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julia-pushkina [17]
2 years ago
13

Write down the dissolution equation for rubidium chromate dissolving in water. (Chromate is a polyatomic ion with the formula Cr

O42-.) If two moles of the ionic compound are dissolved, then how many moles of the CATION are present in the solution?
Chemistry
1 answer:
ycow [4]2 years ago
5 0

Answer:

Four moles of the cation

Explanation:

2Rb2CrO4(s)<--------> 4Rb^+(aq) + 2CrO4^2-(aq)

Now looking at the reaction equation, it can be seen that one mole of rubidium chromate contains two moles of rubidium ions and one mole of chromate ions.

The dissolution of two moles of rubidium chromate should then yield four moles of rubidium ions and two moles of chromate ions since the ratio of ions present is 2:1.

This explains the reaction equation written above for the dissolution of two moles of rubidium chromate as shown.

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At a given set of conditions 241.8 kJ of heat is released when one mole of H2O forms from its elements. Under the same condition
yan [13]

Answer:

44Kj

Explanation:

These are the equations for the reaction described in the question,

Vaporization which can be defined as transition of substance from liquid phase to vapor

H2(g)+ 1/2 O2(g) ------>H2O(g). Δ H

-241.8kj -------eqn(1)

H2(g)+ 1/2 O2(g) ------>H2O(l).

Δ H =285.8kj ---------eqn(2)

But from the second equation we can see that it moves from gas to liquid, we we rewrite the equation for vaporization of water as

H2O(l) ------>>H2O(g)---------------eqn(3)

But the equation from eqn(2) the eqn does go with vaporization so we can re- write as

H2O ------> H2(g)+ 1/2 O2(g)

Δ H= 285.8kj ---------------eqn(4)

To find Delta h of the vaporization of water at these conditions, we sum up eqn(1) and eqn(4)

Δ H=285.8kj +(-241.8kj)= 44kj

5 0
2 years ago
Describe the many different forms of energy involved with stretching and releasing a rubber band. What other processes are simil
Nadya [2.5K]

Answer:

Conversion of kinetic energy to potential energy (chemo mechanical energy)

In the state of rest, the rubber is a tangled mass of long chained cross-linked polymer that due to their disorderliness are in a state of increased entropy. By pulling on the polymer, the applied kinetic energy stretches the polymer into straight chains, giving them order and reducing their entropy. The stretched rubber then has energy stored in the form of chemo mechanical energy which is a form of potential energy

Conversion of the stored potential energy in the stretched to kinetic energy

By remaining in a stretched condition, the rubber is in a state of high potential energy, when the force holding the rubber in place is removed, due to the laws of thermodynamics, the polymers in the rubber curls back to their state of "random" tangled mass releasing the stored potential energy in the process and doing work such as moving items placed in the rubber's path of motion such as an object that has weight, w then takes up the kinetic energy 1/2×m×v² which can can result in the flight of the object.

Explanation:

5 0
2 years ago
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Sixty-five percent of the mass of bone is a compound called hydroxyapatite. sixty-five percent of the mass of bone is a compound
mafiozo [28]
True. Its simple how i remeberd was the bone has 650,000 cells and the bone has 65% hydroxyapatite
5 0
2 years ago
There are many important laws and theories in science. Which of the following is a law of science?
Alenkasestr [34]

Answer: the answer is C

Explanation:

4 0
2 years ago
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a drop of water weighing 0.48 g condenses on the surface of a 55-g block of aluminum that is initially at 25C. if the heat durin
Gwar [14]

Mass of water vapor is 0.48 gms.

Weight of the aluminium block is 55 gms.

Heat of vaporization of water at 25 degree Celsius is 44.0 kJ/mol.

The amount of heat given by the condensation is:

q = Heat of vaporization × Mass of vapor / Molar mass of steam

= 44.0 kJ/mol ×0.48 g / 18 g/mol

= 1.173 kJ

Now the final temperature of the metal block is calculated by the formula:

q = m × c × ΔT

q = m × c × (T₂ - T₁)

Here, q is the amount of heat, m is the mass of the metal, c is the specif heat of the metal, T₁ is initial temperature, and T₂ is the final temperature.

Now, substituting the values we get,

1.173 kJ = 55 g × 0.903 J/g. ° C * (T₂ - 25°C)

1.173 × 10³ J = 55 g × 0.903 J/g. degree C × (T₂ - 25°C)

1.173 × 10³ = 49.665 ° C * (T₂ - 25° C)

T₂ = 49° C.

Thus, the final temperature of the metal block is 49° C.

8 0
2 years ago
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