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Law Incorporation [45]
3 years ago
6

What property of liquid oxygen makes it especially difficult and potentially harmful to work with at home?

Chemistry
1 answer:
irga5000 [103]3 years ago
3 0
The property of liquid oxygen that makes it especially difficult and potentially harmful to work with at home would be its cryogenic temperature. Liquid oxygen is being produced from the compression of oxygen gas to -196 degrees Celsius. As you can see, it has a very cold temperature that is why it used in cryogenics. Although liquid oxygen is non-toxic to humans, it would cause burns that are severe when being touched. Also, it would make certain materials brittle and unstable. Another property that makes it dangerous for use at home would be that it is very flammable. Proper handling is a must for this substance.
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What is wetlands made of
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Wet land isnt it obvious 
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What periodic trends exist for electronegativity?
Anna007 [38]
The highest electronegativity is in the elements in the top left corner of the periodic table, and the lowest in the bottom right corner. Therefore, traveling up or to the left across the periodic table will increase the electronegativity 
5 0
2 years ago
2C4 H10 + 13O2 -→ 8CO2 + 10H2O If 356 moles of O2 are available to react, how many moles of CO2 will fom?
nlexa [21]

Answer:

Answer:

The mole ratio of C₄H₁₀ and CO₂ is 2 : 8, which  simplifies to 1 : 4.

Explanation:

The mole ratio is the relative proportion of the moles of products or reactants that participate in the reaction according to the chemical equation.

The chemical equation given is:

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Once you check that the equation is balanced, you can set the mole ratios for all the reactants and products. The coefficients used in front of each reactant and product, in the balanced chemical equation, tells the mole ratios.

In this case, they are: 2 mol C₄H₁₀ :  13 mol O₂ : 8 mol CO₂ : 10 mol H₂O

Since you are asked about the mole ratio of C₄H₁₀ and CO₂ it is:

2 mol C₄H₁₀ : 8 mol CO₂ , which dividing by 2, simplifies to

1 mol C₄H₁₀ : 4 mol CO₂, or

1 : 2.

Explanation:

5 0
3 years ago
: Citric acid, H3C6H5O7, is a triprotic acid. Consider a buffer system comprising H2C6H5O7 - and HC6H5O7 2- ions. What is the ne
olchik [2.2K]

Answer:

The Net reaction is    

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

Explanation:

From the Question we are told that the buffers are

                H_2C_6H_5O_7^ - and  HC_6H_5O_7 ^{ 2-}

When NaOH is added the Net ionic reaction would be

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

             

8 0
3 years ago
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If an ice cube weighing 25.0 g with an initial
riadik2000 [5.3K]

Answer:

11

∘

C

Explanation:

As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at

0

∘

C

to liquid at

0

∘

C

.

The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have

q

1

+

q

2

=

−

q

3

(

1

)

, where

q

1

- the heat absorbed by the solid at

0

∘

C

q

2

- the heat absorbed by the liquid at

0

∘

C

q

3

- the heat lost by the warmer water sample

The two equations that you will use are

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed/lost

m

- the mass of the sample

c

- the specific heat of water, equal to

4.18

J

g

∘

C

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

and

q

=

n

⋅

Δ

H

fus

, where

q

- heat absorbed

n

- the number of moles of water

Δ

H

fus

- the molar heat of fusion of water, equal to

6.01 kJ/mol

Use water's molar mass to find how many moles of water you have in the

100.0-g

sample

100.0

g

⋅

1 mole H

2

O

18.015

g

=

5.551 moles H

2

O

So, how much heat is needed to allow the sample to go from solid at

0

∘

C

to liquid at

0

∘

C

?

q

1

=

5.551

moles

⋅

6.01

kJ

mole

=

33.36 kJ

This means that equation

(

1

)

becomes

33.36 kJ

+

q

2

=

−

q

3

The minus sign for

q

3

is used because heat lost carries a negative sign.

So, if

T

f

is the final temperature of the water, you can say that

33.36 kJ

+

m

sample

⋅

c

⋅

Δ

T

sample

=

−

m

water

⋅

c

⋅

Δ

T

water

More specifically, you have

33.36 kJ

+

100.0

g

⋅

4.18

J

g

∘

C

⋅

(

T

f

−

0

)

∘

C

=

−

650

g

⋅

4.18

J

g

∘

C

⋅

(

T

f

−

25

)

∘

C

33.36 kJ

+

418 J

⋅

(

T

f

−

0

)

=

−

2717 J

⋅

(

T

f

−

25

)

Convert the joules to kilojoules to get

33.36

kJ

+

0.418

kJ

⋅

T

f

=

−

2.717

kJ

⋅

(

T

f

−

25

)

This is equivalent to

0.418

⋅

T

f

+

2.717

⋅

T

f

=

67.925

−

33.36

T

f

=

34.565

0.418

+

2.717

=

11.026

∘

C

Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be

T

f

=

11

∘

C

Explanation:

3 0
2 years ago
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