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3 years ago

A hydrogen atom contains 1 proton, 1 electron, and 0 neutrons. the atomic number of hydrogen is _____. (enter a numeral.

1 answer:

4 0

The atomic number represents the number of protons present in the nucleus of a certain atom. For a hydrogen atom, which has 1 proton, 1 electron, and 0 neutrons, the atomic number of hydrogen would be 1. The atomic number is the same as the charge number of a nucleus. This number is unique for every element in the periodic table and it distinguishes an element. In a neutral atom, this number is equal to the number of electrons in the atom. Also, the atomic number of an element determines the position of the element in the periodic table which for hydrogen, it is the very first element in the periodic table.

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Which of the following solutions is a good buffer system? Which of the following solutions is a good buffer system? a solution t

Murljashka [212]

Answer:

A solution that is 0.10 M HC2H3O2 and 0.10 M LiC2H3O2 is a good buffer system.

Explanation:

A buffer is defined as the mixture between a weak acid and its conjugate base or vice versa.

For the solutions:

0.10 M HF and 0.10 M LiC2H3O2. HF is a weak acid but LiC2H3O2 is the conjugate base of the weak acid (HC2H3O2, acetic acid).

0.10 M HC2H3O2 and 0.10 M LiC2H3O2. Here, you have a mixture of HC2H3O2, acetic acid, weak acid, and LiC2H3O2 is its conjugate base. Thus, this is a good buffer system

0.10 M LiOH and 0.10 M KOH. LiOH ans KOH are both strong bases.

0.10 M HF and 0.10 M NH4+. Again, HF is a weak acid but NH4+ is the conjugate acid of a weak base (NH3).

4 0

3 years ago

Which change to an equilibrium mixture of this reaction results in the formation of more H2S? Which change to an equilibrium mix

3241004551 [841]

Given question is incomplete. The complete question is as follows.

The decomposition of $NH_{4}HS$ is endothermic:

$NH_{4}HS(s) \rightarrow NH_{3}(g) + H_{2}S(g)$

Which change to an equilibrium mixture of this reaction results in the formation of more $H_{2}S$ ?

a decrease in the volume of the reaction vessel (at constant temperature), an increase in temperature, an increase in the amount of $NH_{4}HS$ in the reaction vessel, all of the above.

Explanation:

As per Le Chatelier's principle, any disturbance caused in an equilibrium reaction will tend to shift the equilibrium in a direction away from the disturbance.

When volume is decreased then there will occur increase in pressure because according to Boyle's law pressure is inversely proportional to volume.

Hence, equilibrium will shift in the direction where there is less pressure. Therefore, in the given reaction a decrease in the volume of the reaction vessel (at constant temperature) will shift the equilibrium in the backward direction.

When we increase the temperature then more number of collisions will take place and product formation will be more. Hence, reaction will shift in the forward direction.

When amount of $NH_{4}HS$ increases then equilibrium this disturbance will shift the equilibrium in forward direction.

Thus, we can conclude that following change to an equilibrium mixture of this reaction results in the formation of more $H_{2}S$ :

an increase in temperature.

an increase in the amount of $NH_{4}HS$ in the reaction vessel.

8 0

3 years ago

Which molecule is butyne?

Ilya [14]

Answer:

H,C CH,

Explanation:

5 0

2 years ago

In Haber’s process, 30 moles of hydrogen and 30 moles of nitrogen react to make ammonia. If the yield of the product is 50%, wha

devlian [24]

Answer:

Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams

Explanation:

Step 1: Data given

Number of moles hydrogen = 30 moles

Number of moles nitrogen = 30 moles

Yield = 50 %

Molar mass of N2 = 28 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate limiting reactant

For 1 mol of N2, we need 3 moles of H2 to produce 2 moles of NH3

Hydrogen is the limiting reactant.

The 30 moles will be completely be consumed.

N2 is in excess. There will react 30/3 =10 moles

There will remain 30 -10 = 20 moles (this in the case of a 100% yield)

In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.

Step 4: Calculate moles of NH3

There will be produced, 30/ (3/2) = 20 moles of NH3 (In case of 100% yield)

For a 50% yield there will be produced, 10 moles of NH3

Step 5: Calculate the mass of NH3

Mass of NH3 = mol NH3 * Molar mass NH3

Mass of NH3 = 20 moles * 17.03

Mass of NH3 = 340.6 grams = theoretical yield ( 100% yield)

Step 6: Calculate actual mass

50% yield = actual mass / theoretical mass

actual mass = 0.5 * 340.6

actual mass = 170.3 grams

Step 7: The mass of nitrogen remaining

There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain

Mass of nitrogen = 25 moles * 28 g/mol

Mass of nitrogen = 700 grams

6 0

3 years ago

Read 2 more answers

What is the definition of relief in science terms

SOVA2 [1]

The apparent topography exhibited by minerals in thin section as a consequence of refractive index.

6 0

3 years ago