What is the compound name for MgCO ?

What effect does salt water have on the strands of green onion?

Semenov [28]

When salt water is added to onion cells, then the cells will lose water due to osmosis, this can be observed.

3 0

3 years ago

copper hydroxide and potassium sulfate are produced when potassium hydroxide reacts with copper sulfate balanced equation

a_sh-v [17]

This problem is requiring the balanced chemical equation that takes place when copper hydroxide and potassium sulfate are produced when reacting potassium hydroxide with copper sulfate.

<h3>Balancing chemical equations:</h3>

In chemistry, balancing chemical equations is based on the law of conservation of mass, which demands us to have equal number of atoms on both sides of the chemical equation. This can be accomplished by inserting coefficients in front of the chemical species.

For this particular case, we have potassium hydroxide with copper sulfate on the reactants side, however, copper can be copper (I) or copper (II) as it has 1+ and 2+ as its possible oxidation numbers. In addition, copper hydroxide and potassium sulfate as the products. Hence, we can assume this is all about copper (II) so we can write:

$KOH+CuSO_4\rightarrow K_2SO_4+Cu(OH)_2$

As we can see, potassium, hydrogen and oxygen have two atoms each on the products side, but just one on the reactants side; drawback we can overcome by putting a 2 in front of KOH so as to balance it:

$2KOH+CuSO_4\rightarrow K_2SO_4+Cu(OH)_2$

Learn more about balancing chemical equations: brainly.com/question/8062886

8 0

2 years ago

A gas sample of 2.31 atm of oxygen gas and 3.75 atm of hydrogen gas that react to form water vapor. Assume the volume of the con

maxonik [38]

Mole fraction of Oxygen=0.381

Mole fraction of Oxygen= (range of moles of oxygen) ÷(general moles)

also, mole fraction of oxygen = (partial stress of oxygen) ÷ (total strain)

consequently , mole fraction of Oxygen= (2.31 atm)÷(2.31 atm + 3.75 atm)

= 0.381

The mole fraction may be calculated by means of dividing the variety of moles of 1 element of a solution by the entire quantity of moles of all the additives of a solution. It is cited that the sum of the mole fraction of all of the components inside the solution should be identical to 1.

Mole fraction is a unit of awareness. in the solution, the relative amount of solute and solvents are measured by way of the mole fraction and it's far represented through “X.” The mole fraction is the variety of moles of a selected aspect inside the answer divided by way of the entire range of moles in the given answer.

Mole fraction is the ratio between the moles of a constituent and the sum of moles of all ingredients in a mixture. Mass fraction is the ratio between the mass of a constituent and the full mass of a mixture.

The question is incomplete. Please read below to find the missing content.

Assuming that only the listed gases are present, what would the mole fraction of oxygen gas be for each of the following situations? A gas sample of 2.31 atm of oxygen gas and 3.75 atm of hydrogen gas react to form water vapor. Assume the volume of the container and the temperature inside the container does not change.

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4 0

2 years ago

(2 x 109) x (5 x 105).

Finger [1]

Answer:

114450

Explanation:

2*109 = 218

5*105 = 525

218*525 = 114450

8 0

2 years ago

Read 2 more answers

Consider the reaction: N2(g) + O2(g) ⇄ 2NO(g) Kc = 0.10 at 2000oC Starting with initial concentrations of 0.040 mol/L of N2 and

IrinaVladis [17]

Answer:

0.011 mol/L

Explanation:

This can be solved with something called an ICE table.

I = initial

C = change

E = equilibrium

Initially, there is 0.04 M of N₂, 0.04 M of O₂, and 0 M of NO.

x amount of N₂ reacts. Since the stoichiometry is 1:1, x amount of O₂ also reacts. This produces 2x of NO.

After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.

Here it is in table form:

$\left[\begin{array}{cccc}&N2&O2&NO\\I&0.04&0.04&0\\C&-x&-x&+2x\\E&0.04-x&0.04-x&2x\end{array}\right]$

Now we can use the equilibrium constant:

Kc = [NO]² / ( [N₂] [O₂] )

Substituting:

0.10 = (2x)² / ( (0.04 - x) (0.04 - x) )

Solving:

0.10 = (2x)² / (0.04 - x)²

√0.10 = 2x / (0.04 - x)

(√0.10) (0.04 - x) = 2x

(√0.10)(0.04) - (√0.10)x = 2x

(√0.10)(0.04) = 2x + (√0.10)x

(√0.10)(0.04) = (2 + √0.10)x

x = (√0.10)(0.04) / (2 + √0.10)

x = 0.0055

At equilibrium, the concentration of NO is 2x. So the answer is:

[NO] = 2x

[NO] = 0.011

The equilibrium concentration of NO is 0.011 mol/L.

3 0

3 years ago