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stira [4]
3 years ago
15

Plz help me guys A B C D

Chemistry
1 answer:
Taya2010 [7]3 years ago
3 0

Answer:

C

Explanation:

I think

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For the gaseous reaction of carbon monoxide and chlorine to form phosgene (COCl₂):(b) Assuming that ΔS° and ΔH° change little wi
maksim [4K]

ΔG° at 450. K is -198.86kJ/mol

The following is the relationship between  ΔG°,  ΔH, and  ΔS°:

ΔH-T ΔS = ΔG

where  ΔG represents the common Gibbs free energy.

the enthalpy change,  ΔH

The temperature in kelvin is T.

Entropy change is  ΔS.

ΔG° = -206 kJ/mol

ΔH° equals -220 kJ/mol

T = 298 K

Using the formula, we obtain:

-220kJ/mol -T ΔS° = -206kJ/mol

220 kJ/mol +206 kJ/mol =T ΔS°.

-T ΔS = 14 kJ/mol

for ΔS-14/298

ΔS=0.047 kJ/mol.K

450K for the temperature Completing a formula with values

ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol

ΔG° = -220 kJ/mol + 21.14 kJ/mol.

ΔG°=198.86 kJ/mol

Learn more about ΔG° here:

brainly.com/question/17214066

#SPJ4

7 0
2 years ago
How many grams of NaCl are needed to prepare 50.0 grams of 35.0% of salt solution?
nevsk [136]

Answer:

17.5 g

Explanation:

Given data

  • Mass of solution to be prepared: 50.0 grams
  • Concentration of the salt solution: 35.0%

The concentration by mass of NaCl in the solution is 35.0%, that is, there are 35.0 grams of sodium chloride per 100 grams of solution. We will use this ratio to find the mass of sodium chloride required to prepare 50.0 grams of a 35.0% salt solution.

50.0gSolution \times \frac{35.0gNaCl}{100gSolution} = 17.5gNaCl

4 0
3 years ago
Calculate the change in entropy when 1.00 kg of water at 100 ∘C is vaporized and converted to steam at 100 ∘C. Assume that the h
andrew11 [14]

Answer : The change in entropy is 6.05\times 10^3J/K

Explanation :

Formula used :

\Delta S=\frac{m\times L_v}{T}

where,

\Delta S = change in entropy = ?

m = mass of water = 1.00 kg

L_v = heat of vaporization of water = 2256\times 10^3J/kg

T = temperature = 100^oC=273+100=373K

Now put all the given values in the above formula, we get:

\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}

\Delta S=6048.25J/K=6.05\times 10^3J/K

Therefore, the change in entropy is 6.05\times 10^3J/K

5 0
3 years ago
2Al + 6HCl → 2AlCl3 + 3H2<br> If 85.0 grams of HCl react, how many moles of H2 are produced?
Murrr4er [49]

Answer:

1.17 mol

Explanation:

Step 1: Write the balanced equation

2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

Step 2: Calculate the moles corresponding to 85.0 g of HCl

The molar mass of HCl is 36.46 g/mol.

85.0 g × 1 mol/36.46 g = 2.33 mol

Step 3: Calculate the number of moles of H₂ produced from 2.33 moles of HCl

The molar ratio of HCl to H₂ is 6:3.

2.33 mol HCl × 3 mol H₂/6 mol H₂ = 1.17 mol H₂

8 0
2 years ago
Will medal!!!
hjlf
Δ H reaction = q / n where q: amount of heat released and n is number of moles of substance.
q = m . C . ΔT where:
m = mass of substance (g)
C = Specific heat capacity (4.18)
ΔT = change in temperature = 24.25 - 23.16 = 1.09
q = 1000 x 4.18 x 1.09 = 4556 J = 4.556 kJ
number of moles (n) = Molarity (M) x Volume (L)
                                 = 0.185 M x 0.07 L = 0.01295 mole
Δ H = q / n = - (4.556 kJ / 0.01295 mole) = -351.8 kJ / mol
Note: it is exothermic reaction (-ve sign)  i.e. temperature is raised

5 0
3 years ago
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