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Salsk061 [2.6K]

2 years ago

6

A liquid's freezing point is -38°C and it's boiling point is 357°C. What is the number of Kelvin between the boiling point and t

he freezing point of the liquid?

1 answer:

laiz [17]2 years ago

7 0

The Kelvin temperature K =℃ + 273.15. ℃ means centigrade temperature. So the centigrade temperature number between freezing point and boiling point is equal to the number of Kelvin. So the answer is 357-(-38)=395.

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A car moves at a speed of 50 kilometers/hour. Its kinetic energy is 400 joules. If the same car moves at a speed of 100 kilomete

lys-0071 [83]

ke prop to v^2

ke1/v1^2=ke2/v2^2

400/50x50=joules/100x100

400x2x2

1600j

8 0

2 years ago

The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s

denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0

2 years ago

You have two 500.0 ml aqueous solutions. solution a is a solution of a metal nitrate that is 8.246% nitrogen by mass the ionic c

almond37 [142]

1) Answer is: the ionic compound in the solution b is K₂CrO₄ (potassium chromate).

Ionic compound in solution b has two potassiums (oxidation number +1), one chromium (oxidation number +6) and four oxygens. Oxidation number of oxygen is -2 and compound has neutral charge:

2 · (+1) + 6 + x · (-2) = 0.

x = 4; number of oxygen atoms.

2) Answer is: the ionic compound in solution a is AgNO₃ (silver nitrate).

ω(N) = 8.246% ÷ 100%.

ω(N) = 0.08246; mass percentage of nitrogen.

M(MNO₃) = M(N) ÷ ω(N).

M(MNO₃) = 14 g/mol ÷ 0.08246.

M(MNO₃) = 169.8 g/mol; molar mass of metal nitrate.

M(M) = M(MNO₃) - M(N) - 3 · M(O).

M(M) = 169.8 g/mol - 14 g/mol - 3 · 16 g/mol.

M(M) = 107.8 g/mol; atomic mass of metal, this metal is silver (Ag).

3) Balanced chemical reaction:

2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq).

Ionic reaction:

2Ag⁺(aq) + 2NO₃(aq) + 2K⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s) + 2K⁺(aq) + 2NO₃⁻(aq).

Net ionic reaction: 2Ag⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s).

Answer is: the blood-red precipitate is silver chromate (Ag₂CrO₄).

4) m(Ag₂CrO₄) = 331.8 g; mass of solid silver chromate.

n(Ag₂CrO₄) = m(Ag₂CrO₄) ÷ M(Ag₂CrO₄).

n(Ag₂CrO₄) = 331.8 g ÷ 331.8 g/mol.

n(Ag₂CrO₄) = 1 mol; amount of silver chromate.

From balanced chemical reaction: n(Ag₂CrO₄) : n(AgNO₃) = 1 : 2.

n(AgNO₃) = 2 · 1 mol.

n(AgNO₃) = 2 mol.

m(AgNO₃) = n(AgNO₃) · M(AgNO₃).

m(AgNO₃) = 2 mol · 169.8 g/mol.

m(AgNO₃) = 339.6 g; mass of silver nitrate.

m(AgNO₃) = m(K₂CrO₄).

m(K₂CrO₄) = 339.6 g; mass of potassium chromate.

n(K₂CrO₄) = m(K₂CrO₄) ÷ M(K₂CrO₄).

n(K₂CrO₄) = 339.6 g ÷ 194.2 g/mol.

n(K₂CrO₄) = 1.75 mol; amount of potassium chromate.

5) Chemical reaction of dissociation of silver nitrate in water:

AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq).

V(solution a) = 500 mL ÷ 1000 mL/L.

V(solution a) = 0.5 L; volume of solution a.

c(AgNO₃) = n(AgNO₃) ÷ V(solution a).

c(AgNO₃) = 2 mol ÷ 0.5 L.

c(AgNO₃) = 4 mol/L = 4 M.

From dissociation of silver nitrate: c(AgNO₃) = c(Ag⁺) = c(NO₃⁻).

c(Ag⁺) = 4 M; the concentration of silver ions in the original solution a.

c(NO₃⁻) = 4 M; the concentration of silver ions in the original solution a.

6) Chemical reaction of dissociation of potssium chromate in water:

K₂CrO₄(aq) → 2K⁺(aq) + CrO₄²⁻(aq).

V(solution b) = 500 mL ÷ 1000 mL/L.

V(solution b) = 0.5 L; volume of solution b.

c(K₂CrO₄) = n(K₂CrO₄) ÷ V(solution b).

c(AgNO₃) = 1.75 mol ÷ 0.5 L.

c(AgNO₃) = 3.5 mol/L = 3.5 M.

From dissociation of silver nitrate: c(K₂CrO₄) = c/2(K⁺) = c(CrO₄²⁻).

c(K⁺) = 7 M; the concentration of potassium ions in the original solution b.

c(CrO₄²⁻) = 3.5 M; the concentration of silver ions in the original solution b.

7) V(final solution) = V(solution a) + V(solution b).

V(final solution) = 500.0 mL + 500.0 mL.

V(final solution) = 1000 mL ÷ 1000 mL/L.

V(final solution) = 1 L.

n(NO₃⁻) = 2 mol.

c(NO₃⁻) = n(NO₃⁻) ÷ V(final solution)

c(NO₃⁻) = 2 mol ÷ 1 L.

c(NO₃⁻) = 2 M; the concentration of nitrate anions in final solution.

8) in the solution b there were 3.5 mol of potassium cations, but one part of them reacts with 2 moles of nitrate anions:

K⁺(aq) + NO₃⁻(aq) → KNO₃(aq).

From chemical reaction: n(K⁺) : n(NO₃⁻) = 1 : 1.

Δn(K⁺) = 3.5 mol - 2 mol.

Δn(K⁺) = 1.5 mol; amount of potassium anions left in final solution.

c(K⁺) = Δn(K⁺) ÷ V(final solution).

c(K⁺) = 1.5 mol ÷ 1 L.

c(K⁺) = 1.5 M; the concentration of potassium cations in final solution.

4 0

2 years ago

Based on the products obtained, rank the functional groups (acetamido, amino, and methoxy) in order of increasing ability to act

I am Lyosha [343]

Answer:

Amino >Methoxy > Acetamido

Explanation:

Bromination is of aromatic ring is an electrophilic substitution reaction. The attached functional group to the benzene ring activates or deactivate the aromatic ring towards electrophilic substitution reaction.

The functional group which donates electron to the benzene ring through inductive effect or resonance effect activates the ring towards electrophilic substitution reaction.

The functional group which withdraws electron to the benzene ring through inductive effect or resonance effect deactivates the ring towards electrophilic substitution reaction.

Among given, methoxy and amino are electron donating group. Amino group are stronger electron donating group than methoxy group. Acetamido group because of presence of carbonyl group becomes electron withdrawing group.

Therefore, decreasing order will be as follows:

Amino >Methoxy > Acetamido

7 0

2 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

2 years ago