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2 years ago

if 0.50 L of a 5.00 M stock solution of HCL is diluted to make 2.0 L of solution, how much HCL, in grams, is in the solution?

1 answer:

3 0

There’s 1,000 grams per liter. I’ll help you out by giving you the formula.

0.50L over 5.00 M= 1,000grams over one liter

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When positively charged particles were radiated onto a gold atom, a few of the particles bounced back. Which of the following ca

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Positively charged protons in the nucleus, hope this helps.

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3 years ago

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CK-12 Boyle and Charles's Laws if Mrs. Pa pe prepares 12.8 L of laughing gas at 100.0 k Pa and -108 °C and then she force s the

nirvana33 [79]

Answer:

The answer to your question is P2 = 2676.6 kPa

Explanation:

Data

Volume 1 = V1 = 12.8 L Volume 2 = V2 = 855 ml

Temperature 1 = T1 = -108°C Temperature 2 = 22°C

Pressure 1 = P1 = 100 kPa Pressure 2 = P2 = ?

Process

- To solve this problem use the Combined gas law.

P1V1/T1 = P2V2/T2

-Solve for P2

P2 = P1V1T2 / T1V2

- Convert temperature to °K

T1 = -108 + 273 = 165°K

T2 = 22 + 273 = 295°K

- Convert volume 2 to liters

1000 ml -------------------- 1 l

855 ml -------------------- x

x = (855 x 1) / 1000

x = 0.855 l

-Substitution

P2 = (12.8 x 100 x 295) / (165 x 0.855)

-Simplification

P2 = 377600 / 141.075

-Result

P2 = 2676.6 kPa

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3 years ago

What molecules can produce more energy per gram than sugars and are used for storing energy in animals?

Andrei [34K]

The answer is lipids

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3 years ago

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John Fang has 5 apples, Jonny ate 3, how many are left!

Doss [256]

Answer:5-3=2

As 5apple eaten 3apple so left 2apple

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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