¿Cuáles son las características de los materiales de laboratorio? Por ejemplo: exactitud, resistencia a la temperatura, etc.

A process used to measure heat change in a system is called ______________________.

Nutka1998 [239]

Measuring heat flow!

6 0

4 years ago

Enter your answer in the provided box. To draw a Lewis structure, first add the number of outer (valence) electrons contributed

DIA [1.3K]

Answer:

14 is this value for a molecule of $CH_3Cl$ .

Explanation:

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

Number of total outer valence electrons is carbon = 4

Number of total outer valence electrons is hydrogen = 1

Number of total outer valence electrons is chlorine = 7

For total number of outer electrons is $CH_3Cl$

$=1\times 4+3\times 1+1\times 7=14$

14 is this value for a molecule of $CH_3Cl$ .

4 0

4 years ago

What is the standard entropy change for the reaction below? 2 CO(g) + 2 NO(g) → N2(g) + 2 CO2(g) S o (CO(g)) = 197.7 J/(mol·K) S

photoshop1234 [79]

Answer:

$\Delta _RS^o=-230J/(mol*K)$

Explanation:

Hello,

In this case, for the given chemical reaction:

$2 CO(g) + 2 NO(g) \rightarrow N_2(g) + 2 CO_2(g)$

The standard entropy change is computed by subtracting the products standard enthalpy and the reactants standard enthalpy considering each species stoichiometric coefficients:

$\Delta _RS^o=S ^o_{N2(g)}+2S ^o_{CO2(g)}-2S ^o_{CO(g)}-2S ^o_{NO(g)}\\\\\Delta _RS^o=191.6J/(mol*K)+2*213.8J/(mol*K) -2*213.8J/(mol*K) -2*210.8J/(mol*K)\\\\\Delta _RS^o=-230J/(mol*K)$

Best regards.

3 0

3 years ago

Read 2 more answers

A certain flexible weather balloon contains 3.5 L of helium gas. Initially, the balloon is in WP at 8500ft, where the temperatur

Advocard [28]

Answer:

$\large \boxed{\text{3.9 L}}$

Explanation:

We can use the Combined Gas Laws to solve this problem

$\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2}}{T_{2}}$

Data

p₁ = 571.2 Torr; p₂ = 400 Torr

V₁ = 3.5 L; V₂ = ?

T₁ = 21.5 °C; T₂ = 6.8 °C

Calculations

(a) Convert the temperatures to kelvins

T₁ = (21.55 + 273.15) K = 294.70 K

T₂ = (6.8 + 273.15) K = 279.95 K

(b) Calculate the new volume

$\begin{array}{rcl}\dfrac{p_{1}V_{1} }{T_{1}} & = & \dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{\text{571.2 Torr $\times$ 3.5 L}}{\text{294.65 K}} & = & \dfrac{\text{400 Torr} \times V_{2}}{\text{279.95 K}}\\\\\text{6.78 L} & = & \text{1.429V}_{2}\\\\V_{2} & = & \textbf{4.7 L}\\\end{array}\\\text{The new volume of the balloon is $\large \boxed{\textbf{4.7 L}}$}$

3 0

3 years ago

Read 2 more answers

How many kilojoules of heat are required to melt a 10-g popsicle is 0°C ? Assume the particle has the same molar mass and heat o

Fantom [35]

<h3><u>Answer;</u></h3>

3.4 kJ

<h3><u>Explanation</u>;</h3>

Taking the mass of water as 10.0 g

Moles of water = Mass /molar mass

Moles of water in 10.0 g;

= 10g H2O x 1mol/18g

=0.56mol

But, 1 mole of water = 6.01 kJ

Hence;

6.01kJ/1mol H2O= x kJ/.56mol,

x = <u>3.4kJ</u>

4 0

4 years ago