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3 years ago

15

Que representa en la reacción química Na2SO4(s)+H2O(1)=

1 answer:

mars1129 [50]3 years ago

7 0

Answer: The given reaction is a double displacement reaction.

Explanation:

A chemical reaction equation in which both positive and negative ions of two different reactants are exchanged then it is called a double displacement reaction.

For example, $Na_{2}SO_{4} + H_{2}O \rightarrow H_{2}SO_{4} + Na_{2}O$

Here, hydrogen ions of $H_{2}O$ are replaced by the sodium ions of $Na_{2}SO_{4}$ . Also, sulfate ion of $Na_{2}SO_{4}$ are replaced by oxygen ion of $H_{2}O$ .

Thus, we can conclude that the given reaction is a double displacement reaction.

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Why is hydronium ion not an electrophile

galina1969 [7]

Answer:

Hydronium ion is not an electrophile because, its positive charge is on oxygen and its octet is filled.

Explanation:

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3 years ago

CaO +<br> HCI →<br> CaCl2 +<br> H20

andrew-mc [135]

Explanation:

To balance this chemical equation-

see

Balanced form:-

CaO + 2 HCl ==. CaCl2 + H2O

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3 years ago

Phosphoric acid, which is commonly used as rust inhibitor, food additive and etching agent for dental and orthopedic use, can be

Sphinxa [80]

Answer:

$P_4_{(I)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}$

Explanation:

The first step is:

$P_4_{(I)}+5O_2_{(g)}\rightarrow 2P_2O_5_{(g)}$

Second step is:

$P_2O_5_{(g)}+3H_2O_{(l)}\rightarrow 2H_3PO_4_{(l)}$

Multiplying second step by 2, and adding both the steps, we get that:

$P_4_{(I)}+5O_2_{(g)}+2P_2O_5_{(g)}+6H_2O_{(l)}\rightarrow 2P_2O_5_{(g)}+4H_3PO_4_{(l)}$

Cancelling common species, we get that:

$P_4_{(l)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}$

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3 years ago

A radioactive sample has a half-life of 5 hours. What fraction of the sample will be left after 15 hours?

DedPeter [7]

Explanation:

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3 years ago

The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati

a_sh-v [17]

Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = $0.80M^{-1}s^{-1}$

t = time taken = 95 s

[A] = concentration of substance after time 't' = ?

$[A]_o$ = Initial concentration = 0.86 M

Now put all the given values in above equation, we get:

$0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)$

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

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3 years ago