How many moles are there in 1.20 x 1025 particles of helium?

How many grams are in 3.8 moles of Li?

Sunny_sXe [5.5K]

In 3.8 moles of li, there are about 0.144071459 grams. I assume that you are talking about lithium.

7 0

3 years ago

Al calentar la mantequilla, esta se transforma en liquido, este cambio se denomina?

BlackZzzverrR [31]

Answer:

cambio fisico

Explanation:

4 0

3 years ago

A sample of pure NH4HS is placed in a sealed 2.0-L container and heated to 550 K at which the equilibrium constant is 3.5 x 10-3

Zolol [24]

Answer:

The mass of $NH_{3}$ in the container is 2.074 gram

Explanation:

Given:

Volume of $NH_{4} HS$ $V = 2$ lit

Equilibrium constant $k _{eq} = 3.5 \times 10^{-3}$

The reaction in which $NH_{3}$ is produced

$NH_{4} HS$ ⇄ $NH_{3} + H_{2}S$

Here equal moles of $NH_{3}$ and $H_{2}S$ is formed.

From the formula of equilibrium constant,

$k_{eq} = (NH_{3})(H_{2}S )$

$x^{2} = 3.5 \times 10^{-3}$

$x = 0.061$ M

Above value shows,

$NH_{3} = 0.061$ $\frac{moles}{L}$

So in 2 L no. moles of $NH_{3}$ = $0.061 \times 2 = 0.122$ moles.

So mass of 0.122 mole of $NH_{3}$ is = $0.122 \times 17 = 2.074$ g

Therefore, the mass of $NH_{3}$ in the container is 2.074 gram

8 0

3 years ago

A gas occupies a volume of 1.00 L at 25.0°C. What volume will the gas occupy at 1.00 x10^2 °C?

Leno4ka [110]

Answer : The volume of gas occupy at $1.00\times 10^2^oC$ is, 1.25 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

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:

$V_1=1.00L\\T_1=25.0^oC=(25.0+273)K=298K\\V_2=?\\T_2=1.00\times 10^2^oC=((1.00\times 10^2)+273)K=373K$

Putting values in above equation, we get:

$\frac{1.00L}{298K}=\frac{V_2}{373K}\\\\V_2=1.25L$

Therefore, the volume of gas occupy at $1.00\times 10^2^oC$ is, 1.25 L

3 0

3 years ago

An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr

Zina [86]

Answer: The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

Explanation:

We are given:

$C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa$

Rate of flow of ideal gas , n = 4 kmol/hr = $\frac{4\times 1000mol}{3600s}=1.11mol/s$ (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)

We know that:

$\Delta U=0$ (For isothermal process)

So, by applying first law of thermodynamics:

$\Delta U=\Delta q-\Delta W$

$\Delta q=\Delta W$ .......(1)

Now, calculating the work done for isothermal process, we use the equation:

$\Delta W=nRT\ln (\frac{P_1}{P_2})$

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = final pressure = 50 kPa

Putting values in above equation, we get:

$\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW$

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

$\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW$

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

4 0

3 years ago