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Zolol [24]
3 years ago
15

At standard pressure, the difference between the freezing point and the boiling point of water, in Kelvin degrees, is D. 373 A.

100 B. 180 C. 273
Chemistry
1 answer:
zepelin [54]3 years ago
6 0
Your answer would be A.100 
 
                                       

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Please please please help me. question is in picture
Vedmedyk [2.9K]

There’s nothing here

7 0
3 years ago
6.02 x 109 = hellllppppppppppppppppppppppppppppppppppppppppppppppppppppp
mojhsa [17]
The answer is 656.18
6.02 x 109 = 656.18
5 0
3 years ago
Read 2 more answers
Complete and balance the following equation: S(s)+HNO3(aq)→H2SO3(aq)+N2O(g)(acidic solution)
meriva

Answer:- The balanced equation is, 2HNO_3(aq)+2S(s)+H_2O(l)\rightarrow N_2O(g)+2H_2SO_3(aq) .

Solution:- Oxidation number of S is increasing from 0 to 4 and so it is oxidation. Oxidation number of N is decreasing from 5 to 1 and so it is reduction.

We write the oxidation and reduction half equations and balance them. The given reaction is taking place in an acidic medium.

First of all we balance all the atoms other than H and O. Then oxygen is balanced by adding H_2O and hydrogen is balanced by adding H^+ . Charge is balanced by adding electrons.

To makes the electrons equal for both the half equations we multiply the equation/equations by appropriate numbers.

Oxidation half equation:

S(s)\rightarrow H_2SO_3(aq)

S is already balanced. To balance O, we need to add three water molecules to the left side:

S(s)+3H_2O(l)\rightarrow H_2SO_3(aq)

For balancing hydrogen, we need to add 4 hydrogen ions to the right side:

S(s)+3H_2O(l)\rightarrow H_2SO_3(aq)+4H^+(aq)

Now to balance the charge we need to add 4 electrons to the right side:

S(s)+3H_2O(l)\rightarrow H_2SO_3(aq)+4H^+(aq)+4e^-

Reduction half equation:

HNO_3(aq)\rightarrow N_2O(g)

To balance N, we need to multiply left side by 2:

2HNO_3(aq)\rightarrow N_2O(g)

For balancing oxygen, we need to add 5 water molecules to the right side:

2HNO_3(aq)\rightarrow N_2O(g)+5H_2O(l)

To balance hydrogen, we need to add 8 hydrogen ions to the left side:

2HNO_3(aq)+8H^+(aq)\rightarrow N_2O(g)+5H_2O(l)

Now, for charge balance, we need to add 8 electrons to the left side:

2HNO_3(aq)+8H^+(aq)+8e^-\rightarrow N_2O(g)+5H_2O(l)

First half equation has 4 electrons and second half equation has 8 electrons.

To make the electrons equal, we need to multiply oxidation half equation by 2:

2S(s)+6H_2O(l)\rightarrow 2H_2SO_3(aq)+8H^+(aq)+8e^-

Now we add both of these two half equations and cancel out common species. What we get on doing this is:

2HNO_3(aq)+2S(s)+H_2O(l)\rightarrow N_2O(g)+2H_2SO_3(aq)



6 0
3 years ago
Read 2 more answers
A (an) _____ is an easier path for electron flow that can cause shocks or an electric fire.
Serga [27]
A short circuit is an easier path for electron flow
6 0
4 years ago
5. A 5.00 g sample of an unknown substance was heated from 25.2 C to 55.1 degrees * C , and it required 133 to do so. Identify t
Leviafan [203]

Answer:

Aluminum

Explanation:

Given

T_1 = 25.2^oC

T_2 = 55.1^oC

m = 5.00g

\triangle Q= 133J

<em>See attachment for chart</em>

Required

Identify the unknown substance

To do this, we simply calculate the specific heat capacity from the given parameters using:

c = \frac{\triangle Q}{m\triangle T}

This gives:

c = \frac{\triangle Q}{m(T_2 - T_1)}

So, we have:

c = \frac{133J}{5.00g * (55.1C - 25.2C)}

c = \frac{133J}{5.00g * 29.9C}

c = \frac{133J}{149.5gC}

c = 0.89\ J/gC

From the attached chart, we have:

Al(s) = 0.89\ J/gC --- The specific heat capacity of Aluminum

<em>Hence, the unknown substance is Aluminum</em>

7 0
3 years ago
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