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Neporo4naja [7]
3 years ago
7

Conditions for an experimental chemistry reaction require a temperature of 300 K. The temperature in the lab is 55°F. Which of t

he following must you do to meet the requirements? (°F= 9/5 (K - 273) + 32
A) Decrease the room temp by 33 F
B) increase the room temp by 26F
C) Increase the room temp by 18 K
D) decrease the room temp by 12 C
Chemistry
2 answers:
slava [35]3 years ago
3 0

300 degrees Kelvin = 300 - 273 = 27 degrees Celsius.

So what you need to do is find out how many degrees Fahrenheit that is.

F = 9/5(27) + 32

F = 48.6 + 32

F = 80.6

You are at 55oF. You need to go from 55 to 80.6 or 25.6 degrees F.

Your closest answer is B. <<<<<< Answer

Maurinko [17]3 years ago
3 0

Answer:

B) increase the room temp by 26°F

Explanation:

From the given information:

Room temperature = 55 °F

Required temperature = 300 K

<u>Conversion from K to °F:</u>

F = 9/5(K-273) + 32\\\\For T = 300 K\\\\F = 9/5(300-273)+32 = 80.6F

The required temperature in °F = 80.6°F

Therefore, the lab temperature must be increased by:

(80.6-55)F = 25.6 F

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At 40°C, how much KNO3 is<br> soluble?
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You have 500.0 ml of a buffer solution containing 0.30 m acetic acid (ch3cooh) and 0.20 m sodium acetate (ch3coona). what will t
Nataly [62]
First, we should get moles acetic acid = molarity * volume

                                                                =0.3 M * 0.5 L

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then, we should get moles acetate = molarity * volume

                                                           = 0.2 M * 0.5L

                                                           = 0.1 mol

then, we have to get moles of OH- which added:

moles OH- = molarity  * volume

                   = 1 M    * 0.02L

                  = 0.02 mol

when the reaction equation is:


                 CH3COOH  +  OH-  → CH3COO-   +  H2O


moles acetic acid after adding OH- = (0.15-0.02) 
                                             
                                                            =  0.13M                                       

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Total volume = 0.5 L + 0.02 L= 0.52 L

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                       = 0.25 M

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by using H-H equation we can get PH:

PH = Pka + ㏒[salt/acid]

when we have Ka = 1.8 x 10^-5

∴Pka = -㏒Ka 

        = -㏒ 1.8 x 10^-5

       = 4.7

So by substitution:

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6 0
3 years ago
NiS2(s) + O2(g) --&gt; NiO(s) + SO2(g) When 11.2 g of NiS2 react with 5.43 g of O2, 4.86 g of NiO are obtained. The theoretical
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Answer:

1. The theoretical yield of NiO is 5.09g.

2. O2 is the limiting reactant.

3. The percentage yield of NiO is 95.5%

Explanation:

Step 1:

The balanced equation for the reaction is given below:

2NiS2(s) + 5O2(g) —> 2NiO(s) + 4SO2(g)

Step 2:

Determination of the masses of NiS2 and O2 that reacted and the mass of NiO produced from the balanced equation. This is illustrated below below:

Molar mass of NiS2 = 59 + (32x2) = 123g/mol

Mass of NiS2 from the balanced equation = 2 x 123 = 246g

Molar mass of o3= 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g

Molar mass of NiO = 59 + 16 = 75g/mol

Mass of NiO from the balanced equation = 2 x 75 = 150g

Summary:

From the balanced equation above, 246g of NiS2 reacted with 160g of O2 to produce 150g of NiO

Step 3:

Determination of the limiting reactant. This can be obtain as follow:

From the balanced equation above, 246g of NiS2 reacted with 160g of O2.

Therefore, 11.2g of NiS2 will react with = (11.2 x 160)/246 = 7.28g of O2.

From the above calculation, we can see that it will take a higher mass of O2 i.e 7.28g than what was given i.e 5.43g to react completely with 11.2g of NiS2.

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1. Determination of the theoretical yield of NiO.

In this case, the limiting reactant will be used as all of it is consumed in the reaction. The limiting reactant is O2.

From the balanced equation above, 160g of O2 reacted to produce 150g of NiO.

Therefore, 5.43g of O2 will react to produce = (5.43 x 150)/160 = 5.09g of NiO.

Therefore, the theoretical yield of NiO is 5.09g.

2. The limiting reactant is O2. Please review step 3 above for explanation.

3. Determination of the percentage yield of NiO. This is illustrated below:

Actual yield of NiO = 4.86g

Theoretical yield of NiO = 5.09g

Percentage yield =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 4.86/5.09 x 100

Percentage yield of NiO = 95.5%

3 0
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vesna_86 [32]

Answer:

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Hence ,

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