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3 years ago

Write a balanced chemical equation for the neutralization reaction between phosphoric acid and rubidium hydroxide.

1 answer:

5 0

Phosphoric acid is a weak acid, while rubidium hydroxide is a strong acid.

H₃PO₄ + RbOH --> Rb₃PO₄ + H₂O

We get Rb₃PO₄ because PO₄ has a charge of 3-, that is PO₄³⁻. Rb has a charge of 1+. You give the subscript of one the charge of the other as this is an ionic compound. So you end up with Rb₃PO₄, a neutral compound.

Now let's balance the equation:
H₃PO₄ + 3RbOH --> Rb₃PO₄ + 3H₂O

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You are using a ruler to measure the width of a test tube. The edge of the test tube is between 3 cm and 4 cm marks. if centimet

Arada [10]

Answer:

A) 3.6 cm

Explanation:

Accuracy comes down to how precisely you can read the length on a given scale. Here since the smallest increment is centimeter, we can go only one decimal beyond to estimate. This is because you can usually estimate to only one decimal place beyond the closest marks on any measuring.

So, the answer should be 3.6 cm.

Here's a document that explains it well: https://www.auburn.wednet.edu/cms/lib03/WA01001938/Centricity/Domain/1360/1_Uncertainty.pdf

Hope that's right!

5 0

1 year ago

An enzyme is discovered that catalyzes the chemical reaction SAD ↔ HAPPY A team of motivated researchers sets out to study the e

Pepsi [2]

Answer:

The $K_{m}$ of a substrate will be "10 μM".

Explanation:

The given values are:

$E_{t} = 20 \ nM$

$[Substract] = 40 \ \mu M$

$K_{cat}=600 \ s^{-1}$

Reaction velocity, $Vo=9.6 \ \mu M s^{-1}$

As we know,

⇒ $Vo=\frac{K_{cat}[E_{t}][S]}{K_{m}+[S]}$

On putting the estimated values, we get

⇒ $9.6=\frac{600\times 20\times 10^{-3}\times 40}{K_{m}+40}$

⇒ $K_{m}+40=\frac{600\times 20\times 10^{-3}\times 40}{9.6}$

⇒ $K_{m}+40=50$

On subtracting "40" from both sides, we get

⇒ $K_{m}+40-40=50-40$

⇒ $K_{m}=10 \ \mu M$

6 0

3 years ago

as the elements period 3 are considered in order of increasing atomic number, the number of principal energy levels in each succ

Diano4ka-milaya [45]

Answer:

stay the same.

Explanation: Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):

Na: 1s2 2s2 2p6 3s1

Ar: s2 2s2 2p6 3s2 3p6

As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:

Na: 3s1

Ar: 3s2, 3p6

All the new electrons electrons filled the third energy level (3s and 3p). So the energy level does not change, just the orbitals.

5 0

3 years ago

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting e

sp2606 [1]

The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

Answer: The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

Explanation:

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

$CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)$