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1 year ago

Which is responsible for the formation of rust on metals?(1 point)

Chemistry

1 answer:

Rainbow [258]1 year ago

5 0

The correct answers about rusting, air pollution and products of water with rock are:

Option A. oxidation

Option D. Pollutants mix with air and water to make acid rain.

Option B. clay minerals and calcium carbonate

<h3>What is rusting?</h3>

Rusting is the process by which a metal especially reacts with oxygen in the atmosphere and water vapor to form a hydrated oxide of the metal known as hydrated iron (iii) oxide. This is known as rust.

The process is an oxidation process; option A.

<h3>How does air pollution impact chemical weathering?</h3>

Air pollution is the presence of substances in air known as pollutants which makes the air impure.

Air pollution impact chemical weathering as the pollutants mix with air and water to make acid rain which weathers rocks; option D.

<h3>What products are produced when water reacts with sodium in rocks?</h3>

The reaction of water with sodium in rocks result in the formation of clay minerals and calcium carbonate also as limestone, marble or chalk; option B.

In conclusion, the presence of pollutants in air results in acid rain and hence rock weathering.

Learn more rock weathering at: brainly.com/question/2341950

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Why is a mass movement of mud called a flow?

wariber [46]

In terms of a deeper scientific reason, I am not sure, but the basic reason is quite simple. "Mud" tends to look like a mix between a solid, dirt, and a liquid, water or some other liquid. Since it is, in fact, a cross between a solid and a liquid, it has properties of both. It has certain physical and visual properties that only a solid would have, such as texture and opaqueness, but it also has physical properties of a liquid. Since it leans more towards the liquid side than the solid side, we say mud "flows" rather than saying that it "rolls" or "bounces".

5 0

3 years ago

How many liters of CO2 gas can be produced at 30.0 °C and 1.50 atm from the reaction of 5.00 mol of C3H8 and an excess of O2 acc

lbvjy [14]

Answer:

249 L

Explanation:

Step 1: Write the balanced equation

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

Step 2: Calculate the moles of CO₂ produced from 5.00 moles of C₃H₈

The molar ratio of C₃H₈ to CO₂ is 1:3. The moles of CO₂ produced are 3/1 × 5.00 mol = 15.0 mol

Step 3: Convert "30.0°C" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 30.0°C + 273.15 = 303.2 K

Step 4: Calculate the volume of carbon dioxide

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T/P

V = 15.0 mol × 0.0821 atm.L/mol.K × 303.2 K/1.50 atm

V = 249 L

5 0

3 years ago

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 $P$ :

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