What is the initials for sodium chloride

Show chemical equation for the reaction of water

cricket20 [7]

Explanation:

2H2(g) + O2(g) → 2H2O(l )

3 0

2 years ago

I’ll give the brainliest

Readme [11.4K]

Answer:

Water gradually degrades them until the metal parts rust and the other materials deteriorate.

Explanation:

8 0

3 years ago

Р + 02 = P4O6 balance the equation

sergejj [24]

To balance a chemical equation, both sides should have equal mass, or in other words both sides should have same number of atoms as to follow the conservation of mass rule.

P + O₂ = P₄O₆

LHS:

Number of Phosphorus atoms = 1 atom

Number of Oxygen atoms = 2 atoms

RHS:

Number of Phosphorous atoms = 4 atoms

Number of Oxygen atoms = 6 atoms

Also since P₄O₆ has the most number of atoms we will make the LHS equalize to P₄O₆.

Difference between Phosphorous atoms in LHS to RHS = 3

Since phosphorous is a monatomic we need 3 phosphorous atoms extra

Difference between Oxygen atoms is LHS to RHS = 4

But Oxygen is diatomic, so we need 4/2 = 2 Oxygen molecules

Now lets see if it is balanced

P + 3P + O₂ + 2O₂ -------> P₄O₆

4P + 3O₂ ------> P₄O₆

LHS:

Phosphorous atoms = 4 atoms

Oxygen atoms = 3 × 2 = 6 atoms

RHS:

Phosphorous atoms = 4 atoms

Oxygen atoms = 6 atoms

LHS = RHS

Therefore the balanced equation is 4P + 3O₂ = P₄O₆

Happy to help :)

If you need more explanation or help in any other question, feel free to ask

3 0

2 years ago

5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf

mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log \frac{base}{acid}$

where, pH = 7.4 and $pK_{a}$ = 7.21

As here, we can use the $pK_{a}$ nearest to the desired pH.

So, 7.4 = 7.21 + $log \frac{base}{acid}$

0.19 = $log \frac{base}{acid}$

$\frac{base}{acid}$ = 1.55

1 mM phosphate buffer means $[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM

Therefore, the two equations will be as follows.

$\frac{HPO_{4}}{H_{2}PO_{4}}$ = 1.55 ............. (1)

$[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM ........... (2)

Now, putting the value of $[HPO_{4}]$ from equation (1) into equation (2) as follows.

1.55 $[H_{2}PO_{4}] + [tex][H_{2}PO_{4}]$ = 1 mM

2.55 $[H_{2}PO_{4}]$ = 1 mM

$[H_{2}PO_{4}]$ = 0.392 mM

Putting the value of $[H_{2}PO_{4}]$ in equation (1) we get the following.

0.392 mM + $[HPO_{4}]$ = 1 mM

$[HPO_{4}]$ = (1 - 0.392) mM

$[HPO_{4}]$ = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0

3 years ago

139%

GaryK [48]

Answer:

The percent composition of this compound is 94%

Explanation:

The reaction can be formed as

$2 \mathrm{Fe}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{FeCl}_{3}$

$\frac{\text { Weight of } \mathrm{Cl}_{2}}{\text { 3* Molar Mass of } \mathrm{Cl}_{2}}=\frac{\text { Weight of } \mathrm{Fe}}{2 * \text { Molar Mass of Fe }}$

$\frac{\text { Weight of } \mathrm{Cl}_{2}}{3 *(2 * 35.5)}=\frac{3.56}{2 * 55.8}$

$\text { Weight of } C l_{2}=\frac{3.56 * 3 * 71}{2 * 55.8}=6.79 \mathrm{g}$

$\mathrm{n}\left(\mathrm{Cl}_{2}\right)=\mathrm{m}\left(\mathrm{Cl}_{2}\right) / \mathrm{M}\left(\mathrm{Cl}_{2}\right)=6.79 / 71=0.1 \mathrm{m}$

$\mathrm{n}(\mathrm{Fe})=\mathrm{m}(\mathrm{Fe}) / \mathrm{M}(\mathrm{Fe})=3.56 / 55.8=0.06 \mathrm{m}$

Based on no. of iron reacted,

$\mathrm{n}(\text { moles of } \mathrm{Fe})=\mathrm{n}\left(\text { moles of } \mathrm{FeCl}_{3}\right)$

n = m/M

$\mathrm{m}\left(\mathrm{FeCl}_{3}\right)=\mathrm{n}^{*} \mathrm{M}=0.06^{*} 162.5=9.75 \mathrm{g}$

% composition of $FeCl_3$

= $(9.75 / 10.39)^{*} 100$

= 94%

6 0

3 years ago