All categories

2 years ago

Write the cations and anions present in NaCl

Chemistry

1 answer:

pochemuha2 years ago

6 0

In table salt, sodium chloride, sodium is the cation (Na+) and chloride is the anion (Cl-).

You might be interested in

Determine the oh of a solution that is 0.220 m in hco3

ICE Princess25 [194]

Answer : The hydroxide ion concentration of a solution is, $5\times 10^{-14}$

Explanation :

As we know that $HCO_3^-$ dissociates in water to give hydrogen ion $(H^+)$ and carbonate ion $(CO_3^{2-})$ .

As, 1 mole of $HCO_3^-$ dissociates to give 1 mole of hydrogen ion $(H^+)$

Or, 1 M of $HCO_3^-$ dissociates to give 1 M of hydrogen ion $(H^+)$

So, 0.200 M of $HCO_3^-$ dissociates to give 0.200 M of hydrogen ion $(H^+)$

Now we have to calculate the hydroxide ion concentration.

As we know that:

$[H^+][OH^-]=1\times 10^{-14}$

$0.200\times [OH^-]=1\times 10^{-14}$

$[OH^-]=5\times 10^{-14}$

Therefore, the hydroxide ion concentration of a solution is, $5\times 10^{-14}$

6 0

3 years ago

What can neutrons change into

Elan Coil [88]

I think they can change into ions

7 0

3 years ago

Awdawdawdadawdanodawdadwadadaw

crimeas [40]

No, actually adawadawada and awawawaw usually addawadadaw but also awawawa so it’s a possibility but very rare.

3 0

2 years ago

What is the balanced equation for H3PO4=H4P2O7+H2O

podryga [215]

Answer:

2H3PO4=H4P2O7+H2O

Explanation:

2H3PO4=H4P2O7+H2O

5 0

2 years ago

Automotive air bags inflate when sodium azide, NaN3, decomposes explosively to its constituent elements. How many moles of nitro

attashe74 [19]

Answer:

2.29 g of N2

Explanation:

We have to start with the chemical reaction:

$NaN_3~->~Na~+~N_2$

The next step is to balance the reaction:

$2NaN_3~->~2Na~+~3N_2$

We can continue with the mol calculation using the molar mass of

$NaN_3$ (65 g/mol), so:

$3.55~g~NaN_3\frac{1~mol~NaN_3}{65~g~NaN_3}=0.054~mol~NaN_3$

Now, with the molar ratio between $NaN_3$ and $N_2$ we can calculate the moles of $N_2$ (2:3), so:

$0.054~mol~NaN_3\frac{3~mol~N_2}{2~mol~NaN_3}=0.0819~mol~N_2$

With the molar mass of $N_2$ we can calculate the grams:

$0.0819~mol~N_2=\frac{1~mol~N_2}{28~g~N_2}=2.29~g~N_2$

I hope it helps!

5 0

2 years ago

Write the cations and anions present in NaCl​

Write the cations and anions present in NaCl