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1 year ago

Write an equation that represents the action in water of rubidium hydroxide as an Arrhenius base.

Chemistry

1 answer:

Anika [276]1 year ago

3 0

Answer:

RbOH → Rb⁺ + OH⁻

As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base

Explanation:

Arrhenius theory states that a compound is considered a base, if the compound can generate OH⁻ ions in aqueous solution.

Our compound is the RbOH.

When it is put in water, i can dissociate like this:

RbOH → Rb⁺ + OH⁻

As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base

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In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be n

bulgar [2K]

Molar mass(C)= 12.0 g/mol
Molar mass (O2)=2*16.0=32.0 g/mol
Molar mass (CO2)=44.0 g/mol

18g C*1mol C/12 g C = 1.5 mol C

C + O2 → CO2

from reaction 1 mol 1 mol 1 mol
from problem 1.5 mol 1.5 mol 1.5 mol

1.5 mol O2*32 g O2/1 mol O2 = 48 g O2

In reality this reaction requires only 48 g O2 for 18 g carbon.
And from 18 g carbon you can get only
1.5 mol CO2*44 g CO2/1 mol CO2=66 g CO2
But these problem has 72g CO2. The best that we can think, it is a mix of CO2 and O2.
So to find all amount of O2 that was added for the reaction (probably people who wrote this problem wanted this)
we need (the mix of 72g - mass of carbon 18 g)= 54 g.
So the only answer that is possible is 2.) 54 g.

3 0

1 year ago

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by No

liraira [26]

Answer:

Molality = 0.066 m

Molar mass = 608.36 g/mol

Explanation:

It seems the question is incomplete. However a web search us shows this data:

" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "

The freezing-point depression is expressed by:

ΔT=Kf * m

We put the data given by the problem and solve for m:

2.63 °C = 40°C·kg/mol * m

m = 0.06575 m

For the calculation of the molar mass: Molality is defined as moles of solute per kilogram of solvent:

0.06575 m = Moles reserpine / kg camphor

25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor

We calculate moles of reserpine:

0.06575 m = Moles reserpine / 0.025 kg camphor

Moles reserpine = 1.64x10⁻³ mol

Finally we use the mass of reserpine and the moles to calculate the molar mass:

1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol

Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.

8 0

1 year ago

Hector claims that a single sodium ion and a single oxygen ion do not bond together to form a stable binary ionic compound. Is h

levacccp [35]

Answer:

Explanation:

Looking at the periodic table, we can see that sodium is in group 1, so a sodium ion would be Na⁺, with a charge of +1. Oxygen is in group 16, so an oxygen ion would be O²⁻, with a charge of -2.

A compound formed only by a single sodium ion and a single oxygen ion would thus have a charge of -1, and in order to have a stable ionic compound its charge must be zero.

8 0

1 year ago

Read 2 more answers

6. The half life for uranium-235 is 7.0x108 years.

kaheart [24]

Answer:

$\large \boxed{4.0}$

Explanation:

$\text{No. of half-lives} = 2.8 \times 10^{9 } \text{ yr} \times \dfrac{\text{1 half-life}}{7.0 \times 10^{8} \text{ yr}} = \textbf{4.0 half-lives}\\\\\text{The sample went through $\large \boxed{\textbf{4.0 half-lives}}$}$

4 0

1 year ago

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

1 year ago