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erastovalidia [21]
3 years ago
6

An iron fence is left unpainted, and it reacts with the oxygen in the air, forming rust. The formation of rust is an oxidation-r

eduction reaction, but it is also an example of a _________ reaction
Chemistry
2 answers:
trapecia [35]3 years ago
5 0
I'm pretty sure it would be a chemical reaction.
IrinaK [193]3 years ago
3 0

Answer: This reaction is a type of synthesis reaction.

Explanation: When iron reacts with oxygen which is available in air, it leads to the formation of a reddish-brown compound known as rust.

This reaction is considered as oxidation-reduction reaction because iron is getting oxidized to Fe^{2+} and oxygen is getting reduced to oxide ion.

Equation of the rust follows:

4Fe(s)+3O_2(g)+nH_2O(l)\rightarrow 2Fe_2O_3.nH_2O

Synthesis reactions are defined as the reactions in which two or more than two reactants combine together to form a single product which can be a molecule or a compound.

The rusting of iron is also considered as a synthesis reaction because iron and oxygen are combining together to form a single compound known as rust.

Both these reactions are a type of Chemical reactions.

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Determine the final temperature of sample with a specific heat of 1.1 J/g°C and a mass of 385 g if it starts out at a temperatur
Assoli18 [71]

Answer:

T2 =21.52°C

Explanation:

Given data:

Specific heat capacity of sample = 1.1 J/g.°C

Mass of sample = 385 g

Initial temperature = 19.5°C

Heat absorbed = 885 J

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )

885 J = 423.5 J/°C× (T2 - 19.5°C )

885 J / 423.5 J/°C = (T2 - 19.5°C )

2.02°C = (T2 - 19.5°C )

T2 = 2.02°C + 19.5°C

T2 =21.52°C

8 0
3 years ago
What coefficient would the O 2 have after balancing C 4 H 10 +O 2 CO 2 +H 2 O ?
nydimaria [60]

Answer: 6

Explanation:

6 0
2 years ago
Consider the reaction of NO and CO to form N2 and CO2, according to the balanced equation: 2 NO (g) + 2 CO (g) → N2 (g) + 2 CO2
Gekata [30.6K]

The image is not given in the question, it is attached below:

<u>Answer:</u> The excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

<u>Explanation:</u>

In the given image:

Red spheres represent oxygen atoms, blue spheres represent nitrogen atoms and black spheres represent carbon atoms

The combination of 1 black and 2 red spheres will represent carbon dioxide (CO_2) compound

The combination of 2 blue spheres will represent nitrogen molecule (N_2)

The combination of 1 blue and 1 red sphere will represent nitrogen monoxide (NO) compound

The combination of 1 black and 1 red sphere will represent nitrogen monoxide (NO) compound

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

We are given:

Given moles of NO = 6 moles

Given moles of CO = 4 moles

For the given chemical equation:

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

By stoichiometry of the reaction:

If 2 moles of CO reacts with 2 moles of NO

So, 4 moles of CO will react with = \frac{2}{2}\times 4=4mol of NO

As the given amount of NO is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, CO is considered a limiting reagent because it limits the formation of the product.

Hence, the excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

3 0
2 years ago
Argon has a pressure of 34.6 atm. It is transferred to a new tank with a volume of 456 L and pressure of 2.94 atm. What was the
NemiM [27]

Answer:

38.75 L

Explanation:

From the question,

Applying Boyles Law,

PV = P'V'....................... Equation 1

Where P = Original pressure of the Argon gas, V = Original Volume of Argon gas, P' = Final pressure of Argon gas, V' =  Final Volume of Argon gas.

make V the subject of the equation

V = P'V'/P.................... Equation 2

Given: P = 34.6 atm, V' = 456 L, P' = 2.94 atm.

Substitute these values into equation 2

V = (456×2.94)/34.6

V = 38.75 L

3 0
3 years ago
Find the mass of one atom of uranium-235. Recall that the mass in atomic mass units is equal to the mass in grams of one mole of
stiv31 [10]

Answer:

3.90*10^{-25}kg/atom

Explanation:

The molar mass of uranium-235 is 235 g/mol. So one mole of uranium-235 has a mass of 235 g. Put differently 6.022×10^23 atoms of uranium-235 have a mass of 235 g. Knowing that, how can we use that to find the mass of one atom?

mass of one atom = \frac{235 g}{1mol} *\frac{1 mol}{6.022*10^{23}atoms } *\frac{1kg}{1000g}= 3.90*10^{-25}kg/atom

7 0
2 years ago
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