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yuradex [85]
2 years ago
12

What can change depending on the number of protons in an atom?​

Chemistry
2 answers:
Rudiy272 years ago
7 0

Answer:

The element of the atom.

Explanation:

The number of protons confirms what element is the atom.

zvonat [6]2 years ago
6 0

Answer:

The type of atom

Explanation:

If you add a proton to a hydrogen atom, then it is no longer hydrogen, it is lithium

Hope this helps

You might be interested in
¿Cuál era su juego favorito cuando eran niños?​
blagie [28]

Answer:

mi juego favorito probablemente sería Mine craft cuando era más joven, aunque sigue siendo uno de mis juegos favoritos

Explanation:

7 0
3 years ago
Use this equation for the following problems: 2NaN3 --> 2Na+3N2
olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

<em>2NaN₃ → 2Na + 3N₂,</em>

  • It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

<em>Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?​ </em>

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

  • Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

  • We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

  • We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

<em><u>using cross multiplication:</u></em>

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

  • Finally, we can get the grams of NaN₃ needed:

<em>mass = no. of moles x molar mass</em> = (0.517 mol)(65.0 g/mol) =<em> 33.6 g.</em>

<em />

<em>Q2: How many mL of N₂ result if 8.3 g Na are also produced?</em>

  • We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

  • We can get the no. of moles of N₂ produced with 0.36 mol of Na:

<em><u>using cross multiplication:</u></em>

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

  • We can get the mass of 0.54 mol of N₂:

mass = no. of moles  x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

  • Now, we can get the mL of 15.12 g of N₂:

<em><u>using cross multiplication:</u></em>

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

<em>∴ The volume of N₂ result </em>= (1.0 L)(15.12 g)/(0.92 g) = <em>16.434 L = 16434 mL.</em>

4 0
3 years ago
A customer experiences worsening side effects in response to a prescription. What do you suggest to them?
o-na [289]

Answer:

Probably stop taking the prescribed durg and contact your pharmacist and your doctor that gave you your prescription asap.

Explanation:

Both of those health professionals will assist the patient in understanding how to go about the next steps for side effect relief.

8 0
2 years ago
_H3PO4 + _HCl ---&gt;_PCl5 + _H2O
Slav-nsk [51]

Dennis g studio and I don't have any money

5 0
2 years ago
The pKapKa of cyclopentane is &gt; 60, which is about what is expected for a hydrogen that is bonded to an sp3sp3 carbon. Explai
aivan3 [116]

Answer and explanation:

cyclopentadiene is more acidic than cyclopentane

hydrocarbon compound are weak acid in nature

This relative acidity is explained by the stability of

cyclopentadienyl anion which is aromatic in nature

(check the attached image file 1)

To answer this question we must look at the stability of the anions that are formed when the compound lose proton.

All the electron in the cyclopentyl anion are localized.

In contrast, the aromatic cyclopentadienyl anion is a stable carbanion as a result of its aromaticity therefore making its conjugate acid a very strong acid compare to other compounds with hydrogen attached to sp³ carbons

4 0
3 years ago
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