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Sergio039 [100]
3 years ago
15

Water boils at what degrees Celsius

Chemistry
2 answers:
Archy [21]3 years ago
4 0
100 degrees Celsius. And apparently this answer needs at least 20 characters to explain it well. But yes, the answer is 100 degrees Celsius
AysviL [449]3 years ago
3 0
100 degrees Celsius :-)
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How many significant figures are in 0.05?
drek231 [11]
There is one significant figure 
3 0
3 years ago
Construct a three-step synthesis of 1,2-epoxycyclopentane from cyclopentanol by dragging the appropriate formulas into the bins.
zubka84 [21]

Answer:

(1) Bromination, (2) E2 elimination and (3) epoxidation

Explanation:

  • In the first step, -OH group in cyclopentanol is replaced by more facile leaving group Br by treating cyclopentanol with PBr_{3}
  • In the second step, E2 elimination in presence of strong base e.g. NaOEt/EtOH produce cyclopentene
  • In the third step, treatment of cyclopentene with mCPBA produces 1,2-epoxycyclopentane
  • Full reaction scheme has been shown below

3 0
3 years ago
NH4Cl + AgNO3 → AgCl + NH4NO3
iVinArrow [24]
<h3><u>A</u><u> </u><u>N</u><u> </u><u>S</u><u> </u><u>W</u><u> </u><u>E</u><u> </u><u>R</u><u> </u><u>:</u><u> </u><u>–</u><u> </u></h3>

  • It's a balanced equation!
  • This is a precipitation reaction.
  • AgCl is the formed precipitate.
7 0
2 years ago
Be sure to answer all parts. The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g) ⇌ H2O(g) +
nikklg [1K]

<u>Answer:</u>

<u>For 1:</u> The value of \Delta G for the chemical equation is -24.636 kJ/mol

<u>For 2:</u> The value of \Delta G for the chemical equation is -20.925 kJ/mol

<u>Explanation:</u>

For the given chemical equation:

H_2(g)+CO_2(g)\rightleftharpoons H_2O(g)+CO(g)

  • <u>For 1:</u>

To calculate the \Delta G for given value of equilibrium constant, we use the relation:

\Delta G=-RT\ln K_p      .....(1)

where,

\Delta G = ? kJ/mol

R = Gas constant = 8.314J/K mol

T = temperature = 2000 K

K_p = equilibrium constant in terms of partial pressure = 4.40

Putting values in above equation, we get:

\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (4.40)\\\\\Delta G=-24636.12J/mol

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -24636.12 J/mol = -24.636 kJ/mol

Hence, the value of \Delta G for the chemical equation is -24.636 kJ/mol

  • <u>For 2:</u>

The expression of K_p for the given chemical equation is:

K_p=\frac{p_{CO}p_{H_2O}}{p_{H_2}p_{CO_2}}

We are given:

p_{CO}=1.18atm\\p_{H_2O}=0.66atm\\p_{CO_2}=0.82atm\\p_{H_2}=0.27atm

Putting values in above equation, we get:

K_p=\frac{1.18\times 0.66}{0.27\times 0.82}\\\\K_p=3.52

Now, calculating the value of \Delta G by using equation 1:

R = Gas constant = 8.314J/K mol

T = temperature = 2000 K

K_p = equilibrium constant in terms of partial pressure = 3.52

Putting values in equation 1, we get:

\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (3.52)\\\\\Delta G=-20925.68J/mol

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -20925.68 J/mol = -20.925 kJ/mol

Hence, the value of \Delta G for the chemical equation is -20.925 kJ/mol

3 0
3 years ago
Read 2 more answers
A certain weak acid, ha, has a ka value of 6.0×10−7. calculate the percent ionization of ha in a 0.10 m solution.
joja [24]
For the purpose, we will use the equation for determining the dissociation constant from concentration and <span>percent of ionization:

Kd = c </span>× α²

α = √(Kd/c) × 100%

Kd = 6.0×10⁻⁷

c(HA) = 0.1M

α = √(6.0×10⁻⁷/0.1)  × 100% =  0.23%

So, in the solution, the acid <span>percent of ionization will be just 0.23%.</span>

5 0
3 years ago
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