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Alekssandra [29.7K]
3 years ago
5

(6.0×10^4)(3.1×10^-1)

Chemistry
1 answer:
yKpoI14uk [10]3 years ago
6 0
(6.0×10^4) = 60,000
(3.1×10^-1) = 0.31

60,000 × 0.31 = 18,600
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Name 5 affects of weathering and erosion
nadezda [96]

Answer Water, wind, and ice can also move pieces of rock or land to new places. The wearing away of a surface of rock or soil is called weathering. ... Weathering and erosion can cause changes to the shape, size, and texture of different landforms (such as mountains, riverbeds, beaches, etc).:

Explanation:

3 0
3 years ago
Consider the reactionI2(g) + Cl2(g)2ICl(g)Using standard thermodynamic data at 298K, calculate the entropy change for the surrou
Georgia [21]

We know,

\Delta H_{I_2(g)}=62.438\ KJ/mol\\\\\Delta H_{Cl_2(g)}= 0.0\ KJ/mol\\\\\Delta H_{ICl(g)}=17.78\ KJ/mol

For given reaction, I_2(g)+Cl_2(g)\ -->\ 2ICl(g)

\Delta H_{rxn}=2\Delta H_{ICl(g)}-\Delta H_{I_2(g)}-\Delta H_{Cl_2(g)}\\\\\Delta H_{rxn}=2(17.78)-0-62.438\ KJ/mol\\\\\Delta H_{rxn}=-26.878\ KJ/mol

For , 2.41 moles of I_2 :

\Delta H_{rxn}=2.41\times (-26.878)\ KJ\\\\\Delta H_{rxn}=-64.78\ KJ

We know :

\Delta S = -\dfrac{\Delta H_{rxn}}{T}\\\\\Delta S = -\dfrac{-64.78}{298}\ KJ/K\\\\\Delta S =-0.21738 \ KJ/K\\\\\Delta S=-217.38\ J/K

Hence, this is the required solution.

7 0
3 years ago
How many grams of Co are needed to react with an excess of fe203 to produce 156.2 g FE? show your work.
Gekata [30.6K]

The reaction is given as

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No.of moles=mass in gram/molar mass

As for Fe mole =156.2g/55.847=2.7969~2.797

The ratio b/w CO and Fe is 3:2

Moles of CO needed= 2.797x3/2=4.1955

Mass of CO needed= 4.195mol x 28.01g/mol= 117.515g

8 0
3 years ago
Balance the 2 chemical equation below: 1) _____H2 + _____02 ---> _____H20 2) _____CH4 + _____Cl2 ---> _____CCl4 + _____HCl
sergij07 [2.7K]

Answer:

1.)    2H2  +  O2  --->  2H2O

2.)   CH4  +  4Cl2 --->  CCl4  +  4HCl

Explanation:

In order to balance the equation, you have to make sure there are the same amount of each element on both sides.

Ex (#1).  By adding the coefficients infront of H2 and H20, You now have 4 hydrogen's and 2 oxygens on the left side as well as 4 hydrogens and 2 oxygens on the right side.

Sorry if that's confusing, but hope this helps! :)

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3 years ago
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Ivahew [28]

Answer:

0.671081 mps

Explanation:

.....................

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