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Dennis_Churaev [7]

2 years ago

7

fill in the blank: 1.)the amount of space that matter fills is its.................. 2.)a state of matter with a definite volume

, but no definite shape is a .................... 3.)A................. will always take the shape and volume of its container. 4.)The ................ is a measure of the average speed of the particles in a substance. 5.)The ................ of a gas is the force of its outward push divided by the area of the walls of its container HELP PLEASE

2 answers:

marshall27 [118]2 years ago

8 0

1) mass

2) gas

3) liquid

4) temperature

5) pressure

katrin [286]2 years ago

4 0

The answers are: Gas,Liquid,Mass,Pressure,Temerature.

I hope this helps you :D

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Answer: Double displacement reaction

Explanation:

Double displacement reaction is defined as the reaction where exchange of ions takes place.

The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The chemical reaction between aqueous magnesium sulfate and aqueous sodium carbonate is represented as:

$MgSO_4(aq)+Na_2CO_3(aq)\rightarrow Na_2SO_4(aq)+MgCO_3(s)$

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Explanation:

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Answer:

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Explanation:

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

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2 years ago

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Ksju [112]

Elements that belong to the same period which is seen horizontally in the periodic table have different properties. Elements with similar properties are seen in columns or groups. Elements in the same period have similar principal quantum number and belong to the same principal shell. Hence the answer is that t<span>hey have different properties that repeat across the next period</span>.

4 0

2 years ago

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