How do you balance this chemical equation?

What is the definition of quanititative observation

sukhopar [10]

Answer:

Quantitative observation is an objective collection of data which is primarily focused on numbers and values.

7 0

3 years ago

For many purposes we can treat ammonia NH3 as an ideal gas at temperatures above its boiling point of −33.°C. Suppose the temper

Keith_Richards [23]

Answer:

The new pressure will be 0.225 kPa.

Explanation:

Applying combined gas law:

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

where,

$P_1\text{ and }V_1$ are initial pressure and volume at initial temperature $T_1$ .

$P_2\text{ and }V_2$ are final pressure and volume at initial temperature $T_2$ .

We are given:

$P_1=0.29 kPa\\V_1=V\\P_2=?\\V_2=V+50\% of V=1.5 V$

$T_1=-25^oC=248.15 K$

$T _2 = 16^oC=289.15 K$

Putting values in above equation, we get:

$\frac{0.29 kPa\times V}{248.15 K}=\frac{P_2\times 1.5V}{289.15 K}$

$P_1=0.225 kPa$

Hence, the new pressure will be 0.225 kPa.

4 0

3 years ago

Be sure to answer all parts. For the complete redox reactions given here, write the half-reactions and identify the oxidizing an

OlgaM077 [116]

Answer : $O_2$ is the oxidizing agent and Fe is the reducing agent.

Explanation :

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The balanced redox reaction is :

$4Fe+3O_2\rightarrow 2Fe_2O_3$

The half oxidation-reduction reactions are:

Oxidation reaction : $Fe\rightarrow Fe^{3+}+3e^-$

Reduction reaction : $O_2+4e^-\rightarrow 2O^{2-}$

In order to balance the electrons, we multiply the oxidation reaction by 4 and reduction reaction by 3 then added both equation, we get the balanced redox reaction.

Oxidation reaction : $4Fe\rightarrow 4Fe^{3+}+12e^-$

Reduction reaction : $3O_2+12e^-\rightarrow 6O^{2-}$

$4Fe+3O_2\rightarrow 2Fe_2O_3$

In this reaction, $'Fe'$ is the reducing agent that loses an electron to another chemical species in a redox chemical reaction and itself gets oxidized and $'O_2'$ is the oxidizing agent that gain an electron to another chemical species in a redox chemical reaction and itself gets reduced.

Thus, $O_2$ is the oxidizing agent and Fe is the reducing agent.

7 0

3 years ago

The pungency of chillies is due to ____________.

Vera_Pavlovna [14]

Answer:

The hotness or pungency of chilli is due to presence of a group of compounds called capsaicinoids. Among capsaicinoids, capsaicin and dihydrocapsaicin are present in major quantities, while homocapsaicin, nordihydrocapsaicin and homo-dihydrocapsaicin are present in very small quantities.

3 0

2 years ago

Read 2 more answers

A 41.1 g sample of solid CO2 (dry ice) is added to a container at a temperature of 100 K with a volume of 3.4 L.A. If the contai

marta [7]

Answer:

Approximately 6.81 × 10⁵ Pa.

Assumption: carbon dioxide behaves like an ideal gas.

Explanation:

Look up the relative atomic mass of carbon and oxygen on a modern periodic table:

C: 12.011;
O: 15.999.

Calculate the molar mass of carbon dioxide $\rm CO_2$ :

$M\!\left(\mathrm{CO_2}\right) = 12.011 + 2\times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

Find the number of moles of molecules in that $41.1\;\rm g$ sample of $\rm CO_2$ :

$n = \dfrac{m}{M} = \dfrac{41.1}{44.009} \approx 0.933900\; \rm mol$ .

If carbon dioxide behaves like an ideal gas, it should satisfy the ideal gas equation when it is inside a container:

$P \cdot V = n \cdot R \cdot T$ ,

where

$P$ is the pressure inside the container.
$V$ is the volume of the container.
$n$ is the number of moles of particles (molecules, or atoms in case of noble gases) in the gas.
$R$ is the ideal gas constant.
$T$ is the absolute temperature of the gas.

Rearrange the equation to find an expression for $P$ , the pressure inside the container.

$\displaystyle P = \frac{n \cdot R \cdot T}{V}$ .

Look up the ideal gas constant in the appropriate units.

$R = 8.314 \times 10^3\; \rm L \cdot Pa \cdot K^{-1} \cdot mol^{-1}$ .

Evaluate the expression for $P$ :

$\begin{aligned} P &=\rm \frac{0.933900\; mol \times 8.314 \times 10^3 \; L \cdot Pa \cdot K^{-1} \cdot mol^{-1} \times 298\; K}{3.4\; L} \cr &\approx \rm 6.81\times 10^5\; Pa \end{aligned}$ .

Apply dimensional analysis to verify the unit of pressure.

4 0

3 years ago