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Alex787 [66]
3 years ago
12

QUICK I NEED HELP in a solution of a carbonated beverage, the dissolved carbon dioxide is the ____

Chemistry
2 answers:
SIZIF [17.4K]3 years ago
5 0
D. solvent because it is the solvent of the mixture

prisoha [69]3 years ago
5 0
I am pretty sure the answer is B for when something is dossolved it is mixed therefore it is a solute
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Describe all the changes a sample of solid water would undergo when heated from -10degrees c to its critical temperature at a pr
masha68 [24]

Answer:Just want to be carefree lately yeah

Explanation: Cradles Profanities

HUSH!

4 0
4 years ago
The enthalpy of vaporization of water at 373 K and 1 bar is 40.7 kJ/mol and the molar heat capacities are 75.3 J/(mol K) for liq
Soloha48 [4]

Answer:

The enthalpy of vaporization of water at 273 K and 1 bar = 44.9 KJ/mol

Explanation:

Enthalpy of vaporization of water at 273 K, ΔHvap(T₂) is given as;

ΔHvap(T₂) = ΔHvap(T₁) + ΔCp * (T₂ - T₁)

where ΔCp = molar heat capacity of gas - molar heat capacity of liquid

Therefore, ΔCp = (33.6 - 75.3) = -41.70 J/(mol K) = 0.0417 kJ/(molK)  

substituting  ΔCp = 0.0417 kJ/(mol K)  in the initial formula

;

ΔHvap(T) = ΔHvap(T1) + ΔCp * (T₂ - T₁)

ΔHvap(T₂)= 40.7 kJ/mol + {-0.0417 kJ/(mol K) * (272 - 373 K)}

ΔHvap(T₂) = 44.9 kJ/mol

Therefore,  enthalpy of vaporization of water at 273 K and 1 bar = 44.9kJ/mol

6 0
4 years ago
Need the following questions answered please and thank you
Ket [755]
Hope this might help u

6 0
3 years ago
Hydrogen iodide decomposes according to the equation: 2HI(g) H 2(g) + I 2(g), K c = 0.0156 at 400ºC A 0.660 mol sample of HI was
Ira Lisetskai [31]

Answer:

[HI] = 0.264M

Explanation:

Based on the equilibrium:

2HI(g) ⇄ H₂(g) + I₂(g)

It is possible to define Kc of the reaction as the ratio between concentration of products and reactants using coefficients of each compound, thus:

<em>Kc = 0.0156 = [H₂] [I₂] / [HI]²</em>

<em />

As initial concentration of HI is 0.660mol / 2.00L = <em>0.330M, </em>the equlibrium concentrations will be:

[HI] = 0.330M - 2X

[H₂] = X

[I₂] = X

<em>Where X is reaction coefficient.</em>

<em />

Replacing in Kc:

0.0156 = [X] [X] / [0.330M - 2X]²

0.0156 = X² / [0.1089 - 1.32X + 4X² ]

0.00169884 - 0.020592 X + 0.0624 X² = X²

0.00169884 - 0.020592 X - 0.9376 X² = 0

Solving for X:

X = - 0.055 → False solution, there is no negative concentrations

X = 0.0330 → Right solution.

Replacing in HI formula:

[HI] = 0.330M - 2×0.033M

<h3>[HI] = 0.264M</h3>

7 0
4 years ago
If 7.84 × 107 J of energy is released from a fusion reaction, what amount of mass in kilograms would be lost? Recall that c = 3
katrin [286]
  

m = 7.84x107/(3x108)2kg  = 7.84x107/9x1016kg = 0.871x10-9 kg = 8.71x10-10 kg





3 0
3 years ago
Read 2 more answers
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