Given the reaction zn(s) + pb(no3)2 (aq) = pb(s) +zn (no3)2 (aq) the oxidation number of zn metal is

When sodium is put in water, the metal floats on the surface and reacts to form sodium

mrs_skeptik [129]

Answer: Sodium also floats on the surface, but enough heat is given off to melt the sodium (sodium has a lower melting point than lithium and the reaction produces heat faster) and it melts almost at once to form a small silvery ball that dashes around the surface.

Explanation:

8 0

3 years ago

Read 2 more answers

A 700.0 mL gas sample at STP is compressed to a volume of 200.0 mL, and the temperature is increased to 30.0°C. What is the new

dolphi86 [110]

Answer: The new pressure of the gas in Pa is 388462

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas at STP = $10^5Pa$

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 700.0 ml

$V_2$ = final volume of gas = 200.0 ml

$T_1$ = initial temperature of gas = 273 K

$T_2$ = final temperature of gas = $30^oC=273+30=303K$

Now put all the given values in the above equation, we get:

$\frac{10^5\times 700.0ml}{273K}=\frac{P_2\times 200.0ml}{303K}$

$P_2=388462Pa$

The new pressure of the gas in Pa is 388462

6 0

3 years ago

A reaction has activation energy of 85kjper mol. What is the effect on the rate of raising the temperature from 20degree to 30 d

DENIUS [597]

Answer: The rate increases 3 times on raising the temperature from 20degree to 30 degree

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

$ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

where $k_2$ = rate constant at temp $T_2$

$k_1$ = rate constant at temp $T_1$

$E_a$ = activation energy

R= gas constant

$T_1$ = temperature = $20^0C=(20+273)K=293K$

$T_2$ = temperature = $30^0C=(30+273)K=303K$

$ln \frac{k_{2}}{k_{1}} = \frac{-85\times 1000J/mol}{8.314J/Kmol}[\frac{1}{303} - \frac{1}{293}]$

$ln \frac{k_{2}}{k_{1}}=1.15$

$\frac{k_{2}}{k_{1}}=3$

Thus rate increases 3 times on raising the temperature from 20degree to 30 degree

3 0

3 years ago

Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.

madreJ [45]

Answer:

$Molar \ solubility=3.12x10^{-5}M$

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

$CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-$

The equilibrium expression is:

$Ksp=[Ca^{2+}][F^-]^2$

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent $x$ is computed as follows:

$3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M$

Thus, the molar solubility equals the reaction extent $x$ , therefore:

$Molar \ solubility=3.12x10^{-5}M$

Regards.

4 0

3 years ago

If the initial temperature of a movable cylinder was 50 degrees Celsius

slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

5 0

3 years ago