 + H2O
Since the conjugate base is the conjugate base of a weak acid it will hydrolyze in water like so
for instance Na+ CH3COO^-1 + HCl---- CH3COOH + NaCl the equivalence point will be way BELOW 7 and in the case of above will be less than 5. So pH of 7 at equivalence point is only reached in strong acid strong base titrations.
hope this helps</span>

5 0

3 years ago

Read 2 more answers

5. Write the names of the following ions.

Talja [164]

A.) Phosphate ion or Orthophosphate
d.) Hydroxide
D.) Ammonium
e.) Iron
C.) Nitrate
f.) Sulfur dioxide

5 0

2 years ago

Which element consists of positive ions immersed in a “sea” of mobile electrons?

denis-greek [22]

Answer:

Calcium

Explanation:

Calcium is the only metal here

hope this helps...

8 0

3 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago

How many grams of phosphorus are in 500.0 grams of calcium phosphide? (i need the work also)

kvv77 [185]

Answer:

$\boxed{\text{170.0 g P}}$

Explanation:

The formula of calcium nitride is Ca₃P₂.

The masses of each element are:

$\begin{array}{lrcr}\text{3Ca:} & 3 \times 40.08&=& \text{120.24 u}\\\text{2P:} & 2\times 30.97&=& \text{61.94 u}\\& \text{TOTAL} & = & \text{182.18 u}\\\end{array}$

So, there are 61.94 g of P in 182.18 g of Ca₃P₂.

In 500 g of Ca₃P₂:

$\text{Mass of P} = \text{500.0 g Ca$_{3}$P$_{2}$} \times \dfrac{\text{61.94 g P}}{\text{182.18 g Ca$_{3}$P$_{2}$}} = \text{170.0 g P}$

There are $\boxed{\textbf{170.0 g P}}$ in 500.0 g of Ca₃P₂.

3 0

3 years ago