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3 years ago

On the diagram below, circle the position of the where the southern hemisphere is experiencing spring. Justify your answer

Chemistry

2 answers:

luda_lava [24]3 years ago

5 0

Answer: D

Explanation:

The sun shines the sputhern hemisphere fully on the top= summer

the sun shines somewhat towards the Southern Hemisphere but itfaces away= fall

sun doesnt show at the bottom= winter

faces it more than example b= D

AVprozaik [17]3 years ago

3 0

Summer so is D yup gn

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When copper (II) chloride reacts with sodium nitrate, copper (II) nitrate and sodium chloride are formed.

anzhelika [568]

Answer:

How many grams of copper (II) nitrate is formed

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3 years ago

The reaction of iron with oxygen to form iron oxide is an example of an oxidation–reduction reaction: 4fe + 3o2 → 2fe2o3. in thi

nadya68 [22]

Answer:-

Oxygen gains electrons and is reduced.

Explanation:-

For this reaction the balanced chemical equation is

4Fe + 3O2 --> 2Fe2O3

When Oxygen is present as oxygen gas, the oxidation number of O is Zero since it is the only element present in Oxygen gas.

Similarly Iron is present in Fe with oxidation number Zero.

In the case of Fe2O3, Oxygen has the oxidation number -2 while Iron has +3.

So the oxidation number of Oxygen goes from Zero to -2.

Since the oxidation number decreases Oxygen is reduced.

Since reduction involves gain of electrons, Oxygen gains electrons.

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3 years ago

Explain how dalton's theory of the atom and the conservation of mass are related

Scorpion4ik [409]

Answer:

Dalton's atomic theory states that atoms may be mixed in certain rations to produce compounds, while the law of conservation of mass states that the mass of reactants equals the mass of products. They are related because in the production of compounds, Dalton made it clear that mass can neither be destroyed or created, which supports the conservation of mass law.

Explanation:^^^

Hoped it helped...

8 0

3 years ago

A wafer of gold measuring 15 cm x 20 cm has a mass of 600 g. How thick is this wafer in mm ?

Gre4nikov [31]

600,000 mm if im not mistaken.

7 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