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2 years ago

Hydrogen can be prepared on a small scale by the hydrolysis of metal hydrides:

Chemistry

1 answer:

lara [203]2 years ago

6 0

Answer:

Ca²⁺ + 2H⁻ + 2(2H⁺ + O²⁻) ----> Ca²⁺ + 2O²⁻ + 2H⁺ + 4H⁰

(2 electrons lost) ----> (2 electrons gained)

Explanation:

The equation of the reaction is as follows: CaH₂ + 2H₂0 —> Ca(OH)₂ + 2H₂

The oxidation numbers of each of the atoms in the molecules are as follows:

In CaH₂, calcium has an oxidation number of +2, while hydrogen has an oxidation number of -1 ; Ca²⁺, 2H⁻

In 2H₂0, hydrogen has an oxidation number of +1 while oxygen has an oxidation number of -2 ; H⁺, O²⁻

In Ca(OH)₂, calcium has an oxidation number of +2 while oxygen has an oxidation number of -2 and hydrogen has an oxidation number of +1 ; Ca²⁺, O²⁻, 2H⁺

In 2H₂, each hydrogen atom has an oxidation number of 0 ; 4 H⁰

Ca²⁺ + 2H⁻ + 2(2H⁺ + O²⁻) ----> Ca²⁺ + 2O²⁻ + 2H⁺ + 4H⁰

On the left hand side, the hydride ion gives up two electrons to become neutral hydrogen atoms.

On the right hand, two hydrogen atoms accept two electrons to form neutral atoms.

These neutral hydrogen atoms pair up to form two hydrogen gas molecules.

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What is the number of grams of an element that is numerically equal to the atomic mass in amu?

bagirrra123 [75]

The wording of your question doesn't quite make sense, but a mole of an element has the same mass in grams as a single atom of that element has in amu. The mole is defined as 6.02 x10^22 things, whether they be atoms or molecules or even moles! 6.02x10^22 atoms of carbon has a mass of 12.01 g, and a single atom of carbon has a mass of 12.01 amu. Hope this helps!

3 0

3 years ago

It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta

Zigmanuir [339]

Answer:

$\lambda=241.9\ nm$

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,

1 mole = $6.023\times 10^{23}\ electrons$

So,

$6.023\times 10^{23}$ electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of $\frac {495.0}{6.023\times 10^{23}}\ kJ$ of energy.

Energy required = $82.18\times 10^{-23}\ kJ$

Also,

1 kJ = 1000 J

So,

Energy required = $82.18\times 10^{-20}\ J$

Also, $E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}$

$\lambda=\frac{19.878}{10^6\times \:82.18}$

$\lambda=2.4188\times 10^{-7}\ m$

Also,

1 m = 10⁻⁹ nm

So,

$\lambda=241.9\ nm$

6 0

3 years ago

6 MARKER FOR CHEMISTRY I NEED THIS!!!

natita [175]

Answer:

when you put the marker in the water the water gets into the marker and it will change the solid color to a plain color witch means that the marker will make the water in the cup the color from the marker.

Hope I can help

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2 years ago

What is the sum of the coefficients of the balanced equation for:

Marianna [84]

Answer:

3 Pb(NO3)2 + Al2(SO4)3 = 2 Al(NO3)3 + 3 PbSO4

Explanation:

I think your equation is incorrect? This is balanced and the sum of coefficients is 9.

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2 years ago

UDP-glucose pyrophosphorylase works by this mechanism: Group of answer choices It adds a phosphate group to glucose from UTP, le

klemol [59]

Answer:

The correct answer is it adds a UMP molecule to glucose-1-phosphate by splitting out pyrophosphate.

Explanation:

UDP glucose pyrophosphorylase enzyme also known as UTP-glucose-1-phsphate uridylyltransferase.

This enzyme plays an important role during synthesis of glycogen also known as glycogenesis.

During the reaction the enzyme UDP glucose pyrophosphorylase transfer uridine monophosphate to glucose-1-phosphate resulting in the formation of UDP glucose along with the release of pyrophosphate moiety.

Glucose-1-phosphate+UTP⇆UDP-glucose+ppi

5 0

2 years ago