Which list contains all pure substance

A gas has a volume of 560 mL at a temperature of –55°C. What volume will the gas occupy at

kap26 [50]

Answer: 778 ml

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=560ml\\T_1=-55^oC=(273-55)K=218K\\V_2=?\\T_2=30.0^oC=(273+30.0)K=303K$

Putting values in above equation, we get:

$\frac{560ml}{218K}=\frac{V_2}{303K}\\\\V_2=778ml$

Thus volume the gas occupy at 30.0°C is 778 ml

7 0

3 years ago

What are the missing coefficients for H3PO4 + HCl = PCl5 + H2o

grigory [225]

Answer:

H3PO4 + 5 HCl → PCl5 + 4 H2O

Explanation:

The given equation is

H3PO4 + HCl = PCl5 + H2O

The above chemical equation has one P atom on both the sides, hence phosphorus is balanced

There are 5 Cl on the RHS but only one Cl on the LHS. On balancing the chlorine, we get -

H3PO4 + 5HCl = PCl5 + H2O

Now, there are 8 hydrogen atom on the LHS but only two on the RHS. On balancing the hydrogen on both the sides, the new equation become

H3PO4 + 5HCl = PCl5 + 4H2O

Let us check for oxygen

Oxygen on LHS = 4 and oxygen on RHS = 4

Thus, the balanced equation is H3PO4 + 5HCl = PCl5 + 4H2O

7 0

3 years ago

Which type of chemical reaction occurs in Ca+CI2 CaCI2

dimaraw [331]

Ca + Cl2 = CaCl₂

A synthesis <span>reaction.</span>

3 0

4 years ago

The compound trimethylamine, (CH3)3N, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibr

GREYUIT [131]

Answer:

The $K_b$ expression for the weak base equilibrium is:

$K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

Explanation:

$(CH_3)_3N(aq)+H_2O(l)\rightlefharpoons (CH_3)_3NH^++OH^-(aq)$

The expression of the equilibrium constant of base $K_c$ can be given as:

$K_c=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N][H_2O]}$

] $K_b=K_c\times [H_2O]=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

As we know, water is pure solvent, we can put $[H_2O]=1$

$K_b=K_c\times 1=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

So, the the $K_b$ expression for the weak base equilibrium is:

$K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

6 0

3 years ago

When considering the relationship among standard free energy change, equilibrium constants, and standard cell potential, the equ

DedPeter [7]

Answer:

ΔG° = - RTLnK is used to find the standard cell potential given the equilibrium constant

Explanation:

for an ideal disolution:

⇒ ΔG = RT∑ni LnXi

∴ ΔG = ( μi - μi*)ni

∴ μ : chemical potential

∴ μ*: chem. potential of the pure component at T and P.

⇒ ΔG = μi - μi* = RT LnXi

for a equilibrium reaction:

⇒ ∑ νi*μi = 0

⇒ ΔGr = ΔG°+ RT Ln Kx = 0

⇒ ΔG° = - RT LnKx

4 0

4 years ago