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4 years ago

How many atoms are in one compound of sodium nitride?

2 answers:

7 0

Https://www.google.com.eg/url?sa=t&source=web&rct=j&url=https://socratic.org/questions/what-is-the-formula-for-sodium-nitride&ved=2ahUKEwjJnpa1mPDeAhUJC-wKHdjPBHcQFjAMegQIBhAB&usg=AOvVaw0UEeq0hYocDWBqSrv3H9Wf

olya-2409 [2.1K]4 years ago

3 0

<em>Explanation: The elements in this ionic compound are sodium and nitrogen (the -ide ending indicates that the anion will be monoatomic (a single element)). Na has a charge of +1, and N generally has a charge of -3.Jan 15, 2018</em>

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Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.35 atm at a

Luden [163]

Answer : The mass of $NaN_3$ at temperature $20^oC$ 28.47 g.

The mass of $NaN_3$ at temperature $10^oC$ 29.51 g.

Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas = $20^oC=273+20=293K$ $(0^oC=273K)$

Volume of gas = $25\times 25\times 20cm=12500cm^3=12.5L$ $(1L=1000cm^3)$

Molar mass of $NaN_3$ = 65 g/mole

Part 1 : First we have to calculate the moles of gas at temperature $20^oC$ . The gas produced in the given reaction is $N_2$ .

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)$

By rearranging the terms, we get the value of 'n'

$n=0.7015moles$

The moles of $N_2$ = 0.7015 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.7015 moles of $N_2$ produced from $\frac{20}{32}\times 0.7015=0.438$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g$

Therefore, the mass of $NaN_3$ needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature $10^oC$ and same volume & pressure.

Using ideal gas equation,

$PV=nRT$

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (283K)$

By rearranging the terms, we get the value of 'n'

$n=0.726moles$

The moles of $N_2$ = 0.726 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.726 moles of $N_2$ produced from $\frac{20}{32}\times 0.726=0.454$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g$

Therefore, the mass of $NaN_3$ needed are 29.51 g.

7 0

4 years ago

11. Which of the following accurately describes properties of valence? A. Nonmetallic elements tend to have a positive valence a

marin [14]

Atoms, the main constituents of matter, consist of positively charged protons and neutral neutrons within a nucleus which are surrounded by a sea of electrons that sit in distinct shells. The electrons on the outer shell are known as valence electrons. The valence can be descibed as the smaller number of electrons an atom has to borrow or to lend, the greater the activity.

The answer is B.

3 0

4 years ago

Which group in the Periodic Table contains elements that form ions which are larger than their atoms?

const2013 [10]

Answer:

Which group in the Periodic Table contains elements that form ions which are larger than their atoms?

The second.. . Luckily for you, I just happen to have a chart in my room.

7 0

4 years ago

A binary ionic compound is known to contain a cation with 51 protons and 48 electrons. The anion contains one-third the number o

7nadin3 [17]

Answer:

$SbCl_3$

Explanation:

The cation contains 51 protons and 48 electrons. The electrons are only lost not protons, so the element which corresponds to atomic number 51 is the cation which is Antimony (Sb).

Also, the charge on the cation is 51 - 48 = 3

This is because neutral atom contains the same number of electrons as protons.

Given the anion has one-third protons as cation which means thta:

Number of protons in anion = $\frac {51}{3}=17$

This corresponds to chlorine atom. The electrons are 1 greater than protons means that it has gain 1 electron.

So,

Sb Cl

3 1

Cross multiply the valency.

So, formula is $SbCl_3$

4 0

3 years ago

Which statement is correct? (

Ronch [10]

A is correct. If an atom has more protons than electrons it is positively charged, if more electrons than protons it is negatively charged. An equal number of each will make it neutral

5 0

3 years ago