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stealth61 [152]
4 years ago
15

If a freezer is not cold how to let it be cold againi mean out of work

Chemistry
1 answer:
Lady_Fox [76]4 years ago
6 0
If a freezer is not cold, maybe the condenser coils are dirty. You should clean these every 6-12 months. If not check the evaporator fan motor, it draws air over the evaporator coil and circulates it through the freezer. Your last option is to start relay, the start relay provides power to the compressor. Hope this helped!
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Natali5045456 [20]
I have no clue honestlly search it up: i found this answer:19<span>.</span>
7 0
3 years ago
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Consider a 0.10 M aqueous benzoic acid, CeHeCOOH. The K benzoic acid. 6.5 x 10 for A) Write a balanced equation that shows the r
7nadin3 [17]

Answer:

a) C6H5COOH + H2O ↔ H3O+  +  C6H5COO-

b) [ H3O+ ] = 2.517 E-3 M

c) pH = 2.599

Explanation:

a) balanced equation:

C6H5COOH + H2O ↔ H3O+  +  C6H5COO-

⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5

mass balance:

0.10 m = [ C6H5COO- ] + [ C6H5COOH ].....(1)

charge balance:

[ H3O+ ] = [ C6H5COO- ] + [ OH- ] .......[ OH- ] : comes from water, it's not significant

⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)

b) (2) in (1):

⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]

⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]

⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5

⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0

⇒ [ H3O+ ] = 2.517 E-3 M

c) pH = - log [ H3O+ ]

⇒ pH = - Log ( 2.517 E-3 )

⇒ pH = 2.599

7 0
3 years ago
Ethanol has a Kb of 1.22 Degrees C/m and usually boils at 78.4 Degrees Celcius. How many mol of an nonionizing solute would need
Gala2k [10]

Answer:

0.3097 moles of an nonionizing solute would need to be added.

Explanation:

Molal elevation constant = k_b=1.22^oC/m

Normal boiling point of ethanol = T_o=78.4^oC

Boiling of solution =T_b=86.30^oC

Moles of nonionizing solute = n

Mass of ethanol (solvent) = 47.84 g

Elevation boiling point:

\Delta T_b=T_b-T_o

\Delta T_b=86.30^oC-78.4^oC=7.9^oC

\Delta T_b=K_b\times  m

m=\frac{\text{Moloes of solute}}{\text{Mass of solvent(kg)}}

7.9^oC=1.22^oC/m\times \frac{n}{0.04784 kg}

n = 0.3097 mol

0.3097 moles of an nonionizing solute would need to be added.

4 0
4 years ago
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Answer:

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Explanation:

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Doss [256]

Answer:

filtration

Explanation:

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4 years ago
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