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3 years ago

Which of these is an example of matter?

Chemistry

2 answers:

wolverine [178]3 years ago

7 0

Answer:

Air

Explanation:

It takes up space/ the rest do not

insens350 [35]3 years ago

3 0

air if you mean elements such as oxygen

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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

The chemical reaction for the formation of syngas is: CH4 + H2O -> CO + 3 H2 What is the rate for the formation of hydrogen,

grin007 [14]

Answer : The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$CH_4+H_2O\rightarrow CO+3H_2$

The expression for rate of reaction :

$\text{Rate of disappearance of }CH_4=-\frac{d[CH_4]}{dt}$

$\text{Rate of disappearance of }H_2O=-\frac{d[H_2O]}{dt}$

$\text{Rate of formation of }CO=+\frac{d[CO]}{dt}$

$\text{Rate of formation of }H_2=+\frac{1}{3}\frac{d[H_2]}{dt}$

The rate of reaction expression is:

$\text{Rate of reaction}=-\frac{d[CH_4]}{dt}=-\frac{d[H_2O]}{dt}=+\frac{d[CO]}{dt}=+\frac{1}{3}\frac{d[H_2]}{dt}$

As we are given that:

$+\frac{d[CO]}{dt}=0.35M/s$

Now we to determine the rate for the formation of hydrogen.

$+\frac{1}{3}\frac{d[H_2]}{dt}=+\frac{d[CO]}{dt}$

$+\frac{1}{3}\frac{d[H_2]}{dt}=0.35M/s$

$\frac{d[H_2]}{dt}=3\times 0.35M/s$

$\frac{d[H_2]}{dt}=1.05M/s$

Thus, the rate for the formation of hydrogen is, 1.05 M/s

6 0

4 years ago

Give me 10 examples of entropy.

Mrac [35]

Answer:

A campfire is an example of entropy. The solid wood burns and becomes ash, smoke and gases, all of which spread energy outwards more easily than the solid fuel. Ice melting, salt or sugar dissolving, making popcorn and boiling water for tea are processes with increasing entropy in your kitchen.

Explanation:

If this helped you, the please mark me the brainliest.

6 0

3 years ago

The rusting of iron is due to its reaction with oxygen. An iron nail weighing 8.531 g was left outdoors for a period of one year

Annette [7]

Answer:

m(O₂) = 3.666 g

Explanation:

The rusting of iron is due to its reaction with oxygen. The balanced equation is:

4 Fe + 3 O₂ → 2 Fe₂O₃

According to Lavoisier's law of conservation of mass, the sum of the masses of the reactants is equal to the sum of the masses of the products.

m(reactants) = m(products)

m(Fe) + m(O₂) = m(Fe₂O₃)

8.531 g + m(O₂) = 12.197 g

m(O₂) = 3.666 g

4 0

3 years ago

Look at the images of Mars below. choose three features from the images and explain why they provide evidence thatt there is or

asambeis [7]

Answer:

Spot (4) Has evidence that there were waves here at some point. It looks as if the sand/surface debris was moved back and forth, creating this pattern in the sand. Spots (5) and (7) show evidence of water erosion. The smooth edges of both hills mean that water once flowed through that area, carrying little pebbles and rocks that gradually shaped and smoothed those mounds. An example for this would be how there are smooth areas in the Grand Canyon where scientists have concluded that some, if not all, of the Grand Canyon was formed by flowing water. "Rapid uplift, cracking, and surface drainage of receding floodwaters provide both the path and the necessary volume of water to quickly carve out Grand Canyon," says scholars on icr.org.

Explanation:

Water Erosion exists.

3 0

3 years ago