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3 years ago

1. Write equations for the reaction of sodium with water and with dilute acids. (2 marks)

Chemistry

1 answer:

alexandr402 [8]3 years ago

3 0

Answer:

See below, please.

Explanation:

With water

2Na+2H2O=2NaOH+H2 (gas)

With dilute acids

for example, HCl

2Na+2HCl=2NaCl+H2 (gas)

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Calculate the mass in grams of carbon dioxide produced from 11.2 g of octane (C8H18) in the reaction above.

Ray Of Light [21]

Answer:

34.6g

Explanation:

Given parameters:

Mass of Octane = 11.2g

Reaction expression;

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Mass of octane = 11.2g

Unknown:

Mass of carbon dioxide produced = ?

Solution:

From the balanced reaction equation;

2 mole of octane produced 16 moles of carbon dioxide

From the given specie, let us find the number of moles;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of C₈H₁₈ = 8(12) + 18(1) = 114g/mole

Number of moles of octane = $\frac{11.2}{114}$ = 0.098mole

2 mole of octane produced 16 moles of carbon dioxide

0.098 mole of octane will produce $\frac{0.098 x 16}{2}$ = 0.79mole of CO₂

Mass of CO₂ = number of moles x molar mass

Molar mass of CO₂ = 12 + 2(16) = 44g/mol

Mass of CO₂ = 0.79 x 44 = 34.6g

8 0

3 years ago

How many moles are there in 3.612 x 1024 formula units of Na2SO4?

vitfil [10]

Answer:

<h3>Theanswer is 6 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.612 \times {10}^{24} }{6.02 \times {10}^{23} } \\$

We have the final answer as

<h3>6 moles</h3>

Hope this helps you

5 0

3 years ago

Write the concentration equilibrium constant expression for this reaction.

Akimi4 [234]

In a chemical reaction, the equilibrium constant refers to the value of its reaction quotient at chemical equilibrium, that is, a condition attained by a dynamic chemical system after adequate time has passed, and at which its composition has no measurable capacity to undergo any kind of further modification.

The given reaction is: HCN (aq) + OH⁻ = CN⁻ (aq) + H2O (l)

The equilibrium constant = product of concentration of products / product of concentration of reactants

(Here, H2O is not considered as its concentration is very high)

So, Keq = [CN⁻] / [HCN] [OH⁻]

8 0

3 years ago

For 13 points plz. Plz

Ludmilka [50]

Answer:

Individual versus nature

Explanation:

The type of conflict is individual versus nature because the cat faced many problems in nature such as facing the mouse, grass, people, etc. These are all considered as nature

PLEASE DO MARK ME AS BRAINLIEST IF MY ANSWER IS HELPFUL ;) 

4 0

3 years ago

Sulfonation of benzene has the following mechanism: (1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3− [slow]

ziro4ka [17]

Question is incomplete, complete question is as follows :

Complete Question : .Sulfonation of benzene has the following mechanism:

(1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3

[fast]

(2) SO3 + C6H6 → H(C6H5+)SO3−

[slow]

(3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4

[fast]

(4) C6H5SO3− + H3O+ → C6H5SO3H + H2O

[fast]

write the overall rate law for the initial rate of the reaction as a fraction.

Rate=k(________/_________)

Answer:

The overall rate law for the initial reaction is = $k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]$

Explanation :

Frist of all, all the common terms are cancelled out and written the overall reaction.

As we know that the rate depednant step is the slowest step of the reaction, rate law is :

rate = $k_{2} [SO_{3}][C_{6}H_{6}]$

But the problem is that SO3 cannot be written in the overall rate law because it is an intermediate.

Rate law for synthesis of S03 is as follows :

rate = $k_{1}[H_{2}SO_{4}]^{2}$

Hence when we substitute equation 2 in equation one,

Rate comes out to be = $k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]$

6 0

3 years ago

1. Write equations for the reaction of sodium with water and with dilute acids. (2 marks)​

1. Write equations for the reaction of sodium with water and with dilute acids. (2 marks)