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2 years ago

Consider the reaction:

Chemistry

1 answer:

murzikaleks [220]2 years ago

7 0

<u>Answer:</u> The formation of given amount of oxygen gas results in the absorption of 713 kJ of heat.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of oxygen gas = 83 g

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

$\text{Moles of oxygen gas}=\frac{83g}{32g/mol}=2.594mol$

For the given chemical equation:

$2Fe_2O_3\rightarrow 4Fe+3O_2;\Delta H^o_{rxn}=+824.2kJ$

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

By Stoichiometry of the reaction:

When 3 moles of oxygen gas is formed, the amount of heat absorbed is 824.2 kJ

So, when 2.594 moles of oxygen gas is formed, the amount of heat absorbed will be = $\frac{824.2kJ}{3mol}\times 2.59mol=713kJ$

Hence, the formation of given amount of oxygen gas results in the absorption of 713 kJ of heat.

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Plant cells and animal cells have many of the same organelles, but only plant cells have

Ierofanga [76]

Answer: chlorophyll

Explanation:

7 0

2 years ago

A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?

Tanzania [10]

<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume = 0.50 cm × 10.0 cm × 2.0 cm

= 10 cm³

Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

= 19.3 g/cm³

Thus, the density of the gold bar is 19.3 g/cm³

3 0

2 years ago

State the five the five basic assumptions of the kinetic-molecular theory.

Ivan

Answer:

The primary assumptions are as follows:

Any gas is a collection of innumerable number of minuscule particles which are known as molecules according to Avogadro’s law.

There are no forces of attraction or repulsion among the particles or between the molecules and the surroundings.

The gas particles are always at straight, rapid, fast & random motion resulting in inevitable collisions with other particles and the surroundings that changes direction of motion.

Since the particle are spherical, solid and elastic the collisions involving them are elastic in nature as well i.e their kinetic energy is conserved even after collisions.

The total kinetic energy of the particles is proportional to the absolute temperature.

In some books two other assumptions are given as well:

1. The size or area of each particle is negligible compared to that of the container.

2. Pressure of gas is result of the continuous clash of the particles with the wall of the container.

The simplest kinetic model is based on the assumptions that: (1) the gas is composed of a large number of identical molecules moving in random directions, separated by distances that are large compared with their size; (2) the molecules undergo perfectly elastic collisions (no energy loss) with each other and with the walls of the container, but otherwise do not interact; and (3) the transfer of kinetic energy between molecules is heat. These simplifying assumptions bring the characteristics of gases within the range of mathematical treatment.

Such a model describes a perfect gas and is a reasonable approximation to a real gas, particularly in the limit of extreme dilution and high temperature. Such a simplified description, however, is not sufficiently precise to account for the behaviour of gases at high densities.

Based on the kinetic theory, pressure on the container walls can be quantitatively attributed to random collisions of molecules the average energy of which depends upon the gas temperature. The gas pressure can therefore be related directly to temperature and density. Many other gross properties of the gas can be derived, such as viscosity, thermal and electrical conductivity, diffusion, heat capacity, and mobility. In order to explain observed deviations from perfect gas behaviour, such as condensation, the assumptions must be appropriately modified. In doing so, considerable insight has been gained as to the nature of molecular dynamics and interactions.

7 0

1 year ago

How many alcohols and ethers share the molecular formula c2h6o?

Nadya [2.5K]

Answer is: there are two compounds with molecular formula C₂H₆O, one alcohol and one ethar.

Alcohol with molecular formula C₂H₆O is ethanol or ethyl alcohol (C₂H₅OH).

Ethanol is a volatile, flammable, colorless liquid with characteristic odor.

Ether with molecular formula C₂H₆O is dimethyl ether (CH₃OCH₃).

Dimethyl ether is a colorless gas.

4 0

2 years ago

Read 2 more answers

During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

8 0

2 years ago