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never [62]
2 years ago
8

A metal M forms the oxide M2O. When 0.890 grams of M reacts with pure oxygen, 0.956 grams of M2O form. Write the balanced equati

on for the reaction between M and O2, find the molar mass of M and identify M.
Chemistry
2 answers:
Svet_ta [14]2 years ago
8 0

Answer:

Good Luck!

Explanation:

Alla [95]2 years ago
5 0

The molar mass of M is  0.225g/mol and the element M is Hydrogen

If a metal M combines with an oxygen element to form the oxide, M_2O then the chemical reaction will be expressed as:

4M + O_2 -> 2M_2O\\

This shows that 4 moles of an unknown element M react with the oxygen element to produce the oxide M_2O

Given the following parameters

Mass of M = 0.890 grams

Mass of M_2O = 0.956 grams

Get the molar mass of M:

Molar mass = Mass/number of moles

Molar mass = 0.890/4

Molar mass = 0.225g/mol

Hence the molar mass of M is  0.225g/mol and the element M is Hydrogen

Learn more here: brainly.com/question/6996520

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Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
2 years ago
2. The pressure of the oxygen gas inside a
Flauer [41]

Answer: 4.41 atm

Explanation:

Given that,

Original pressure of oxygen gas (P1) = 5.00 atm

Original temperature of oxygen gas (T1) = 25°C

[Convert 25°C to Kelvin by adding 273

25°C + 273 = 298K

New pressure of oxygen gas (P2) = ?

New temperature of oxygen gas (T2) = -10°C

[Convert -10°C to Kelvin by adding 273

-10°C + 273 = 263K

Since pressure and temperature are given while volume is held constant, apply the formula for Charle's law

P1/T1 = P2/T2

5.00 atm /298K = P2/263K

To get the value of P2, cross multiply

5.00 atm x 263K = 298K x V2

1315 atm•K = 298K•V2

V2 = 1315 atm•K / 298K

V2 = 4.41 atm

Thus, the new pressure inside the canister is 4.41 atmosphere

4 0
3 years ago
What is the molarity of a solution that contain 1.1 miles of lithium in 0.5 liters of solution
Solnce55 [7]
1.1 Moles / 0.5 Liters = 0.22 Molarity
4 0
3 years ago
Why are nutrient molecules important?
ycow [4]

Answer : Option A) The body breaks down the molecules into molecules they need.

Explanation : Nutrient molecules are important because the body breaks the nutrient molecules and convert it into the absorb-able molecule that can generate energy for doing the activities of the body.

Nutrients are taken in the body in the form of carbohydrates which later on gets broken into glucose molecules and generates energy, proteins which is decomposed into amino acids, fats breaks down into fatty acids, minerals and salts are also nutrients which is essential nutrient for body.

4 0
3 years ago
Read 2 more answers
What types of substances would definitely be good conductors?
serious [3.7K]

Answer: Ionic substances would be good conductors since they dissolve in water and form ions.

Explanation:

Conductors can be defined as those materials which can allow the electricity to flow through them. The conductors also allow the transmission of heat and light from one to another source. They conduct electricity by allowing the electrons to flow through them easily.

The ionic substances can be good conductors because these can be dissolved in water and they will produce ions. The positively-charged ions will be attracted towards the negative electrode whereas the negatively charged ions to the positive electrode.

7 0
3 years ago
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