Which group of the periodic table had the elements with the lowest first ionization energies

Given a fixed amount of gas held at constant pressure, calculate the volume it would occupy if a 2.00 L sample were cooled from

aliina [53]

Answer:

1.82 L

Explanation:

We are given the following information;

Initial volume as 2.0 L
Initial temperature as 60.0°C
New volume as 30.0 °C

We are required to determine the new volume;

From Charles's law;

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

Where, $V_1 and V_2$ are initial and new volume respectively, while $T_1 and T_2$ are initial and new temperatures respectively;

$T_1= 333 K$

$T_2=303K$

$V_1 =2.0L$

Rearranging the formula;

$V_2=\frac{V_1T_2}{T_1}$

$= \frac{(2.0L)(303K)}{333K} \\=1.820 L$

Therefore, the new volume that would be occupied by the gas is 1.82 L

7 0

3 years ago

What type of rock is this?

zloy xaker [14]

Answer:

I'm gonna say igneous

Explanation:

The texture looks rough and the rock looks like it's composed of different types of minerals making it an igneous rock

I hope this helps :)

6 0

3 years ago

Read 2 more answers

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

Pls help 20020200202

Ronch [10]

tell me what to do dudette

6 0

3 years ago

Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calcu

stealth61 [152]

Answer:

Explanation:

Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. slope=-12070, Ea=100kJ/mol, k= 0.000717(45C), 0.00284(55C), 0.00492(65C), 0.0165(75C), 0.0396(85C)

Explanation;

According to Arrhenius equation:

i.e. ln(k2/k1) = -Ea/R (1/T2 - 1/T1)

Where, k1 = 0.000717, T1 = 45 oC = (45+273) K = 318 K

T2 = 25 oC = (25 + 273) K = 298 K

i.e. ln(k2/0.000717) = -12070 (1/298 - 1/318)

i.e. ln(k2/0.000717) = -2.54738

i.e. k2/0.000717 = $e^{-2.54738}$

= 0.078286

Therefore, the required constant (k2) = 0.078286 * 0.000717 = $5.61*10^-^5$

6 0

3 years ago