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omeli [17]
3 years ago
7

4.0 L sample of methane gas is collected at 30.0c, what is the volume of the sample at -8c?

Chemistry
1 answer:
Vedmedyk [2.9K]3 years ago
8 0
This particular law is a gas law, called Charle's Law. The formula is:

V1    V2
---- = ----
T1     T2

So we know our original volume is 4.0L, so we would plug that into our V1. We know T1 is the 30 degrees, since it relates to our original volume. However, we need to convert it to kelvin. We do this simply by adding 273 degrees to the 30 degrees, since 273 is the constant for kelvin.

We do not know our second volume, however we know out T2. It is -8 degrees, and don't forget to convert it to Kelvin. 

So, when we plug all of these numbers into the equation, we are left with V2 to find. To do this we cross multiply (V1 x T2) and then divide by T1. That leaves us with the number for V2. Don't forget to round to the least # of sig figs! And you can divide V1 by T1, and then divide V2 by T2, to ensure your answers are the same, since they are directly porportional and need to be equal to each other. 

Hope I could help!
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Please help me! 45 points each!!
Sveta_85 [38]

Answer:

CH4+ 2Cl2→CH2Cl2+ 2HCl

Explanation:

mass is 132.8`125

as methane and dichlormethane has the same molar ration 1 to 1

mol of ch4 is 1.5625

so 1.5625 times mr of ch2cl2 is 132.8125

8 0
3 years ago
Convert 5.8 km to the unit mm. <br> 0.0000058 mm <br> 0.0058 mm <br> 5,800 mm <br> 5,800,000 mm
spin [16.1K]
Its going to b 5,800,000 due to each Km being 1,000,000 Mm
Hope this helps :D
5 0
3 years ago
The chemical equation for a reaction between K2Cr2O7 and HCl is shown.
deff fn [24]

K2Cr2O7 + 14HCl → 2CrCl3 + 2KCl + 3Cl2 + 7H2O

the correct option is :

K2Cr2O7, because the oxidation number of Cr changes from +6 to +3.


<u>Oxidation number of Cr in K2Cr2O7 is:</u>

K2Cr2O7 = 2K + 2 Cr + 7 O

= 2(+1) + 2Cr + 7(-2)

= 2 + 2Cr -14

[total charge on K2Cr2O7 = 0], Hence;

2 + 2Cr -14 = 0

2Cr -12 = 0

2Cr = 12

Cr = 12/2

<u>Cr = +6</u>


<u>Oxidation number of Cr in CrCl3 is:</u>

CrCl3 = Cr + 3Cl = 0

Cr + 3(-1) = 0

Cr -3 = 0

<u>Cr = +3</u>

Hence Cr is changing its oxidation number from

+6 in K2Cr2O7 to +3 in CrCl3.

Since the oxidation number of Cr [ +6 → +3] is decreasing here,

Cr is getting reduced.

 

so K2Cr2O7 is an oxidizing agent,as it is getting itself reduced and oxidizes others.

3 0
3 years ago
Read 2 more answers
What would happen if there no decomposers
AleksAgata [21]
Producers would not have enough nutrients .
3 0
3 years ago
Read 2 more answers
Use the standard enthalpies of formation for the reactants and products to solve for the ΔHrxn for the following reaction. (The
Rudiy27
<span>
1. Remember (sum of products) - (sum of reactants)
So ΔHrxn = 2 ΔHf [H2(g)] + ΔHf [Ca(OH)2(s)] - 2 ΔHf [H2O(l)] - ΔHf [Ca(s)]
= 2*0 + -986.09 kJ/mol - 2*(-285.8 kJ/mol) - 0

Do the math and you'll have the answer. BTW the ΔHf [H2(g)] and ΔHf [Ca(s)] were 0 because these are elements in their standard states.

</span>HOPE THIS HELPS ;)
8 0
3 years ago
Read 2 more answers
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