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1 year ago

10 mL of 2(M) NaOH solution is added to 200 mL of 0.5 (M) of NaOH solution. What is the final concentration?

Chemistry

1 answer:

ratelena [41]1 year ago

3 0

Hello its me Sanya yadav

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Which choice below would affect the rate of reaction in the opposite way from the other four?

bulgar [2K]

Adding a catalyst as this would speed up the reaction and the rest would slow it down

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3 years ago

C (graphite) is used as a lubricant, whereas C (diamond) is used as an abrasive. Why is this?

zysi [14]

Answer:

Carbon atoms in graphite and diamond are arranged in different ways. Hence, the two allotropes of carbon have different physical properties.

Explanation:

Both graphite and diamond are both made of only carbon atoms. However, their physical properties differ from each other. Hence, they are called allotropes. Think about how these carbon atoms are arranged in each of the allotropes.

<h3>Graphite</h3>

In graphite, each carbon atom is bonded to three other carbon atoms. These carbon atoms will be located in the same plane. A chunk of graphite can contain many of these planes.

Each carbon atom has four valence electrons. Three of these electrons will be used in the bonds. The other electron will be delocalized. These electrons would flow between the sheets of carbon atoms. That keeps the sheets separate and allow them to slide on top of each other.

<h3>Diamond</h3>

In diamond, each carbon atom is bonded to four other carbon atoms. These carbon atoms will form a tetrahedral network.

In graphite, there's a significant separation between two adjacent sheets of carbon atoms. The force between the two sheets is rather weak. When a piece of graphite is between two objects that move over one another, the layers in the graphite would also slide over one another. Since the attraction between two adjacent sheets isn't very strong, there wouldn't be much resistance. Hence the graphite acts as a lubricant.

In contrast, most of the carbon atoms in a piece of diamond would be connected to each other. Unlike the sheets in graphite, in a diamond there are almost no moving parts. Also, the forces between neighboring carbon atoms are very strong. When an external force acts on a chunk of diamond, the carbon atoms would barely move. Hence, the structure appears to be very rigid. That gives diamond its abrasive properties.

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3 years ago

How many electrons will a metal generally have in its outer energy level?

Zarrin [17]

Answer:

Metals have one or two electrons in their outermost shell

C. 1-2

Explanation:

Metals have low ionisation energy because they easily looses the outermost electrons

They have only one- two electrons in the outer most shell.

They loose these electron to form charged species called cation.

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3 years ago

How much heat is required to convert 20.0 g of ice at 50.0⁰C to liquid water at 0.0⁰C? The specific heat of ice is 2.06 J/(g∙⁰C)

Alex777 [14]

Answer:

8740 joules are required to convert 20 grams of ice to liquid water.

Explanation:

The amount of heat required ( $Q$ ), measured in joules, to convert ice at -50.0 ºC to liquid water at 0.0 ºC is the sum of sensible heat associated with ice and latent heat of fussion. That is:

$Q = m\cdot [c\cdot (T_{f}-T_{o})+L_{f}]$ (1)

Where:

$m$ - Mass, measured in grams.

$c$ - Specific heat of ice, measured in joules per gram-degree Celsius.

$T_{o}$ , $T_{f}$ - Temperature, measured in degrees Celsius.

$L_{f}$ - Latent heat of fussion, measured in joules per gram.

If we know that $m = 20\,g$ , $c = 2.06\,\frac{J}{g\cdot ^{\circ}C}$ , $T_{f} = 0\,^{\circ}C$ , $T_{o} = -50\,^{\circ}C$ and $L_{f} = 334\,\frac{J}{g }$ , then the amount of heat is:

$Q = (20\,g)\cdot \left\{\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot [0\,^{\circ}C-(-50\,^{\circ}C)]+334\,\frac{J}{g} \right\}$

$Q = 8740\,J$

8740 joules are required to convert 20 grams of ice to liquid water.

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2 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

10 mL of 2(M) NaOH solution is added to 200 mL of 0.5 (M) of NaOH solution. What is the final concentration?​

10 mL of 2(M) NaOH solution is added to 200 mL of 0.5 (M) of NaOH solution. What is the final concentration?