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3 years ago

Which correctly describes the reaction between potassium and excess water?

1 answer:

7 0

Balanced chemical reaction: 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g).

KOH is inorganic compound p<span>otassium hydroxide, a strong base.
H</span>₂ is hydrogen gas.
In balanced chemical reaction number of atoms on both side of chemical reaction must be same. There are two potassium atoms, four hydrogen atoms and two oxygen atoms on both side of reaction.

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If the titrant has a molarity of 0.1000 m and there are 45.00 ml of analyte present, what is the molarity of the analyte?

marysya [2.9K]

Given:

Concentration of titrant = 0.1000 M
Volume of titrant = 45 mL

The molarity of analyte depends on the amount of the analyte present in the titrated solution. If the amount of analyte is 20 mL, then its concentration is:

45ml * 0.10 M = C analyte * 20 ml

C analyte = 0.225 M

4 0

2 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-

grin007 [14]

Answer:

Total pressure 5.875 atm

Explanation:

The equation for above decomposition is

$2N_2O \rightarrow 2N_2 + O_2$

rate constant $k = 1.94\times 10^{-4} min^{-1}$

Half life $t_{1/2} = \frac{0.693}{k} = 3572 min$

Initial pressure $N_2 O = 4.70 atm$

Pressure after 3572 min = P

According to first order kinematics

$k = \frac{1}{t} ln\frac{4.70}{P}$

$1.94\times 10^{-4} = \frac{1}{3572} \frac{4.70}{P}$

solving for P we get

P = 2.35 atm

$2N_2O \rightarrow 2N_2 + O_2$

initial 4.70 0 0

change -2x +2x +x

final 4.70 -2x 2x x

pressure of $O_2$ after first half life = 2.35 = 4.70 - 2x

x = 1.175

pressure of $N_2$ after first half life = 2x = 2(1.175) = 2.35 ATM

Total pressure = 2.35 + 2.35 + 1.175

= 5.875 atm

5 0

3 years ago

Read 2 more answers

After a polypeptide chain has been synthesized, certain amino acids in the peptide may become modified. For each modified amino

AnnZ [28]

Answer:

The modification that yields the first amino acid depicted is lysine

Explanation:

Lysine being an amino acid used in preparation of various medicine. It is used to prevent cold as well as sore throat, which is found to be caused by the herpes simplex labialis. It can be used orally or can be directly applied over the skin. various foods have lysine in it and they are meat, cheese, certain fishes, eggs, soybeans, spirulina and also fenugreek seed. They are the building block for the protein particles.

5 0

3 years ago

Read 2 more answers

What is Y? 222/86 Rn A/z + 4/2He

kondaur [170]

$^{222}_{\phantom{2}86}\text{Rn} \to ^{218}_{\phantom{2}84}\text{Po}+ ^{4}_{2}\text{He}$

Y is Po-218.

A = 218
Z = 84.

<h3>Explanation </h3>

$^{222}_{\phantom{2}86}\text{Rn} \to ^{A}_{Z}\text{Y}+ ^{4}_{2}\text{He}$

Here's the symbol of a particle in a nuclear reaction. $^{A}_{Z}\text{Y}$ .

A stands for mass number. Z stands for atomic number. Both numbers shall conserve in a nuclear reaction.

The mass number on the left hand side is 222.
The two mass numbers on the right hand side add up to A + 4.
222 = A + 4.
A = 218

So is the case for the atomic number. Try figure out the atomic number of Y using the same approach.

The atomic number on the left hand side is 86.
The two atomic number on the right hand side add up to __ + __.
86 = __ + __.
Z = 84.

What element is Y? The atomic number of Y is 84. Refer to a periodic table. Element 84 corresponds to Po (polonium). Y is Polonium-218. The symbol of Y should be written as $^{218}_{\phantom{2}84}\text{Po}$ . Hence the equation:

$^{222}_{\phantom{2}86}\text{Rn} \to ^{218}_{\phantom{2}84}\text{Po}+ ^{4}_{2}\text{He}$ .

3 0

2 years ago