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3 years ago

Water boils at what degrees Celsius

2 answers:

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 :-)

You might be interested in

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]

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]

Answer:

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

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

Explanation:

For the given chemical equation:

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

For 1:

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

For 2:

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 percent of ionization:

Kd = c × 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 percent of ionization will be just 0.23%.

5 0

3 years ago